PLZ ANSWER THE QUESTION

A train travels north at a speed of 50 m/s.

Gala2k [10]

The apparent velocity is B) 48 m/s north

Explanation:

Here we have a problem of relativity of velocities.

In fact, the train is travelling north at a speed of

$v_t = 50 m/s$

where this velocity is measured with respect to the ground.

At the same time, a passenger on the train is walking towards the rear (so, south) at a velocity of

$v'=2 m/s$

where this velocity is measured with respect to the train, which is in motion in the opposite direction.

Therefore, the apparent velocity of the passenger with respect to an observer standing on the ground is:

$v=V_t - v' = 50 - 2 = 48 m/s$

And the direction is north, since this number is positive.

Learn more about velocity:

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8 0

3 years ago

When does constructive interference occur?

Anvisha [2.4K]

I think it occurs whenever waves come together so that they are in phase with each other.

5 0

3 years ago

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

7 0

3 years ago

Read 2 more answers

When is the average velocity of an object equal to the instantaneous velocity?

atroni [7]

Answer:

Average velocity of an object is equal to the instantaneous velocity when it's acceleration is zero.

Explanation:

4 0

3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

PLZ ANSWER THE QUESTION ​

PLZ ANSWER THE QUESTION