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andreev551 [17]
3 years ago
7

Explain to your friend, who is willing to accept that light moves at the same speed in any frame, why clocks on a passing train

are not synchronized. If it helps, assume diat Anna is at tire middle of the train.
Physics
1 answer:
vovikov84 [41]3 years ago
7 0

First we start from the definition of synchronization itself. This is known as the rate of change in the occurrence of an event to occur at the same time. If the light clocks of a train are verified it will be possible to observe that there are different verification times between each of them different. This is because the light of the clocks at each end of the train does not appear instantaneously over them, but travels from one direction to another; the light at the rear must have traveled a distance that will cause the first clock to have changed position.

In this way, if a clock of these is observed at the same time, the light will generate that the first one is out of sync with the last one, which would cause the clocks not to be synchronized.

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View this and help out??
Zina [86]
I would say b as well. I’m sorry if it’s wrong
5 0
3 years ago
A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o
Thepotemich [5.8K]

Answer:

The value is  KE_b =0.710 \ J

Explanation:

From the question we are told that

   The mass  of the bullet is m_b  = 5.00 \ g  = 0.005 \  kg

  The mass  of the wood is  m_w =  900 \  g  =  0.90\  kg

   The height attained by the combined mass is  h =  8.0 \ cm  =  0.08 \ m

Generally according to the law of energy conservation    

    KE _b  =  PE_c

Here KE_b is the kinetic energy of the bullet before collision.

and PE_c is the  potential  energy of the combined mass of bullet and wood at the height h which is mathematically represented as

      PE_m  =  [m_b  + m_w] *  g *  h

So

   KE_b =PE_c   = [0.005  + 0.90] * 9.8 *0.08

=> KE_b =0.710 \ J

3 0
3 years ago
What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af
sladkih [1.3K]

Answer:

L=0.0045\ kg-m^2/s

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, \omega=4300\ rpm=450.29\ rad/s

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

L=I\omega

Where I is moment of inertia

For sphere, I=\dfrac{2}{5}mr^2

L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

So, the magnitude of the angular momentum of the sphere is 0.0045\ kg-m^2/s.

4 0
2 years ago
6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity
Svetlanka [38]

The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v

where:

m_1 = 6000 kg is the mass of the first train

u_1 = 5 m/s is the initial velocity of the first train (we take east as positive direction)

m_2 = 5000 kg is the mass of the second train

u_2 = -3 m/s is the initial velocity of the second train

v is the final combined velocity of the two trains

Re-arranging the equation and substituting the values, we find:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(5)+(5000)(-3)}{6000+5000}=1.37 m/s

And the positive sign indicates their final direction is east.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

8 0
3 years ago
A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

t = time

With the data we have we can find the time as well

v = \frac{x}{t}

t = \frac{x}{v}

t = \frac{18.6}{34}

t = 0.547s

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

y = \frac{1}{2} gt^2+0+0

y = \frac{1}{2} gt^2

y = \frac{1}{2} 9.8*0.547^2

y = 1.46m

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0
3 years ago
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