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bogdanovich [222]

1 year ago

5

A 40,000 kg subway train is brought to a stop from a speed of 0.700 m/s in 0.250 m by a large spring bumper at the end of its tr

ack. What is the force constant k of the spring?

1 answer:

lozanna [386]1 year ago

5 0

Given,

The mass of the train, m=40000 kg

The initial velocity of the train, u=0.700 m/s

The compression in the spring bumper that stopped the train, x=0.250 m

The final velocity of the train, v=0 m/s

From the equation of motion,

$v^2-u^2=2ax$

Where a is the acceleration of the train.

On substituting the known values,

$\begin{gathered} 0-0.700^2=2a\times0.250 \\ \Rightarrow a=\frac{-0.700^2}{2\times0.25} \\ =-0.98\text{ m/s}^2 \end{gathered}$

The magnitude of the force applied by the train will be equal to the magnitude of the restoring force of the spring.

Therefore,

$\begin{gathered} m|a|=kx \\ \Rightarrow k=\frac{m|a|}{x} \end{gathered}$

Where k is the spring constant of the spring.

On substituting the known values,

$\begin{gathered} k=\frac{40000\times0.98}{0.250} \\ =156800\text{ N/m} \end{gathered}$

Therefore the spring constant of the spring is 156800 N/m

Thus the correct answer is option C.

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Since everything in the circuit is in series .. .

-- The total resistance is (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

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(2/5) of (9 volts) = 18/5 = 3.6 volts .

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3 years ago

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Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

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Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :

$C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1$

The center of mass in y direction is :

$C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5$

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

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3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

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A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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Here is your answer

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The direction of current will be just opposite to the direction of electron(negative charge) because current moves from positive to negative terminal whereas electron moves from negative to positive terminal.

So, direction of current- North to South

Now applying Fleming's Left hand rule we get the direction of force in downward direction, i.e. towards the floor.

HOPE IT IS USEFUL

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3 years ago

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Suppose a 1.2 m antenna is installed on top of a building that is 225 m tall,

vazorg [7]

Answer: 226.2 m

Explanation:

If we want to find the new total height $H$ of the building, we have to add to its initial height $h_{B}=225m$ the height of the antenna $h_{A}=1.2m$ , as follows:

$H=h_{B}+h_{A}$

$H=225m+1.2m$

Then:

$H=226.2m$ This is the new total height of the building.

7 0

3 years ago