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A supercluster is 100 million light-years across. How long would it take light to travel from one edge of the supercluster to the center of the supercluster?
Answer:
50 million years
Explanation:
light years is the distance light travels in one year given that the supercluster is 100 million light years across the the distance to the center will be half that amount therefore the answer is 50 million years
Why is Energy, Work and Power all Scalar Quantity?
Answer:
Explanation:
We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. ... since energy, work done and time are scalar, Power is a scalar quantity
A satellite of mass m is orbiting Earth in a stable circular orbit of radius R. The mass and radius of Earth are ME and RE , respectively. Express your answers to parts (a), (b), and (c) the following in terms of m, R, ME , RE , and physical constants, as appropriate.
a. Derive an expression for the speed of the satellite in its orbit.
b. Derive an expression for the total mechanical energy of the satellite-Earth system in its orbit.
c. Derive an expression for the period of the satellite’s orbit.
Answer:
a) v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex], b) Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex], c) T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]
Explanation:
a) For this exercise we must use Newton's second law with the gravitational force
F = ma
[tex]G \frac{m M_e}{(R+R_e)^2 }[/tex] = m a
the acceleration of the satellite is centripetal
a = [tex]\frac{v^2}{(R+R_e)}[/tex]
we substitute
[tex]G \frac{m M_e}{(R+R_e)^2 } = m \frac{v^2}{ (R+R_e)}[/tex]
[tex]G \frac{M_e}{(R+R_e)}[/tex] = v²
v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex]
the distance is from the center of the earth
b) mechanical energy is the sum of kinetic energy plus potential energy
Em = K + U
Em = ½ m v² - G m M / (R + R_e)
we substitute the expression for the velocity
Em = ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex] - [tex]G \frac{M_e}{(R+R_e)}[/tex]
Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex]
c) as the orbit is circulating, the velocity modulus is constant
v = d / t
in a complete orbit the distance traveled of the circle is
d = 2π (R + R_e)
where time is called period
v = 2π (R + R_e)
T = 2π (R + R_e) / v
we substitute the speed value
T = 2π (R + R_e) [tex]\sqrt{\frac{(R+R_e) }{G M_e } }[/tex]
T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]
(a) An expression for the speed of the satellite in its orbit.
[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]
(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.
[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]
(c) An expression for the period of the satellite’s orbit.
[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]
What are satellites?A satellite is a moon, planet or machine that orbits a planet or star. For example, Earth is a satellite because it orbits the sun
a) For this exercise we must use Newton's second law with the gravitational force
F = ma
[tex]ma =G\sqrt{\dfrac{mM_e}{(R+R_e)}[/tex]
the acceleration of the satellite is centripetal
[tex]a=\dfrac{v^2}{R+R_e}[/tex]
we substitute
[tex]G\dfrac{mM_e}{(R+R_e)}=m\dfrac{v^2}{(R+R_e)}[/tex]
[tex]G\dfrac{M_e}{(R+R_e)}=v^2[/tex]
[tex]v=\sqrt{G\dfrac{M_e}{(R+R_e)}[/tex]
b) mechanical energy is the sum of kinetic energy plus potential energy
Em = K + U
[tex]Em =\dfrac{1}{2}m v^2 - \dfrac{G m M} {(R + R_e)}[/tex]
we substitute the expression for the velocity
[tex]E_m=\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}-G\dfrac{M_e}{(R+R_e)}[/tex]
[tex]E_m=-\dfrac{1}{2}G\dfrac{M_e}{(R+R_e)}[/tex]
c) as the orbit is circulating, the velocity modulus is constant
[tex]v=\dfrac{d}{t}[/tex]
in a complete orbit the distance traveled of the circle is
[tex]d = 2\pi (R + R_e)[/tex]
where time is called period
[tex]v = 2\pi (R + R_e)[/tex]
[tex]T = \dfrac{2\pi (R + R_e)} { v}[/tex]
we substitute the speed value
[tex]T = 2\pi (R + R_e) . \sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]
[tex]T=2\pi\sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]
(a) An expression for the speed of the satellite in its orbit.
[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]
(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.
[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]
(c) An expression for the period of the satellite’s orbit.
[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]
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In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing the centripetal force required to keep the object moving in a circle: a. A car driving around a track. b. A ball being swung on the end of a string. c. The moon orbiting the earth. d. A rotating wheel.
Answer:
Explanation:
Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.
Hence;
1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.
2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.
3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.
4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.
The force that provides the centripetal force in each of the given situations are;
A) Friction Force
A) Friction ForceB) Tension Force
A) Friction ForceB) Tension ForceC) Force of gravity
A) Friction ForceB) Tension ForceC) Force of gravity D) No centripetal force
When an object is in circular motion, the force that keeps it moving round the circle while centrifugal force is the one that tries to pull the object away from the center.
A) When a car is driving around a track, there is a frictional force between the tires of the car and the track that acts on the vehicle to keep it in that circular motion. This frictional force is the centripetal force required to keep the vehicle in circular motion.
B) When a ball is swing on the end of a string, there is an upward force called tension force that acts on the ball to keep it swinging in circular motion. Thus, the centripetal force here is provided by the tension force.
C) When the moon is orbiting the earth, there is a force of gravity exerted by the earth on the moon that keeps the moon in a circular motion about Earth instead of moving in a straight line.
D) For a rotating wheel, the centripetal force does not do any work. The reason for that is because the centripetal force points toward the center of the circle, and as a result it means that the velocity of the rotating wheel is tangent to the circle.
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3 ) find the electrical force between the two charges Q1=3mc ,Q2=-6mc when they are 0.3 m parted ?
find the amount of the force when Q1 is doubled ?
Answer:
F = 3.6 10⁶ N
Explanation:
The expression for the electric force is
F = [tex]k \ \frac{q_1 q_2}{r^2}[/tex]
in this case it indicates that the charge q1 is doubled
q₁ = 2 3 10⁻³ C
q₁ = 6 10⁻³ C
let's reduce the magnitudes to the SI system
q₂ = 6 10⁻³ C
r = 0.3 m
let's calculate
F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²
F = 3.6 10⁶ N
1. Three confused sled dogs are trying to pull a sled across the Alaskan snow. Tim pulls east with a force of 42 N; Sam also pulls east, but with a force of 53 N; and big Ethan pulls west with a force of 67 N. What is the net force on the sled? then find the acceleration?
2. A 944-kg dragster, starting from rest, attains a speed of 31,2 m/s in 0.670 s. a. Find the average acceleration of the dragster.
b. What is the magnitude of the average net force on the dragster during this time?
c. What horizontal force does the seat exert on the driver if the driver has a mass of 68.0 kg?
Please try to atleast answer one
Answer:
Explanation:
According to the statement, three confused sleigh dogs are trying to pull a sled across the Alaskan snow.
Forces in same direction gets added , so 35N + 42N=77N and the Net Force is 77N -53N as it is acting in opposite direction.
Net force is 25N in east to the maximum without any hassle.
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A bullet traveling at 5.0 x10^2 meters per is brought to rest by an impulse of 50 Newton*seconds. Find the mass of the bullet.
The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.
What is impulse?Impulse in physics is the change in momentum. It is the product of the force and change in time.
hence, impulse = f. dt
When the bullet is travelling with a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.
Therefore, f.dt = m. v
f.dt = 50 N s
v = 500 m/s
m = 50 N s/500 m/s = 0.1 Kg
Therefore, the mass of the bullet is 0.1 Kg.
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what energy transformation is preformed by a radio
Answer:
Chemical energy to sound energy to heat energy
Part A
What is the radius of the hydrogen-atom Bohr orbit shown in the figure? (Figure 1)
r = ____ nm
The radius of the hydrogen-atom Bohr orbit shown in the figure is 5.3 nm.
What is Bohr orbit?The path that hypothetical electrons take around the nucleus is known as Bohr's orbit.
These orbits are described by Bohr in his hypothesis of the structure of an atom as energy levels or shells where electrons move in a fixed circle around the nucleus.
These orbits resemble solar system orbits, with the exception that they are attracted by electrical forces rather than gravity. The term "ground state" refers to the amount of energy that an electron typically occupies.
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What is the resulting acceleration when a net force of 18 N is applied to a 3.0 kg mass?
O A. 6.0 m/s2
C. 54 m/s2
O B. 15 m/s2
O D. 9.8 m/s2
Answer:
F =ma
a=f/m =18/3= 6m/m²
Explanation:
the resulting acceleration will 6 m/s² because the question asking us to solve for the acceleration.
with what speed will a clock have to be moving in order to run at a rest that is one half the rate of clock at rest
The speed at which a clock would have to be moving in order to run at half the rate of a clock at rest depends on the theory of relativity that you are using.
What is the clock speed about?In special relativity, time dilation is the phenomenon where time appears to pass differently for objects in motion relative to an observer at rest.
According to the theory, time appears to slow down for an object as it approaches the speed of light. The rate at which time appears to pass for an object is given by the equation:
T' = T / [tex]\sqrt{(1 - (v^2 / c^2))}[/tex]
Where T is the time as measured by an observer at rest, T' is the time as measured by an observer moving relative to the object, v is the velocity of the object, and c is the speed of light.
In addition, this is a theoretical scenario, practically in order to measure time dilation in a laboratory, it is required a very high precision of measurements, that are currently not possible.
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QUESTION 4
A student lifts a 400 N sandbag 2 meters off the ground. How much work, in joules, did the student perform?
Answer:
800J
Explanation:
W = Fs, Work equals force times displacement
in this case, the force is 400N and the displacement is 2 meters.
The regular SI unit for work is joules
Electromagnetic waves are commonly referred to as _________
O electricity
O magnetism
O light
A hair dryer uses 1200 watts of power. Current flow through
the dryer is 10 amperes. At what potential difference does the hair dryer operate
Answer:
did any of this help
Explanation:
y = (-2/3)x - 1
y-(-5)= -2/3(x-6)
y-y1=m(x-x1)
2x-3y=11
A iguana runs back and forth along the ground. The horizontal position of the iguana in meters over time is shown
What is the displacement of the iguana between 3 s and 6 s?
m
What is the distance traveled by the iguana between 3 s and 6 s?
The displacement of the iguana between 3 s and 6 s is 6.71 meters.
The distance traveled by the iguana between 3 s and 6 s is 8.08 meters.
What are distance and displacement?Distance is the sum of an object's movements, regardless of direction.
The term "displacement" refers to a shift in an object's position.
According to the graph:
The displacement of the iguana between 3 s and 6 s
= √{ (3-6)²+(6-0)²} meters
= 6.71 meters.
The distance traveled by the iguana between 3 s and 6 s
= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters
= [2+ 6.08] meters
= 8.08 meters.
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three way to calculate average are
Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which radiates with a peak wavelength of about 970 nm? K (b) Rigel, a bluish-white star in Orion, radiates with a peak wavelength of 145 nm. Find the temperature of Rigel's surface. K
Answer:
(a) T = 2987.6 k
(b) T = 19986.2 k
Explanation:
The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}[/tex]
where,
T = Radiated surface temperature
[tex]\lambda_{max}[/tex] = peak wavelength
(a)
here,
[tex]\lambda_{max}[/tex] = 970 nm = 9.7 x 10⁻⁷ m
Therefore,
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}[/tex]
T = 2987.6 k
(b)
here,
[tex]\lambda_{max}[/tex] = 145 nm = 1.45 x 10⁻⁷ m
Therefore,
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}[/tex]
T = 19986.2 k
Grandma Sue (mass 80 kg) and her grandson James (mass 40 kg) are on a smooth icy surface. As Grandma Sue whizzes around the icy surface at 3 m/s in a straight line, she is suddenly confronted with scared James standing at rest directly in her path. Rather than knock him over, she picks him up and continues her uniform motion in a straight line without braking. Find the speed of Grandma Sue and James after the collision.
Answer:
v = 2 m/s
Explanation:
Here, we will use the law of conservation of momentum to solve this problem:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of grandma = 80 kg
m₂ = mass of James = 40 kg
u₁ = initial speed of grandma = 3 m/s
u₂ = initial speed of James = 0 m/s
v₁ = v₂ = v = final speed of grandm and James = ?
Therefore,
[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]
v = 2 m/s
A 4.51 kg object is placed upon an inclined plane which has an incline angle of 23.0*. The object slides down the inclined plane with a constant speed. Find the normal force, friction force and the coefficient of sliding friction
To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).
How to find the normal force ?The weight of the object is (4.51 kg) * (9.8 m/s^2) = 44.398 NTo find the friction force, we can use the equation: friction force = coefficient of friction * normal force.We can assume that the friction force is equal to the force of gravity acting against the object because it is moving down the inclined plane at a constant pace. As a result, the friction force is equal to the product of the object's weight and sin (incline angle)Friction force is equal to (9.927 N)*sin(23.0)*(44.398 N)We can use the following equation to determine the coefficient of sliding friction:friction coefficient is calculated as friction force divided by normal force.coefficient of sliding friction = 9.927 N /44.398 N = 0.224Therefore, the normal force is 44.398 N, the friction force is 9.927 N, and the coefficient of sliding friction is 0.224.To know more about normal force , check out :
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You are riding your bicycle down the street at a speed of 14 m/s. Your bicycle frame has a mass of 6.6 kg, and each of its two wheels has mass 2.2 kg and radius 0.35 m. Each wheel can be thought of as a hollow hoop (assuming that the rim has much larger mass than the spokes). What is the total kinetic energy of the bicycle (in Joules), taking into account both the translational and rotational motion
Answer:
1078 Joules
Explanation:
The computation of the total kinetic energy of the bicycle is shown below:
Given that
mass of bicycle's frame (m) = 6.6 kg
mass of each wheel (M) = 2.2 kg
radius of each wheel (r) = 0.35 m
And, the linear speed (v) = 14 ms
Now
As we know that
Angular velocity (ω) = v ÷ r
= 140 ÷ .35
= 40 rads
Total kinetic energy = translational kinetic energy + rotational kinetic energy
= (1 ÷2 × m × v^2) + (2 × 1 ÷ 2×I × ω^2)
= (0.5 × 6.6 × [14]^2) + (M × r2 × ω^2)
= 646.8 + (2.2 × 0.35 × 0.35 × [40]^2)
= 646.8 + 431.2
= 1078 Joules
just give me the right answer!
If the distance between two objects decreased, what would happen to the force of gravity between them?
It would increase.
It would stay the same.
It would depend on the speed.
It would decrease.
It would increase................
A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect friction between car and road. Answer in units of m/s.
Answer:
1.66 m/s
Explanation:
Work or kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex]
[tex]4864=\frac{1}{2} (3540)v^{2}[/tex]
v = 1.66 m/s
HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!
Answer:
a. 25000 J
b. 2500 J/s
Explanation:
Given,
Distance ( s ) = 50 m
Force ( f ) = 500 N
a.
To find : -
Work done ( W ) = ?
Formula : -
W = fs
W
= 500 x 50
= 25000 J
Therefore,
the work done by the force the horse exerts is
25000 J.
b.
To find : -
Power ( P ) = ?
Formula : -
W = Pt
P = W / t
P
= 25000 / 10
= 2500 J/s
Therefore,
the power produced if the movement took 10 s
is 2500 J/s.
Answer:
2500j/s
Explanation:
a defender running away from a goalkeeper at 5m/s is hit in the back by the goal kick. the ball stops dead and the players speed increases to 5.5m/s if the ball had a mass of 500g and the player mass is 70k how fast was the ball moving?
The goal kick strikes a defender racing at 5 m/s away from a goalkeeper in the back. The ball was moving at a speed of 70 m/s if it weighed 500g and the player's mass was 70k.
Why does the goalie in a football game pull his hands back after holding the ball that has been shot towards the goal?The goalkeeper extends the amount of time he has to hold the ball by pulling his hands back. He lessens the force (rate of change of momentum) the football exerts on him by lengthening the time.
Write the formula for momentum and define it.A vector quantity, momentum. A body's momentum is equal to P=m.v if it is travelling at a speed of v while having mass m. P=mv describes the magnitude of the momentum. Because momentum is a function of mass and velocity, its unit is the product of those two quantities.
By using conservation of momentum,
Initial momentum = Final momentum
(5 x 70) + (v x 0.5) = (5.5 x 70) + (0 x 0.5)
350 + 0.5v = 385 + 0
0.5v = 35
v = 70 m/s
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Kind of energy a piece of radioactive metal contains
Answer:
Radioactive materials give off a form of energy called ionizing radiation.
A boat has a speed of 20m/s in still waters. what is the speed in a river flowing due east with a velocity of 5m/s if the boat is heading north. How far from a point directly north on the other side of the river will the boat arrive if the width of the river is 10m.
The boat will arrive 2.5 meter in east far from a point directly north on the other side of the river.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction.
A boat has a speed of 20m/s in still waters.
The velocity of river is 5m/s in east.
The width of the river is = 10 meter.
Hence, time taken by the board to cross the river = 10/20 second
= 0.5 second.
In this time, the board will move in east direction = 5×0.5 meter = 2.5 meter.
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A 6 kg bowling ball is lifted 1.2 m into a storage rack. The acceleration of gravity is 9.8 m/s². Calculate the increase in the ball's potential energy. Answer in units of J.
Answer:
70.56 J
Explanation:
The increase in potential energy of the ball can be calculated using the formula:
PE = mgh
where:
PE is the increase in potential energy
m is the mass of the ball (6 kg)
g is the acceleration of gravity (9.8 m/s²)
h is the height the ball is lifted (1.2 m)
Substituting in the values, we get:
PE = (6 kg)(9.8 m/s²)(1.2 m)
This simplifies to:
PE = 70.56 J
So the increase in the ball's potential energy is 70.56 J.
Please give me the answer to this question
There is not enough information to determine the work done. Option iv
What is the work done?Let us note that we say that there is work done when the force that has been applied moves a distance in the direction of the force. In this case, we have been told that there is the combination of the works that is done by the object.
Now, we also have to note that we do not have other information to determine the work done such as the magnetic feild and the mass of the electron. All these are lacking in the question.
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Answer:
a. 6000J or 6KJ
b. Force =600
Some bat species have auditory systems that work best over a narrow range of frequencies. To account for this, the bats adjust the sound frequencies they emit so that the returning, Doppler-shifted sound pulse is in the correct frequency range. As a bat increases its forward speed, should it increase or decrease the frequency of the emitted pulses to compensate?
Answer:
As a bat increases its forward speed, it should decrease the frequency of the emitted pulses to compensate.
decrease
Explanation:
Decreasing the frequency of the emitted pulse will help the bat reduce its frequency caused by its forward motion. The forward motion shifts the bat's auditory frequency to a higher frequency; consequently, the bat should adjust downwards the frequency of the emitted pulse so the reflected pulse will be in the correct frequency range.