A person on a rocket traveling at 0.47 c (with respect to the Earth) observes a meteor come from behind and pass her at a speed she measures as0.47 c.How fast is the meteor moving with respect to the Earth?

Answers

Answer 1
Answer:

The concept to solve this problem is related to the relativistic physics for which the speed of the object in different frames of reference is related. This concept is called Velocity-addition formula

and can be written as,

u = (v+u')/(1+(vu')/(c^2))

Where,

u = Velocity of a body within a Lorentz Frame

v = Velocity of a second frame

u'= The transformed velocity of the body within the second frame

c = speed of light

Replacing we have to

u = (v+u')/(1+(vu')/(c^2))

u = (0.47c+0.47c)/(1+(0.47c*0.47c)/(c^2))

u = 0.769c

u \approx 230'700.000m/s

Therefore the meteor moving with respect to the Earth to 230'700.000m/s


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A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.​

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

\Sigma F = F - m\cdot g = m\cdot a(1)

Where:

F - Buoyant force, measured in newtons.

m - Mass of the plastic ball, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration, measured in meters per square second.

If we know that F = 5\,N, m = 0.408\,kg and g = 9.807\,(m)/(s^(2)), then the net acceleration of the plastic ball is:

a = (F)/(m) - g

a= 2.448\,(m)/(s^(2))

The acceleration is 2.448 meters per square second and is vertically upward.

A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.

Answers

The motion of sand is due to the movement of conveyor belt. The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

What is equation of motion?

The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.

The second equation of the motion for distance can be given as,

y=ut+(2)/(2)gt^2

Here, u is the initial body, g is the acceleration of the body due to gravity and t is the time taken by it.


Given information-

The conveyor is tilted at an angle of 18° above the horizontal.

The Sand is moved without slipping at the rate of 2 m/s.

The sand is collected in a big drum 5 m below the end of the conveyor belt.

The horizontal component of the velocity is given as,

v_y=2\cos 18

The vertical component of the velocity is given as,

v_y=2\sin18

Put the value in the above equation as,

y-y_0=v_yt+(1)/(2)gt^2

0-5=2\sin18 (t)+(1)/(2)*9.8\tiems t^2\nt=1.075\rm sec

The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is,

d=v_xt\nd=2\cos18*1.075\nd=2.044\rm m

Thus, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

Learn more about the equation of motion here;

brainly.com/question/13763238

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

y = yo+ V_y t +(1)/(2)gt^2

0 = 5 + 2sin18+ (1)/(2)*9.8t^2

solving for t

t = 1.075 sec

for horizontal motion

x = V_x t

x = 2cos18*1.075

x = 2.044 m

A person is making homemade ice cream. She exerts a force of magnitude 26 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.26 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 2.0 s, what is the average power being expended?

Answers

Answer:

P = 31.83 W

Explanation:

Our data are,

Magnitude of the force F = 26 N

Radius of the circular path r = 0.26 m

The angle between force and handle \theta = 0°

Time t = 2 s

We know that the formula to find the velocity is given by

Velocity v = (2\pi r)/(t)

v= (2\pi r)/(t)

v=(2 \pi 0.26)/(2)

v= 0.8168m/s

We know also that the formula to find the power is given by,

P = F*v

P = (26)(0.8168)

P = 31.83 W

A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?

Answers

Answer:

315.5 N/m

Explanation:

m = 500 g = 0.5 kg

T = 0.25 second

Total energy, E = 4 J

Let K be the spring constant.

The formula for the time period is given by

T = 2\pi \sqrt{(m)/(K)}

0.25 = 2* 3.14* \sqrt{(0.5)/(K)}

0.0398=\sqrt{(0.5)/(K)}

1.585* 10^(-3)={(0.5)/(K)}

K = 315.5 N/m

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.

Answers

Answer:

r = 7.1 × 10⁻¹⁵

Explanation:

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

I=\Delta p = m\Delta v

where

m = 62.0 kg is the mass of the passenger

\Delta v is the change in velocity of the car (and the passenger), which is

\Delta v = 5.30 m/s - 0 = 5.30 m/s

So, the linear impulse experienced by the passenger is

I=(62.0 kg)(5.30 m/s)=328.6 kg m/s

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

I=F \Delta t

where in this case

I=328.6 kg m/s is the linear impulse

\Delta t = 0.812 s is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N