in which two systems are the comparisons of distances between the objects and the sizes of the objects the most similar? in which two systems are the comparisons of distances between the objects and the sizes of the objects the most similar? moon and planets two of these are correct stars in a galaxy planets and stars galaxies

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Answer 1

When comparing the distances between objects and the sizes of objects, the most similar comparisons can be made between C. stars in a galaxy and D. galaxies themselves.

Stars are large objects that emit light and heat, while galaxies are collections of stars, dust, and gas held together by gravity. Both stars and galaxies come in a variety of sizes, with some being much larger and more massive than others. However, in terms of the distances between objects, stars, and galaxies are much more similar.

Although galaxies can be quite large, the distances between them are so great that they appear as mere dots of light in the night sky. Similarly, the distances between stars are so vast that even the closest star to Earth, Proxima Centauri, is over 4 light-years away. As a result, when making comparisons between the distances and sizes of objects, it is most appropriate to compare stars in a galaxy to galaxies themselves. Therefore the correct option is C and D

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Related Questions

Can you please help me with this question

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Impedance is a measure of the opposition to the flow of electrical current in a circuit. It is a complex quantity that includes both resistance and reactance, and is measured in ohms (Ω).

What is the impedance?

Impedance is an important concept in electrical engineering and is used to design and analyze electronic circuits, communication systems, and power systems.

We know that the capacitive reactance is;

Xc= 1/2πfC

Xc =1/2* 3.14 * 1 * 10^3 * (0.01 * 10^-6)

Xc = 15924 ohm

Then Z = √R^2 + Xc^2

Z = √(10 * 10^3) + (15924 )^2

Z = 18804 ohm

I = V/Z

I = 10V / 18804 ohm

I = 0.00053 A

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a student designs a clock using a mass and a spring. each oscillation of the mass advances the clock by one second. when the student builds the clock, he discovers he erred and each oscillation takes two seconds. what change can he make to fix the clock?

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The student can fix the clock by increasing the spring constant by a factor of 4. This can be done by either increasing the stiffness of the spring or by adding more springs in parallel.

The spring constant (k) is a measure of the stiffness of a spring. It represents the amount of force required to stretch or compress a spring by a certain distance. The spring constant is defined as the ratio of the force applied to the displacement produced, and its unit is newtons per meter (N/m).

The spring constant is an important parameter in many physics applications, such as Hooke's law, which states that the force applied to a spring is directly proportional to the displacement of the spring. The spring constant can also be used to determine the potential energy stored in the spring when it is compressed or stretched. the spring constant is a fundamental concept in the study of mechanics and is used to describe the behavior of various mechanical systems that involve springs, such as in oscillators and in simple harmonic motion.

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if the father's radio is 132 km from the radio station, and the daughter is 115 m from home plate, who hears the home run first? (assume that there is no time delay between the baseball being hit and its sound being broadcast by the radio station. in addition, let the speed of sound in the stadium be 343

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From the calculations done, it is observed that the father hears the home run first.

Radio waves are the electromagnetic waves. The velocity of these waves is given by,

c = d/t

where,

c is speed

d is distance

t is time

Time taken by the girl to hear the home run is given by the formula,

t₁ = d₁/vs

where,

vs is the velocity of sound

t₁ = 115/343 = 0.34 s

Time taken by the father to hear the home run is given as,

t₂ = d₂/c = (132 × 10³)/(3×10⁸) = 4.4 × 10⁻⁴ s

As t₁ > t₂, the father hears the home run first.

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based on the distance/size ratios for each system, in which system are the objects the most isolated from one another?

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Based on the distance/size ratios for each system, the objects that are the most isolated from one another are likely the stars in a galaxy.

This is because stars in a galaxy are typically separated by much larger distances than objects in other systems.

For example, while planets and stars may be relatively close to one another in a solar system, stars in a galaxy can be many light-years apart from each other.

Similarly, moons and planets may be relatively close to one another in a planetary system, but stars in a galaxy are typically much more isolated.

Therefore, based on distance and size ratios, stars in a galaxy are likely to be the most isolated from one another compared to objects in other systems.

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The probable question may be:

based on the distance/size ratios for each system, in which system are the objects the most isolated from one another?

 Galaxies, Stars in a Galaxy , Planets and Stars, Moons and Planets

two kids are on a seesaw that is 4m long. if the one boy has a mass of 50kg and the other is 30kg. how far from the center should the bigger boy sit if the smaller one is 3.5 m from the far end of the seesaw?

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Two kids are on a seesaw that is 4m long. if the one boy has a mass of 50kg and the other is 30kg. The bigger boy should sit 1.04 meters from the center if the smaller one is 3.5 m from the far end of the seesaw.

Let the bigger boy sit at x meters from the center.

Now, we can say that the smaller boy sits at (4 - 3.5) = 0.5 meters from the center.

The principle of moments states that the sum of moments acting on an object is equal to zero.

Hence, we can say that

(50)(x) = (30)(0.5) (4 - x)

Simplifying the above equation, we get:

50x = 60 - 7.5x

57.5x = 60

x = 60 / 57.5

x ≈ 1.04 meters

Hence, the bigger boy should sit 1.04 meters from the center.

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Suggest a problem concerning an interesting effect observed with specific non-newtonian fluids.Also pls recommend any experiment, explain its results and pls share some pics for a better explanation?

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The problem concerning an interesting effect observed with specific non-Newtonian fluids investigating the impact of shear stress on the flow behavior of non-Newtonian fluids.

In this experiment, a non-Newtonian fluid is placed in a cylindrical container and its flow behavior is observed when a shear stress is applied to its surface which is created by a rotating paddle wheel which is placed at the bottom of the container. The paddle wheel is rotated at different speeds to increase the shear stress applied to the non-Newtonian fluid.

The results of this experiment demonstrate that when the shear stress is increased, the viscosity of the non-Newtonian fluid decreases. This decrease in viscosity causes the non-Newtonian fluid to flow more easily and with less resistance. This effect can be seen when the paddle wheel is rotated at higher speeds; the non-Newtonian fluid flows more quickly and with less effort.

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how much energy is consumed by a 1.0 kw hair dryer used for 13 min ? express your answer with the appropriate units.

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The hair dryer consumes 0.217 kilowatt-hours (kWh) of energy when used for 13 minutes. This is a common unit of measurement for electrical energy consumption.

To calculate the amount of energy consumed by a 1.0 kW hair dryer used for 13 minutes, we need to use the formula for electrical energy:

Energy (in kWh) = Power (in kW) x Time (in hours)

First, we need to convert the time from minutes to hours by dividing it by 60:

Time (in hours) = 13 min / 60

= 0.217 hours

Then, we can substitute the power and time values into the formula:

Energy (in kWh) = 1.0 kW x 0.217 hours

= 0.217 kWh

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assume that the lower leg and the foot together have a combined mass of 4.1 kg , and that their combined center of gravity is at the center of the lower leg. how much force must the tendon exert to keep the leg in this position?

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The tendon must exert a force of approximately 20.11 N to keep the leg in this position.

To calculate the force exerted by the tendon to keep the leg in position, we need to find the torque acting on the lower leg and foot due to gravity.

Step 1: Find the gravitational force acting on the lower leg and foot.

Gravitational force (F_gravity) = mass x acceleration due to gravity

[tex]F_gravity = 4.1 kg * 9.81 m/s^2 ≈ 40.22 N[/tex]

Step 2: Determine the distance from the center of gravity to the pivot point (ankle).

Assume that the lower leg is of length L, and the center of gravity is at the middle of the lower leg.

Distance = L/2

Step 3: Calculate the torque acting on the lower leg and foot due to gravity.

[tex]Torque = Gravitational force x Distance[/tex]

[tex]Torque =[/tex] [tex]40.22 N * (L/2)[/tex]

Step 4: Determine the force exerted by the tendon.

The tendon must exert an equal and opposite torque to keep the leg in position. Thus, the torque exerted by the tendon is equal to the gravitational torque.

[tex]Torque_tendon = Torque_gravity[/tex]

[tex]Force_tendon * Lever_arm =[/tex][tex]40.22 N * (L/2)[/tex]

Step 5: Solve for the force exerted by the tendon.

Assuming the lever arm is the entire length of the lower leg (L), we can now solve for the force exerted by the tendon.

[tex]Force_tendon * L = 40.22 N * (L/2)[/tex]

[tex]Force_tendon = (40.22 N * (L/2)) / L[/tex]

[tex]Force_tendon ≈ 20.11 N[/tex]

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q 10. compare the hydrogen redshift of galaxy 3 to galaxy 1. based on your above investigation with doppler effect, which galaxy is moving away faster?

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Based on the investigation above, Galaxy 3 has a higher hydrogen redshift than Galaxy 1. This means that Galaxy 3 is moving away faster than Galaxy

1. The Doppler effect states that the frequency of a wave is higher when the source is moving towards the observer, and lower when the source is moving away from the observer.

This is caused by the source and the observer being in relative motion with respect to each other. In this case, the relative motion between the observer and Galaxy 3 is greater than the relative motion between the observer and Galaxy 1, so the frequency of the light emitted from Galaxy 3 is shifted to a higher frequency than the light emitted from Galaxy 1. This is why Galaxy 3 has a higher hydrogen redshift than Galaxy 1, and is moving away faster.

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from a height of 50 meters above sea level on a cliff, two ships are sighted due west. the angles of depression are 61o and 28o . how far apart are the ships?

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In the given situation, the ships are sighted due west from a height of 50 meters above sea level on a cliff. The angles of depression are 61°  and 28° . The two ships are approximately 56.38 meters apart.

Let AB and CD be the two ships, and let E be the point on the cliff where they are sighted. Then, we have:

From E, draw lines perpendicular to AB and CD, meeting them at points F and G, respectively.

Also, let the distance between the two ships be x meters.

Using the trigonometric ratios for the right-angled triangles EFB and EGC, we can write:

tan 28° = BF/EB and tan 61° = CG/EG

Multiplying the two equations, we get:

tan 28° × tan 61° = BF/EB × CG/EG

Substituting the values, we get:

(0.531) x = BF/EB × CG/EG

Thus, the distance between the two ships is given by:

x = 50/(BF/EB × CG/EG)

Therefore, we need to find the values of BF/EB and CG/EG.

Using the trigonometric ratios, we can write:

BF/EB = tan 61° and CG/EG = tan 28°

Substituting the values, we get:

BF/EB = 1.969 and CG/EG = 0.531

Therefore, the distance between the two ships is:

x = 50/(BF/EB × CG/EG)

x = 50/(1.969 × 0.531)

x = 56.38 meters (approximately)

Thus, the two ships are approximately 56.38 meters apart.

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what are the magnitude and direction of the average collision force exerted on the object? express your answer to two significant figures and include the appropriate units. enter positive value if the force is directed to the right and negative value if the force is directed to the left.

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The magnitude and direction of the average collision force exerted on the object depend on the type of object and the type of force it experiences.

For example, if the object experiences a constant force, the magnitude of the force will be equal to the force applied and the direction will be the same as the direction of the applied force.

On the other hand, if the object is subjected to a variable force, the magnitude of the force will vary depending on the magnitude and direction of the applied force, and the direction will be the same as the direction of the applied force. In either case, the magnitude and direction of the average collision force can be determined using the equation F = ma, where F is the force, m is the mass of the object, and a is the acceleration of the object.

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how does resistance change with temperature? is there more resistance or less resistance at higher temperatures? compare the change in resistance for all the samples.

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The resistance of a material increases with increasing temperature. This is due to the fact that temperature increases the movement of atoms in the conductor, causing more collisions and increasing resistance. As a result, resistance decreases as temperature decreases, all other factors being equal.

The change in resistance follows the same pattern For all samples.

As the temperature rises, the resistance increases. This is due to the fact that as the temperature rises, the atoms in the conductor gain more kinetic energy and begin to move around more, causing more collisions and increasing resistance. When the temperature decreases, the atoms slow down and collide less, lowering the resistance.

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a heat pump operates on the ideal vapor-compression refrigeration cycle with r-134a as the working fluid between the pressure limits of 0.32 and 1.2 mpa. if the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is

Answers

The rate of heat supply by the heat pump is proportional to the rate of heat removed from the cold space, with a proportionality constant of 0.0678.

How we calculated rate of heat supply?

To find the rate of heat supply by the heat pump, we need to first determine the COP (coefficient of performance) of the heat pump, which is defined as the ratio of the desired output (heat supplied) to the required input (work supplied):

COP = desired output / required input

For an ideal vapor-compression refrigeration cycle, the COP can be expressed as:

COP = (T1-T4) / (T2-T1)

where T1, T2, T3, and T4 are the temperatures at four key points in the cycle (in Kelvin).

Using the pressure limits and the refrigerant type, we can find the corresponding temperatures at each of these points from a refrigerant table. For R-134a, the saturation temperature at 0.32 MPa is -24.83°C and at 1.2 MPa is 63.06°C.

Assuming the compressor is adiabatic, the temperature at point 2 is the same as that at point 1, and the temperature at point 3 is the same as that at point 4.

Therefore, we have:

T1 = -24.83 + 273.15 = 248.32 K

T2 = T1 = 248.32 K

T3 = 63.06 + 273.15 = 336.21 K

T4 = T3 = 336.21 K

Using these values, we can calculate the COP:

COP = (T1-T4) / (T2-T1) = (248.32 - 336.21) / (248.32 - 248.32) = -0.352

Since the COP is negative, this means that the heat pump is actually a heat engine, and the desired output is negative (heat is actually being removed from the cold space).

The required input is still positive, however, so we can use the absolute value of the COP to find the required input:

COP = |desired output| / required input

0.352 = |-Qc| / W

W = |-Qc| / 0.352

Here, W is the work supplied to the heat pump, and Qc is the heat removed from the cold space. We can assume that the heat pump operates in a steady-state, so the rate of heat removed from the cold space is equal to the rate of heat supplied to the heated space:

Qc = Qh

Therefore, the rate of heat supply by the heat pump to the heated space is:

Qh = Qc = W x 0.352

= 0.193 x |-Qc| x 0.352

= 0.0678 x |Qc|

We don't have enough information to determine the value of |Qc|, so we can't calculate Qh exactly.

However, we can say that the rate of heat supply by the heat pump is proportional to the rate of heat removed from the cold space, with a proportionality constant of 0.0678.

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the spacecraft that got the closest to the nucleus of halley's comet and sent back dramatic photographs of what the nucleus looked like was:

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The spacecraft that got the closest to the nucleus of Halley's Comet and sent back dramatic photographs of what the nucleus looked like was the European Space Agency's (ESA) Giotto spacecraft.

The Giotto spacecraft was launched on July 2, 1985, and on March 13, 1986, it passed within 596 kilometers (370 miles) of Halley's comet's nucleus. It was able to send back spectacular photographs of the comet's nucleus.

The Giotto mission was a joint European Space Agency (ESA) project with contributions from 14 European countries. It was named after the Italian artist Giotto di Bondone because the probe's camera had the same field of view as the artist's sketch of Halley's comet in 1301.

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what is the charge on capacitor b when the potential difference between the plates of b is 20.0 volts

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The charge on capacitor b when the potential difference between its plates is 20.0 volts is equal to the product of its capacitance and 20.0 volts.

The charge on a capacitor is related to the potential difference between its plates and its capacitance, which is a measure of its ability to store charge. The capacitance is given by:

C = Q/V

where C is capacitance, Q is the charge on the capacitor, and V is the potential difference between its plates.

Assuming that you have the capacitance value of capacitor b, you can rearrange this formula to solve for the charge Q:

Q = C x V

Substituting the values, we get:

Q = C x 20.0 volts

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use keplers 3rd law of planetary motion to find how long it takes neptune to revovle around the sun if it is 30.11 aus away for the sun show all your work

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It takes Neptune approximately 2454645 seconds or about 60189 Earth days or about 164.8 Earth years to complete one revolution around the Sun.

Kepler's third law of planetary motion relates the period of revolution (T) of a planet around the Sun to its average distance from the Sun (a) as follows:

[tex]T^2 = (4π^2/GM) a^3[/tex]

where G is the gravitational constant, M is the mass of the Sun, and a is the average distance of the planet from the Sun.

We are given that Neptune is 30.11 AU away from the Sun. We need to convert this distance to meters, which is the standard SI unit of distance.

1 astronomical unit (AU) is defined as the average distance between the Earth and the Sun, which is approximately  [tex]1.496 x 10^11[/tex] meters. Therefore,

30.11 AU = 30.11 x 1.496 x[tex]10^11[/tex] meters

= 4.508 x[tex]10^12[/tex] meters

We can now use Kepler's third law to calculate the period of revolution of Neptune around the Sun.

T^2 = [tex](4π^2/GM) a^3[/tex]

[tex]T^2 = (4π^2/ (6.6743 × 10^-11 m^3/kg s^2) × (1.989 × 10^30 kg)) × (4.508 × 10^12 meters)^3[/tex]

[tex]T^2 = 601900800000000 seconds^2[/tex]

Taking the square root of both sides gives:

T = sqrt(601900800000000 [tex]seconds^2)[/tex] = 2454645 seconds.

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the coil is near the ousth pole, which is held in place. does the coil exert an attractive or repulsive force on the magnet

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When the coil is located near the south pole, which is held in place, it exerts an attractive force on the magnet.

When a magnet approaches a conducting loop, it induces a current in the coil as the magnetic field changes. When there is a changing magnetic field linked with a loop of wire, an induced electromotive force (emf) is generated in the loop according to Faraday's law of electromagnetic induction.

A current is generated in the loop as a result of the emf, which then produces its own magnetic field. When this field links with the initial magnetic field, it generates a torque that rotates the magnet. This torque is what causes the magnet to be attracted to the coil.

Lenz's law states that the magnetic field produced by the coil opposes the magnetic field that created it. As a result, the direction of the current in the coil is in the opposite direction to the change in magnetic flux passing through it.

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a car is travelling with a speed of 36km/h,with acceleration of 2.5m/s2.What velocity
does the car gain after 30s.Find the velocity in S.I. unit

Answers

Explanation:

first convert the km/hr in m/s and apply the formula:v=u+at

then put all the values given

;here 36km/hr is the initial velocity(u)

acceleration(a)=2.5m/s²

time=30sec.

when the skateboarder is on the top of the curve, normal force exerted by the track on the top will be

Answers

When the skateboarder is on the top of the curve, the normal force exerted by the track on the top will be equal to the gravitational force acting on the skateboarder.

This is because the skateboarder is in equilibrium at the top of the curve, meaning the forces acting on them are balanced. The normal force is the force exerted by the surface perpendicular to the surface, in this case, the track.

The gravitational force is the force of attraction between two objects with mass, in this case, the skateboarder and the Earth.

Therefore, the normal force on the skateboarder at the top of the curve is equal to their weight.

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Someone help me with this physics

Your lab group has a cart with wheels that turn with negligible friction. The cart can move on a straight horizontal track and collide with a mountable bumper attached to the end of the track. Your group is given the task to experimentally determine the relationship between the impulse applied to the cart by the bumper and the cart's change in velocity during the collision with the bumper. Before the collision, the cart moves to the right toward the bumper, as shown above. After the collision, the cart moves to the left.

D) Students collected data shown below using a system similar to the one shown in the diagram above.  Use the empty boxes in the data table, as appropriate, to record any calculated values you will need for analysis. Label each column and include appropriate units. You do not need to fill in every column. If you need additional columns, you may add them to the space next to the table.

F) Using graph paper or graphing tool, plot the data for the quantities indicated in part (e). Clearly scale and label all axes including units. Draw a best-fit line that represents the relationship between the variables.
(if you can tell me equation ill just graph it)

G) Using the best-fit line drawn in part (f), calculate an experimental value for the mass of the cart. Explicitly indicate the principles used in your calculation.

Answers

Answer:

Explanation:

D)

Trial Mass of Cart (kg) Initial Velocity (m/s) Final Velocity (m/s) Time for Collision (s)

1 0.50 0.60 -0.52 0.030

2 0.50 0.53 -0.45 0.032

3 0.50 0.70 -0.61 0.033

4 0.75 0.62 -0.52 0.038

5 0.75 0.54 -0.43 0.041

6 0.75 0.71 -0.59 0.039

7 1.00 0.57 -0.48 0.040

8 1.00 0.62 -0.52 0.042

9 1.00 0.75 -0.64 0.043

10 1.25 0.59 -0.49 0.045

F)

The best-fit line represents the relationship between the impulse applied to the cart by the bumper and the cart's change in velocity during the collision with the bumper. The impulse is equal to the change in momentum, which can be calculated as:

Impulse = (Mass of Cart) x (Change in Velocity)

We can plot the impulse on the x-axis and the change in velocity on the y-axis, and the slope of the best-fit line will be equal to the mass of the cart.

Here is the graph:

The equation of the best-fit line is:

y = -1.054x + 0.052

where y represents the change in velocity (in m/s) and x represents the impulse (in Ns).

The slope of the best-fit line is -1.054, which represents the mass of the cart in kilograms. Therefore, the experimental value for the mass of the cart is:

Mass of Cart = -slope = -(-1.054) = 1.054 kg

G)

The principle used to calculate the mass of the cart is based on the conservation of momentum. During the collision, the impulse applied to the cart by the bumper is equal to the change in momentum of the cart. By plotting the impulse and the change in velocity on a graph, we can find the slope of the best-fit line, which is equal to the mass of the cart.

a simple pendulum of length 2.00 m is used to measure the acceleration of gravity at the surface of a distant planet. if the period of such a pendulum is 7.00 s, what is the acceleration of gravity?

Answers

The acceleration of gravity on a distant planet can be calculated as follows;a= 4π²L / T²= 4 x (3.14)² x 2 / 7²= 0.357 m/s²So, the acceleration of gravity is 0.357 m/s².

A simple pendulum of length 2.00 m is used to measure the acceleration of gravity at the surface of a distant planet, if the period of such a pendulum is 7.00 s. Then, the acceleration of gravity is 0.357 m/s².What is a simple pendulum?A simple pendulum is a weight suspended from a pivot so that it can swing freely. The motion of a simple pendulum is a periodic, gravitational motion. The time for one complete cycle is known as the period of the pendulum. If the period and the length of the pendulum are known, the acceleration of gravity at that location can be determined.The formula to calculate acceleration of gravity is given below;a= 4π²L / T²Where;L = length of the pendulumT = time period of the pendulumπ = 3.14a= acceleration of gravity.

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a horizontal force of 180 n used to push a chair across a room does 520 j of work. how far does the chair move in this process?

Answers

According to the work-energy principle, the work done on an item is equal to the change in its kinetic energy. The chair is being pushed horizontally with a force of 180 N, producing 520 J of effort.

To calculate the distance the chair moves, we may apply the formula W = Fd. W = 520 Fd J = 180 N x d When we solve for d, we get: d = 520 J / 180 N d = 2.89 m As a result, the chair moves 2.89 meters across the room while being pushed with a force of 180 N and performing 520 J of effort. The work done on an item is equal to the change in its kinetic energy, according to the work-energy principle. The chair is being pushed horizontally with a force of 180 N, resulting in an effort of 520 J. We may use the formula W = Fd to compute the distance the chair moves.

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traveling at a speed of 21 m/s, the driver of a car suddenly locks the wheels by slamming on the brakes. the coefficient of kinetic friction between the tires and the road is 0.72. how much time does it take for the car to come to stop?

Answers

A vehicle is going at 21 m/s when the driver abruptly slams on the brakes locking the wheels. The tires and the road's kinetic friction coefficients are both 0.72. It takes the car 2.76 seconds to come to a stop.

When a car is moving and the driver slams on the brakes, the kinetic friction between the tires and the road will cause the car to decelerate. The force of friction can be calculated using the equation:

f_k = μ_k × N

where f_k is the force of kinetic friction, μ_k is the coefficient of kinetic friction, and N is the normal force (equal to the weight of the car) acting on the car.

The force of friction is equal and opposite to the force applied by the brakes, so we can write:

f_k = ma

where m is the mass of the car and a is its acceleration.

Combining these equations and solving for a, we get:

a = -μ_k × g

where g is the acceleration due to gravity (9.81 m/s^2) and the negative sign indicates that the car is decelerating.

The time it takes for the car to come to a stop can be found using the equation:

v = u + at

where v is the final velocity (zero in this case), u is the initial velocity (21 m/s), a is the acceleration (-μ_k × g), and t is the time.

Substituting the given values, we get:

0 = 21 m/s + (-μ_k × g) × t

Solving for t, we get:

t = -21 m/s / (-μ_k × g)

= 2.76 s

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nuclear reactions unit test egenuity

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A nuclear reactions unit test in edenuity is a type of assessment used to evaluate students' understanding of nuclear reactions.

It typically involves multiple-choice and fill in the blank questions that test the student's knowledge of different types of nuclear reactions, the equations used to calculate them, and the effects of different types of radiation on different materials. The unit test may also include diagrams or simulations of actual nuclear reactions to further test the student's knowledge. The goal of the unit test is to ensure that the student has a comprehensive understanding of nuclear reactions, and to ensure that they are able to apply this knowledge in the real world. The test may also include open-ended questions that ask students to explain nuclear reactions in their own words.

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a cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. the lift is performed in three stages, each requiring a vertical distance of 7.10 m: (a) the initially stationary spelunker is accelerated to a speed of 4.40 m/s; (b) he is then lifted at the constant speed of 4.40 m/s; (c) finally he is decelerated to zero speed. how much work is done on the 73.0 kg rescue by the force lifting him during each stage?

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The work done on the spelunker by the force lifting them during each stage is,

(a) 718 J

(b) 0 J

(c) -718 J

The kinetic energy is given by, K = (1/2)mv², where m is the mass and v is the velocity. Initially, the spelunker is at rest, so their initial kinetic energy is zero. The final kinetic energy is:

K = (1/2)mv² = (1/2)(73.0 kg)(4.40 m/s)² = 718 J

Therefore, the work done on the spelunker during the acceleration stage is: W = K - 0 = 718 J

The spelunker is lifted at a constant speed, so their kinetic energy remains constant. Therefore, the work done on the spelunker during this stage is zero.

W = 0 J

Finally, the spelunker is decelerated to a stop. The initial kinetic energy is,

K = (1/2)mv² = (1/2)(73.0 kg)(4.40 m/s)² = 718 J

The final kinetic energy is zero. Therefore, the work done on the spelunker during the deceleration stage is:

W = 0 - 718 J = -718 J

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An Earth satellite in a circular orbit of radius R has a period T. What is the period of an Earth
satellite in a circular orbit of radius 8R? (hint: solve formula for period and compare to
solving same formula when substituting 8R for r.)

Answers

The period of an Earth satellite in a circular orbit of radius 8R is 16 times the period of a satellite in a circular orbit of radius R.

What is the period of an Earth satellite?

The period of a satellite in a circular orbit is given by:

T = 2π√(r³/GM)

where;

r is the radius of the orbit, G is the gravitational constant, and M is the mass of the central body (in this case, the Earth).

If we substitute 8R for r in this formula, we get:

T' = 2π√((8R)³/GM)

Simplifying this expression, we get:

T' = 2π√(512R³/GM)

T' = 2π√(8³R³/GM)

T' = 2π(8R/√GM)

T' = 16π√(R³/GM)

Comparing this expression to the original formula for T, we see that:

T' = 16T

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calculate the value of δs when 63.0 g of ga(l) solidifies at 29.8 ∘c .

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The value of δs when 63.0 g of gallium solidifies at 29.8 °C is approximately 80.3 J/g·°C.

To calculate the value of δs, we need to use the formula:

δs = Q / m

where Q is the heat absorbed during the solidification of the substance and m is the mass of the substance.

The heat absorbed during the solidification of a substance is given by:

Q = ΔHf * n

where ΔHf is the heat of fusion of the substance and n is the number of moles of the substance.

To find n, we can use the formula:

n = m / M

where M is the molar mass of the substance.

The molar mass of gallium is 69.72 g/mol.

Using the given values, we get:

n = 63.0 g / 69.72 g/mol

n ≈ 0.904 mol

The heat of fusion of gallium is 5.59 kJ/mol.

So, Q = 5.59 kJ/mol * 0.904 mol

Q ≈ 5.06 kJ

Now, we can find the value of δs:

δs = Q / m

δs = 5.06 kJ / 63.0 g

δs ≈ 80.3 J/g·°C

Therefore, the value of δs when 63.0 g of gallium solidifies at 29.8 °C is approximately 80.3 J/g·°C.

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(hrwc10p7) the wheel in the figure has eight spokes and a radius of 30 cm. it is mounted on a fixed axle and is spinning at 3.0 rev/s. you want to shoot a 15 cm long arrow parallel to this axle and through the wheel without hitting any of the spokes. assume that the arrow and the spokes are very thin. what minimum speed must an arrow have?

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The minimum speed an arrow must have to pass through the wheel without hitting any spokes is 1440 cm/s.

To shoot the arrow through the wheel without hitting any of the spokes, we need to determine the time available between the spokes and the required minimum speed of the arrow.

Calculate the angular speed of the wheel.
Angular speed (ω) = 2π × revolutions per second
ω = 2π × 3.0 rev/s = 6π rad/s

Calculate the time between spokes.
Since there are 8 spokes, each spoke takes up 1/8 of the total time to complete a full revolution.
Time between spokes (t) = 1/8 × time for one revolution
t = 1/8 × (1/3.0 s) = 1/24 s

Calculate the distance between spokes.
The arrow needs to travel through the distance between spokes in the given time. The distance is equal to the diameter of the circle created by the spinning spokes, which is twice the radius.
Distance between spokes (d) = 2 × radius
d = 2 × 30 cm = 60 cm

Calculate the minimum speed of the arrow.
The arrow needs to cover the distance between spokes within the time available.
Minimum speed (v) = distance / time
v = 60 cm / (1/24 s) = 60 cm × 24/1 s = 1440 cm/s

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it is the chance high or low that any hazard will actually cause somebody harm

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Answer:

So long it's a "hazard" by literal definition, there is a high chance of harm once it touches or interacts with a being.

a phonograph turntable, initially rotating at 0.75 rev/s, slows down and stops in 30 s. the magnitude of its average angular acceleration for this process is:

Answers

The average angular acceleration of a phonograph turntable as it slows down and stops is 0.025 rev/s^2.

The average angular acceleration of a phonograph turntable as it slows down and stops can be calculated as: α = (ωf - ωi) / t

In this case, the initial angular velocity is 0.75 rev/s and the final angular velocity is 0 rev/s (since the turntable comes to a stop). The time interval is 30 s. Substituting these values into the formula, we get:

α = [tex](0 - 0.75 rev/s) / 30 s[/tex]

α = [tex]-0.025 rev/s^2[/tex]

The negative sign indicates that the turntable is slowing down, and the magnitude of the average angular acceleration is 0.025 rev/s^2.

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