how much tension must a rope withstand if it is used to accelerate a 1410- kg k g car horizontally along a frictionless surface at 1.40 m/s2 m / s 2 ?

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Answer 1

The amount of tension a rope must withstand if it is used to accelerate a 1410 kg car horizontally along a frictionless surface at 1.40 m/s² is 1974 N.

What is tension in physics?

Tension is the force that stretches or pulls something tightly. In physics, the term "tension" refers to the force that is transmitted through a string, cable, chain, or wire when it is pulled taut by forces acting on either end. The acceleration of the car is given as 1.40 m/s². This means that the force required to move the car is given as F = m × a. Where, F = force acting on the car M = mass of the car = 1410 kg.a = acceleration = 1.40 m/s²Therefore, F = 1410 × 1.40= 1974 N. The rope or cable used to pull the car must generate enough force to overcome the weight of the car and move it forward. The force of tension in the rope or cable required to move the car at the given acceleration is Ft = F= 1974 N.

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Related Questions

a) What is the y component of the vector given in the diagram? (120 N, 50 °) 120 N 50⁰ (C) 60 N (B) 92 N (D) 120 N (A) 77N

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Answer: B

Explanation:

The y component of the vector is given by the formula:

y component = magnitude of the vector x sin(angle between the vector and the y-axis)

In this case, the magnitude of the vector is 120 N and the angle between the vector and the y-axis is 50 degrees. So we have:

y component = 120 N x sin(50°) ≈ 92 N

Therefore, the answer is (B) 92 N.

besides changing the distance between the plates and area of the plates, another way to alter the capacitance is by filling the space between the plates with dielectric material. doing this will reduce the electric field between the plates. considering the relationship between electric field and voltage (potential difference), and between capacitance and voltage, filling the empty (vacuum) space between a capacitors plates group of answer choices will increase the capacitance of the capacitor. will have no effect on the capacitance of the capacitor. will reduce the capacitance of the capacitor. may increase or reduce the capacitance of the capacitor.

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Filling the space between a capacitor's plates with a dielectric material will increase the capacitance of the capacitor.


1. A dielectric material is inserted between the plates of the capacitor.

This material has the property of reducing the electric field between the plates.
2. The relationship between electric field (E) and voltage (V) is given by E = V/d, where d is the distance between the plates.

Since the electric field is reduced by the presence of the dielectric material, the voltage (potential difference) between the plates also decreases.

3. The relationship between capacitance (C), voltage (V), and charge (Q) is given by C = Q/V.

As the voltage decreases due to the presence of the dielectric material, the capacitance increases for a given charge on the plates.

4. The dielectric material has a property called dielectric constant (K), which is a measure of how effectively it reduces the electric field between the plates.

The capacitance of the capacitor with the dielectric material is given by C = K * C0,

where C0 is the capacitance without the dielectric material.

Since K is always greater than 1 for dielectric materials, the capacitance with the dielectric material is always higher than without it.
In conclusion, filling the empty (vacuum) space between a capacitor's plates with dielectric material will increase the capacitance of the capacitor.

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suppose you wish to make a solenoid whose self-inductance is 2.4 mh. the inductor is to have a cross-sectional area of 1.80 10-3 m2 and a length of 0.045 m. how many turns of wire are needed?

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We need approximately 369 turns of wire to make the solenoid.

The self-inductance (L) of a solenoid is given by the formula:

L = (μ₀ * N² * A * l) / l

where:

μ₀ = permeability of free space (4π × 10^-7 H/m)

N = number of turns of wire

A = cross-sectional area of the solenoid

l = length of the solenoid

We can rearrange this formula to solve for N:

N = [tex]\sqrt{L * l) / (M_0 * A))}[/tex]

Substituting the given values, we get:

N = [tex]\sqrt{2.4(10^-^3 H * 0.045 m) / (4\pi (10^-^7 H/m * 1.80(10^-^3 m^2))}[/tex]

N ≈ 369.25

Therefore, the turn of wire that we are needed are 369.25 turns.

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calculate the efficiency of an electric motor which uses 7.4kJ of energy to lift a 34kg object 11m​

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The electric motor's efficiency is 51.06%.

What is the electric motor's efficiency?

The majority of electric motors are made to operate between 50% and 100% of rated load. Typically, maximum efficiency is within 75% of rated load. Hence, the allowable load range for a 10-horsepower (hp) motor is between 5 and 10 hp; its peak efficiency is at 7.5 hp. Below roughly 50% load, a motor's efficiency tends to decline significantly.

To calculate the effort required to raise the object, use the formula:

Work = Force x Distance

= m x g x h (where m is the mass of the object, g is the acceleration due to gravity, and h is the height lifted)

= 34 kg x 9.81 m/s² x 11 m

= 3,769.34 J

The energy consumed by the electric motor is given as 7.4 kJ.

Therefore, the input power is:

Input power = Energy consumed / time taken

= 7.4 kJ / t

Efficiency=(Output power/Input power) x 100%

Output power = Work done/time taken

= 3,769.34 J / t

As a result, the electric motor's efficiency is:

Efficiency=(Output power/Input power)x 100%

= [(3,769.34 J / t) / (7.4 kJ / t)] x 100%

= 51.06%.

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according to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have ? the current of the second resistor.

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According to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have 1/3 the current of the second resistor.

A parallel circuit is an electrical circuit in which the various components are linked together in such a way that the current can pass through multiple branches. All of the elements in a parallel circuit are linked in parallel to one another. This implies that the voltage across each element is the same, but the current through each element is different because of the different resistance values.

Total current is determined using the equation I = V/R, where V is voltage and R is resistance. The total resistance in a parallel circuit is determined using the equation 1/Rt = 1/R1 + 1/R2 + 1/R3 +…1/Rt = 1/R1 + 1/R2 + 1/R3 +… 1/Rt = 1/R1 + 1/R2 + 1/R3 +… The total resistance of a parallel circuit is always less than the resistance of the smallest resistor. According to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have 1/3 the current of the second resistor.

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a light plane must reach a speed of 35 m/s for takeoff. how long a runway is needed if the (constant) acceleration is 3.0 m/s^2?

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The length of runway needed for a light plane to take off with a constant acceleration of 3.0 m/s2 and a speed of 35 m/s is 203.7 meters.

The time taken for a light plane to reach 35 m/s from 0 m/s is given by using the formula,

v = u + at

where, v = final velocity, u = initial velocity, a = acceleration, and t = time

Here, the initial velocity of the plane is u = 0 m/s.

The final velocity of the plane is v = 35 m/s.

The acceleration of the plane is a = 3.0 m/s².

The time taken to reach the final velocity can be calculated as,

35 = 0 + (3.0)t

t = 35 / 3.0

t = 11.67 s

Therefore, the plane takes 11.67 s to reach a speed of 35 m/s for takeoff.

The distance traveled by plane during this time can be calculated as,

s = ut + 1/2 at²

s = 0 x 11.67 + 1/2 x 3.0 x (11.67)²

s = 203.7 m

Therefore, the runway should be at least 203.7 meters long if the acceleration is constant at 3.0 m/s².

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what is the speed of a proton that has been accelerated from rest through a potential difference of -800 v ?

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A proton that has been accelerated from rest through a potential difference of -800 V has a speed of: 2.60 x 10⁶ m/s.

When a particle is accelerated through a potential difference, its potential energy is transformed into kinetic energy, resulting in an increase in velocity.

To calculate the velocity of a proton that has been accelerated through a potential difference of -800 V, we may use the equation: v = √(2qV/m)

where: v is the speed of the proton,

q is the charge of the proton (1.6 x 10⁻¹⁹ C),

V is the potential difference (-800 V)

m is the mass of the proton (1.67 x 10⁻²⁷ kg)

Using these values, we may calculate the velocity:

v = √(2(1.6 x 10⁻¹⁹C)(-800 V)/(1.67 x 10⁻²⁷kg))= 2.60 x 10⁶ m/s

Therefore, the velocity of a proton that has been accelerated from rest through a potential difference of -800 V is 2.60 x 10⁶ m/s.

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what is the magnitude of the upward acceleration of the load of bricks? express your answer with the appropriate units.

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The magnitude of the upward acceleration of a load of bricks is 2.77 m/s².

What is tension in the rope?

Tension is the pulling force cаrried by flexible mediums like ropes, cаbles аnd string. Tension in а body due to the weight of the hаnging body is the net force аcting on the body.

The tension in the string when the body cаn be given аs,

T = m(a +g)

Here, (m) is the mаss of the body, (а) is the аccelerаtion аnd (g) is the аccelerаtion due to grаvity.

The mаss of the bricks is 15.2 kg.The mаss of the counterweight is 27.2 kg аnd the system is releаsed from the rest.

The tension due to the bricks with mаss of 15.2 kg is,

T = 15.2(a + 9.80)

The tension due to the bricks with mass of 27.2 kg is,

T = 27.2(9.80 - a)

Equate both the equation as,

15.2(a + 9.80) = 27.2(9.80 - a)

15.2a + 148.96 = 266.56 - 27.2a

42.4a = 117.6

a = 2.77 m/s²

Thus, the magnitude of the upward acceleration of a load of bricks is 2.77 m/s².

Your question is incomplete, but most probably your full question can be seen in the Attachment.

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a 300 n/c uniform electric field points perpendicularly towardthe left face of a large neutral conducting sheet. the area chargedensity on the left and right faces, respectively, are:

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The question given is a 300 N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet.

The formula of the electric field. [tex]E=\frac{F}{q}[/tex]

The formula of area charge density.[tex]$$\ \sigma = \frac {q} {A} $$[/tex]

where E is the electric field.

F is the force of the electric charge.

q is the charge.

σ is the area charge density.

A is the area.

The electric field is given as E=300 N/C.

As the area is neutral and conductive, thus, there is no net charge and so σ = 0. A neutral conductor sheet doesn't have a charge on its face. Therefore the area charge density on the left and right faces is zero.

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the wreck skids along the ground and comes to a stop. the coefficient of kinetic friction while the wreck is skidding is 0.55. assume that the acceleration is constant. how far does the wreck skid?

Answers

The given coefficient of kinetic friction is 0.55. Assuming that the acceleration is constant, so the wreck skids a distance of 0 meters.

The distance that the wreck skids while coming to a stop is calculated below.

Data Coefficient of kinetic friction = 0.55

Conversion of acceleration to m/s²0.55 coefficient of kinetic friction can be written as 0.55 times acceleration to calculate the distance that the wreck skids. We know that the acceleration due to gravity is 9.8 m/s². Hence the acceleration due to gravity can be written as follows.

a = 9.8 m/s² × 0.55a

= 5.39 m/s²

Calculation of the distance that the wreck skids is calculated by using the formula below:

d = (v² - u²)/2as = distance = initial velocity = final velocity a = acceleration

The wreck is coming to stop, so the final velocity is 0. Hence the formula can be written as:

d = (v² - u²)/2a

= (0 - u²)/2×5.39d

= -u²/10.78d

= -0.093u²

Calculation of velocity can be calculated by using the following formula below.

v² = u² + 2asv²

= u² - 2u²/10.78v²

= (8.78u²)/10.78v²

= (2u²)/2.45v

= (u²)/1.56

The final velocity is zero. Hence we can write the formula as :

0 = (u²)/1.56u² = 0

The initial velocity of the wreck is zero. Hence the wreck is moving from rest condition.

Calculation of the distance that the wreck skids is calculated by using the formula below:

d = -u²/10.78d

= 0 meters.

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on june 9, 1988, sergei bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m. how long did it take bubka to return to the ground from the highest part of his vault?

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On june 9, 1988, Sergei Bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m. It took Bubka 1.11 seconds to return to the ground from the highest part of his vault.

Sergei Bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m on June 9, 1988. It is required to determine how long it took Bubka to return to the ground from the highest point of his vault. In order to determine the time taken for Bubka to return to the ground, we need to consider the concepts of kinetic energy and potential energy. The pole vaulter gains potential energy during the ascent phase of the vault as he gains altitude. When he reaches the highest point, he has the maximum potential energy. During the descent phase of the vault, the potential energy is converted into kinetic energy.

Based on this principle, we can use the conservation of energy equation to find the time taken by Bubka to return to the ground. The equation for conservation of energy is given as: Potential energy (P.E) = Kinetic energy (K.E)

P.E = mgh where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.

K.E = 1/2 mv² where v is the velocity of the object.

The velocity of Bubka when he reached the highest point can be assumed to be zero since he had to come to a stop before starting his descent. Therefore, the initial kinetic energy is zero.

P.E at the highest point = K.E at the lowest point

Let t be the time taken by Bubka to return to the ground. We can assume that Bubka moves with uniform acceleration. Using the kinematic equation, we have: v = u + at where u is the initial velocity and a is the acceleration.

When Bubka reaches the ground, his final velocity is zero.

Therefore, we have: v = 0u = at

Substituting the value of u in the equation for K.E, we have: K.E = 1/2 mv² = 1/2 ma²t²

Substituting the value of P.E and K.E in the equation for conservation of energy, we have:

mgh = 1/2 ma²t²

Simplifying, we get: t = sqrt(2h/g)

Substituting the values of h and g, we have:

t = sqrt(2 x 6.10 / 9.81)t = sqrt(1.240)t = 1.11 seconds

Therefore, it took Bubka 1.11 seconds to return to the ground from the highest part of his vault.

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why do the phases of venus show that the solar system is in a heliocentric model instead of a geocentric model?

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The phases of Venus show that the solar system is in a heliocentric model instead of a geocentric model because the heliocentric model states that the Sun is at the center of the solar system, while the geocentric model states that Earth is at the center of the universe.

The phases of Venus can only be explained in the heliocentric model because the planet is orbiting the Sun.The phases of Venus are an important piece of evidence supporting the heliocentric model proposed by Nicolaus Copernicus. The geocentric model was the widely accepted model of the universe until the 16th century when Copernicus proposed the heliocentric model, which suggested that the Sun is at the center of the solar system and the Earth and other planets orbit around it.

The phases of Venus show that it orbits the Sun and not the Earth because, as it orbits the Sun, different portions of the planet's sunlit side are visible from Earth. This can only occur in a heliocentric model because Venus is between the Earth and the Sun in its orbit, which causes it to pass through phases. Therefore, the phases of Venus are not consistent with a geocentric model, which suggests that Venus orbits the Earth.

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two objects collide and stop. their kinetic energy becomes sound energy. when does the energy stop being sound energy?

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Answer: Two objects collide and stop. Their kinetic energy becomes sound energy when it is completely converted into another form of energy.

Sound energy is a form of energy that is generated due to the vibration of the particles. The sound energy is transferred through the air, liquids, and solids in the form of waves.

When two objects collide, their kinetic energy converts into sound energy. This sound energy is due to the collision of the objects. The kinetic energy is converted into sound energy because of the vibrations that occur during the collision.

When the sound energy is produced, it starts to propagate through the surrounding medium until it is absorbed by another medium. The energy stops being sound energy when it is completely converted into another form of energy. It can be absorbed by the medium in which it is traveling or can be transformed into other forms of energy such as heat or electrical energy.

Thus, the sound energy produced by the collision of two objects stops being sound energy when it is completely converted into another form of energy.



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suppose an asteroid had an orbit with a semimajor axis of 4 au. how long would it take for it to orbit once around the sun? question 28 options: 2 years 4 years 8 years 16 years

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It would take approximately 19.2 years for the asteroid to orbit once around the sun. But that none of the answer choices match the calculated value of approximately 19.2 years.

The period (T) of an orbit of a celestial body with semimajor axis (a) around the sun can be calculated using Kepler's third law:

T² = (4π² / GM) * a³

where G is the gravitational constant and M is the mass of the sun.

Plugging in the given value for the semimajor axis (a = 4 AU), we get:

T² = (4π² / (6.674 × 10⁻¹¹ m³/(kg s²) * 1.989 × 10³⁰ kg)) * (4 AU)³

T² = 3.652 × 10¹⁶ s²

Taking the square root of both sides, we get:

T = 6.04 × 10⁸ s

We can convert this time to years by dividing by the number of seconds in a year:

T = (6.04 × 10⁸ s) / (31,536,000 s/year)

T ≈ 19.2 years

Therefore, it would take approximately 19.2 years for the asteroid to orbit once around the sun. The closest answer choice is 16 years.

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when you look into a mirror, what is happening to the light?

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When we look into a mirror, a process called reflection occurs with the light.

What happens to the light:
1. Light source: The process begins with a light source, such as the sun or a light bulb, emitting light waves in all directions.
2. Light traveling: The light waves travel through the air and reach the mirror.
3. Mirror's surface: Mirrors have a smooth, reflective surface made of glass with a thin layer of metal, usually aluminum or silver, on the back.
4. Incident light: The light waves that strike the mirror's surface are called incident light.
5. Reflection: The mirror's smooth surface causes the incident light waves to bounce off, or reflect, at the same angle at which they arrived.

This is known as the law of reflection, which states that the angle of incidence is equal to the angle of reflection.

6. Reflected light: The light waves that bounce off the mirror are called reflected light.
7. Image formation: As the reflected light waves travel away from the mirror, they converge at a point and form an image of the object you see in the mirror.
8. Observing the image: Your eyes detect the reflected light waves, and your brain processes this information to create the perception of the image you see in the mirror.
In summary, when you look into a mirror, the light emitted from a source travels towards the mirror, strikes its reflective surface, and bounces off at the same angle, following the law of reflection.

The reflected light then forms an image, which you observe as the reflection in the mirror.

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time dilation. a. a clever student, after learning about the theory of relativ?ity, decides to apply his knowledge in order to prolong his life. he decides to spend the rest of his life in a car, trave?ling around the freeways at 55 miles per hour (89 km/hr). suppose he drives for a period of time during which 70 years pass in his house. how much time will pass in the car? (hint: if you are unable to find a difference, be sure to explain why.) b. an even more clever student decides to prolong her life by cruising around the local solar neighborhood at a speed of 0.95c (95% of the speed of light). how much time will pass on her spacecraft during a period in which 70 years pass on earth? will she feel as if her life span has been extended? explain. c. suppose you stay home on earth while your twin sister takes a trip to a distant star and back in a spaceship that travels at 99% of the speed of light. if both of you are 25 years old when she leaves and you are 45 years old when she returns, how old is your sister when she gets back?

Answers

The time that passes in the car is approximately 70 years. when the sister returns from the trip, she would be approximately 319.15 years old according to Earth's reference frame.

a. To calculate the time that passes in the car, we can use the concept of time dilation in special relativity. The formula for time dilation is:

[tex]\[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \][/tex]

where [tex]\( \Delta t' \)[/tex] is the time experienced in the moving frame (car), [tex]\( \Delta t \)[/tex] is the time experienced in the stationary frame (house), [tex]\( v \)[/tex] is the velocity of the car, and [tex]\( c \)[/tex] is the speed of light.

Given:

Velocity of the car, [tex]\( v = 89 \, \text{km/hr} = \frac{89}{3.6} \, \text{m/s} \)[/tex]

Time experienced in the stationary frame (house), [tex]\( \Delta t = 70 \, \text{years} \)[/tex]

Converting the velocity to meters per second:

[tex]\[ v = \frac{89}{3.6} = 24.72 \, \text{m/s} \][/tex]

Substituting the given values into the time dilation formula:

[tex]\[ \Delta t' = \frac{70}{\sqrt{1 - \left(\frac{24.72}{299792458}\right)^2}} \][/tex]

Simplifying the expression:

[tex]\[ \Delta t' = \frac{70}{\sqrt{1 - 8.7203 \times 10^{-17}}} \][/tex]

Since the value inside the square root is extremely close to 1, we can approximate the square root as 1:

[tex]\[ \Delta t' \approx \frac{70}{\sqrt{1}} = 70 \, \text{years} \][/tex]

Therefore, the time that passes in the car is approximately 70 years.

b. Using the same formula for time dilation, we can calculate the time that passes on the spacecraft. Given that the speed of the spacecraft is 0.95c, where c is the speed of light, we have:

[tex]\[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \][/tex]

Given:

The velocity of the spacecraft, [tex]\( v = 0.95c \)[/tex]

Time experienced on Earth, [tex]\( \Delta t = 70 \, \text{years} \)[/tex]

Substituting the values:

[tex]\[ \Delta t' = \frac{70}{\sqrt{1 - (0.95)^2}} \][/tex]

Simplifying the expression:

[tex]\[ \Delta t' = \frac{70}{\sqrt{1 - 0.9025}} \\\\= \frac{70}{\sqrt{0.0975}} \]\\\\\ \Delta t' = \frac{70}{0.31224} \approx 224.33 \, \text{years} \][/tex]

Therefore, during a period of 70 years on Earth, approximately 224.33 years pass on the spacecraft. The clever student will perceive that her life span has been extended because more time has passed for her relative to the observers on Earth.

c. In this scenario, the twin sister is traveling to a distant star and back at a speed of 0.99c. The time experienced in the stationary frame (Earth) is given by [tex]\( \Delta t \)[/tex], which is 45 years. The time experienced in the moving frame (spaceship) is given by [tex]\( \Delta t' \)[/tex].

Using the time dilation formula:

[tex]\[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \][/tex]

Given:

The velocity of the spaceship, [tex]\( v = 0.99c \)[/tex]

Time experienced on Earth, [tex]\( \Delta t = 45 \, \text{years} \)[/tex]

Substituting the values:

[tex]\[ \Delta t' = \frac{45}{\sqrt{1 - (0.99)^2}} \][/tex]

Simplifying the expression:

[tex]\[ \Delta t' = \frac{45}{\sqrt{1 - 0.9801}} \\\\= \frac{45}{\sqrt{0.0199}} \]\\\\\ \Delta t' = \frac{45}{0.141} \approx 319.15 \, \text{years} \][/tex]

Therefore, when the sister returns from the trip, she would be approximately 319.15 years old according to Earth's reference frame.

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if the frequency of the incoming light is decreased, will the energy of the ejected electrons increase, decrease, or stay the same?

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If the frequency of the incoming light is decreased, the energy of the ejected electrons will decrease.

The frequency of the incoming light will affect the energy of the ejected electrons. This is because the energy of the ejected electrons is proportional to the frequency of the incoming light.

The energy of the electrons can be determined using the equation:

E = h * f,

where E is the energy, h is Planck’s constant, and f is the frequency of the incoming light. This equation shows that the energy of the electrons is directly proportional to the frequency of the incoming light.


Therefore, if the frequency of the incoming light is decreased, the energy of the ejected electrons will also decrease.

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how close would the masses 0.510 kg and 0.108 kg have to be in order for the gravitational force between them to have a magnitude of 1.03 n?

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The gravitational force between two masses is inversely proportional to the square of the distance between them. This means that the two masses must be much closer together for the force to be 1.03 N. The masses 0.510 kg and 0.108 kg have to be 0.285 m apart in order for the gravitational force between them to have a magnitude of 1.03 N.

The equation for gravitational force is F=G*m1*m2/d^2, where G is the gravitational constant, m1 and m2 are the two masses, and d is the distance between them.

Assuming G=6.67*10^(-11) Nm^2/kg^2, m1=0.510 kg, and m2=0.108 kg, then d=0.285 m. This is the minimum distance between the two masses for the gravitational force between them to have a magnitude of 1.03 N.

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how much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 45 j of work? express your answer to two significant figures and include the appropriate units.

Answers

The amount of charge transferred is 4.5C (coulombs). This is calculated by dividing 45 J (joules) of work by 9.0 V (volts) of voltage.

We can use the equation W = qV, where W is the work done, q is the charge transferred, and V is the potential difference (voltage) across the battery.

Rearranging the equation to solve for q, we get q = W/V.

Plugging in the values given, we have:

q = 45 J / 9.0 V

q = 5.0 C

Therefore, the battery transfers 5.0 coulombs of charge from the negative to the positive terminal while doing 45 J of work.

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europa, one of the moons of jupiter, was discovered by galileo in 1610. europa has a circular orbit of radius 6.708 105 km and period 3.551 days. find the mass of jupiter.

Answers

Therefore, the mass of Jupiter is approximately 1.898 × 1027 kg.

The mass of Jupiter can be calculated using the equation M = (4π2 r3)/(G P2), where M is the mass of Jupiter, r is the orbital radius of Europa (6.708 105 km), G is the gravitational constant (6.674 × 10-11 m3 kg-1 s-2), and P is the orbital period of Europa (3.551 days).

The circular orbit of Europa is given as, r = 6.708 × 105 km. The period of Europa is given as, T = 3.551 days are supposed to calculate the mass of Jupiter. In order to calculate the mass of Jupiter, we need to use Kepler's 3rd law. Kepler's 3rd law is given as, T2 = (4π2/GM) × r3 where T is the period of orbit, G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit.

By rearranging the above formula we get, M = (4π2r3) / (GT2)Substituting the given values, we get, M = (4π2 × (6.708 × 105)3) / ((6.67430 × 10-11) × (3.551 × 24 × 60 × 60)2) ≈ 1.898 × 1027 kg. Therefore, the mass of Jupiter is approximately 1.898 × 1027 kg.

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a hard drive rotates at 7200 rpm. the disk has a diameter of 5.1 in 13 cm. what is the speed of a point 6.0 cm. from the center axle? what is the acceleration of this point on the disk.

Answers

The speed of a point 6.0 cm from the center axle is approximately 4.524 cm/s, and the acceleration of this point on the disk is approximately 3.408 cm/s².

The first step to solving this problem is to convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s):

ω = (7200 rpm) * (2π rad/rev) / (60 s/min) ≈ 753.98 rad/s

The speed of a point 6.0 cm from the center axle can be found using the formula:

v = r * ω

where r is the distance from the center axle to the point of interest. Substituting the given values, we get:

v = (6.0 cm) * 0.75398 rad/s ≈ 4.524 cm/s

To find the acceleration of this point on the disk, we can use the formula for centripetal acceleration:

a = r * ω²

where r is the distance from the center axle to the point of interest, and ω is the angular velocity in radians per second. Substituting the given values, we get:

a = (6.0 cm) * (0.75398 rad/s)² ≈ 3.408 cm/s²

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what fraction of the engine power is being used to make the airplane climb? (the remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

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The fraction of the engine power used to make the airplane climb is 83% to two significant figures.

The power generated by the engine is given by the equation P=W/t, where P is power, W is work, and t is time.

The work done in lifting the airplane can be calculated as W=mgh, where m is mass, g is the gravitational constant, and h is the altitude gained.

Therefore, the power used to make the airplane climb can be calculated as P=mgh/t, where t is the time taken to gain the altitude.

Since the rate of climb is given as 2.5 m/s, the time taken to gain the altitude can be calculated as t=h/2.5, where h is the altitude gained.

Substituting the values into the equation, the power used to make the airplane climb can be calculated as P=750*9.8*h/2.5, where h is the altitude gained.

Therefore, the fraction of the engine power used to make the airplane climb is

P/(80*1000)=750*9.8*h/(2.5*80*1000).

Finally, the fraction of the engine power used to make the airplane climb expressed as a percentage is

(750*9.8 h/(2.5*80*1000))*100=83%.

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will the car's propulsion increase, decrease or remain the same when the static friction is reduced, as if there is an icy/slippery road?

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The car's propulsion will decrease when the static friction is reduced, as if there is an icy/slippery road. Static friction is the force that resists the movement of two surfaces in contact. When the static friction is reduced, the available friction for the car's propulsion is decreased, and the car's propulsion will be affected.

In order to understand this better, we need to understand friction. Friction is a force which acts in a direction opposite to the direction of motion and resists any change in the state of motion. Static friction is a force which acts in the direction opposite to the direction of motion and opposes any change in the state of rest of the body.

When there is a decrease in the static friction, the car will not be able to accelerate as quickly as it would on a normal road. This is because the static friction is the force which helps to accelerate the car and push it forward. On an icy/slippery road, the friction between the car and the ground will be greatly reduced and the car will be unable to move as quickly as it would on a normal road.

To summarize, the car's propulsion will decrease when the static friction is reduced, as if there is an icy/slippery road. This is because static friction is the force which helps to accelerate the car and push it forward and when there is a decrease in the static friction, the car will not be able to accelerate as quickly as it would on a normal road.

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in another universe where the speed of light is only 100 m/s, an airplane that is 45 m long at rest and flies at 320 km/h will appear to be how long (in m) to an observer at rest?

Answers

Answer:

320 km/hr = 320000 / 3600 = 88.9 m/s

(1 - v^2  / c^2) = (1 - 88.9^2 / 100^2)^1/2 = .46

Since L = L0 (1 - v^2  / c^2)^1/2

L = .46 L0 = 20.7 m

The airplane will appear to be only 20.64 m long to an observer at rest in this universe, even though its actual length is 45 m when at rest.

In this universe, the speed of light is only 100 m/s, which is much slower than in our universe, where the speed of light is approximately [tex]3 \times 10^8 m/s[/tex]. This means that the effects of special relativity will be much more noticeable in this universe.

We can use the formula for length contraction to calculate the apparent length of the airplane as seen by an observer at rest:

[tex]L' = L / \gamma[/tex]

where L is the length of the airplane at rest, L' is the apparent length of the airplane as seen by the observer, and γ is the Lorentz factor given by:

[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

where v is the speed of the airplane relative to the observer, and c is the speed of light in the given universe.

Converting the airplane's speed from km/h to m/s, we have:

[tex]v = (320 \ km/h) \times (1000 \ m/km) / (3600 \ s/h) = 88.89 \ m/s[/tex]

Substituting this value and c = 100 m/s into the expression for γ, we get:

[tex]\gamma = \frac{1}{\sqrt{1 - (88.89 m/s)^2 / (100 m/s)^2}} = 2.18[/tex]

Substituting this value of γ and L = 45 m into the expression for L', we get:

[tex]L' = L / \gamma = 45 \ m / 2.18 = 20.64 \ m[/tex]

Therefore, the length of the plane will appear to be 20.64 m. This significant length contraction is due to the low speed of light in this universe.

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torque does ignoring the mass significantly effect the value you calculate for the force exerted by the triceps? explain why or why not. triceps

Answers

When calculating the force exerted by the triceps, ignoring the mass significantly affects the torque value.

The torque is the product of the force and the distance from the force application point to the axis of rotation.

Torque= force*distance (N m)

The torque calculation for a muscle depends on the point of attachment of the muscle. Muscle mass is related to its force production capacity, and it is necessary to consider it when calculating the force applied by the triceps.

However, the force exerted by the triceps muscle would be affected by the mass of the object being lifted or moved. The force required to move an object increases with the mass of the object. Therefore, ignoring the mass of the object would result in an underestimate of the force required to move the object, and thus an underestimate of the force exerted by the triceps.

In summary, ignoring the mass of the object being lifted or moved would not significantly affect the calculated value of torque, but it would affect the calculated value of force.

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suppose the frequency of a note on an organ is 15 hz. what is the shortest organ pipe with both ends open that will resonate at this frequency? the velocity of sound in air is 343 m/s. answer in units of m.

Answers

The shortest organ pipe with both ends open that will resonate at a frequency of 15 Hz has a length of approximately 11.43 meters.

The wavelength of a sound wave is related to its frequency by the formula,

λ = v/f

where λ is the wavelength, v is the velocity of sound, and f is the frequency.

For a pipe with both ends open, the fundamental frequency (the lowest resonant frequency) is given by,

f = v/2L

where L is the length of the pipe.

We can combine these two equations to find the shortest length of an open pipe that will resonate at a frequency of 15 Hz,

λ = v/f = v/(v/2L) = 2L

L = λ/2 = v/(2f) = 343/(2*15) = 11.43 m

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a constant direct current is passing through a loop of wire. the loop is free to rotate about an axis that is parallel to and passes through the plane of the loop. under what circumstance is the maximum torque produced on the loop by the magnetic force?

Answers

The maximum torque produced on a current-carrying loop of wire by a magnetic field occurs when the plane of the loop is perpendicular to the direction of the magnetic field.

This can be explained using the formula for the torque on a current-carrying loop in a magnetic field,

τ = N * A * B * sin(θ)

where τ is the torque, N is the number of turns in the loop, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the normal to the plane of the loop and the direction of the magnetic field. If the loop is parallel to the magnetic field, then θ = 0, and the sin(θ) term in the formula is zero. Therefore, there is no torque produced on the loop.

On the other hand, if the loop is perpendicular to the magnetic field, then θ = 90°, and the sin(θ) term in the formula is maximum, which results in the maximum torque on the loop. Therefore, to obtain the maximum torque on the loop, the plane of the loop should be perpendicular to the direction of the magnetic field.

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How does the power switch on a computer work?

Answers

The power button has a cable, which is connected to two pins on the motherboard. By pressing the power button, a circuit is closed on the mainboard. At that moment, the power supply receives the signal to supply the computer with power and thus start up.

c. what will be the charges of the spheres in fractions of after connection? how does the total charge of the two spheres after the connection compare to the initial charge of the left sphere?

Answers

The charges of the spheres after connection will be the same as the charge of the left sphere. The total charge of the two spheres after connection is equal to the initial charge of the left sphere.

To understand this, it is important to know that electric charge is a conserved quantity. This means that the net charge of a system cannot change. Therefore, if two objects with opposite charges (like the two spheres) are connected, the charges of the two objects will become equal and the total charge of the two spheres will remain the same as the initial charge of the left sphere.
To further understand this concept, consider two spheres with opposite charges. If the two spheres are not connected, then the total charge of the two spheres is equal to the sum of the charges of each sphere. However, if the two spheres are connected, the net charge of the system cannot change. Therefore, the charge of each sphere will become equal and the total charge of the two spheres after the connection will remain the same as the initial charge of the left sphere.

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4. An object experiences an acceleration of 6.8 m/s². As a result, it accelerates from rest to 24 m/s. How
much distance did it travel during that acceleration?

Answers

The distance traveled by the object moving with an acceleration of 6.8 m/s² is 42.35 m.

What is distance?

Distance is the length between two points.

To calculate the distance traveled by the object, we use the formula below.

Formula:

v² = u²+2as.................. Equation 1

Where:

v = Final velocity of the objectu = Initial velocity of the objecta = Acceleration of the objects = Distance traveled by the object

From the question,

Given:

v = 24 m/su = 0 m/s (from rest)a = 6.8 m/s²

Substitute these values into equation 1 and solve for s

24² = 0²+(2×6.8×s)576 = 13.6ss = 576/13.6s = 42.35 m

Hence, the distance traveled by the object is 42.35 m.

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