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3 years ago

PLEASE PLEASE PLEASE HELP FAST!!!!!!!!!!!!!!!

Physics

1 answer:

aleksandr82 [10.1K]3 years ago

4 0

Answer:

I have examined the different parts of blub and I assume the answer is (D)T and U

Hope this helps you

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A _____ map provides information about how and when rocks formed in a particular area.A.contourB.geologicC.topographic

givi [52]

I believe the answer is B

6 0

3 years ago

The total mass of the train and its passengers is 750000kg. The train is traveling at a speed of 84m/s. The driver applies the b

fiasKO [112]

Answer:

|F| = 393750 N

Explanation:

Given that,

Total mass of the train, m = 750000 kg

Initial speed, u = 84 m/s

Final speed, v = 42 m/s

Time, t = 80 s

We need to find the net force acting on the train. The formula for force is given by :

F = ma

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{750000\times (42-84)}{80}\\\\F=-393750\ N$

So, the magnitude of net force is 393750 N.

4 0

3 years ago

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$