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3 years ago

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A train is moving west with an initial velocity of 20m/s accelerates at 4m/s for 10 seconds during this time the train moves a d

istance

1 answer:

alex41 [277]3 years ago

4 0

Answer:

400m

Explanation:

From Newton's law of motion;

S = ut + 1/2 at2

Where U is initial velocity

a is acceleration

t is velocity hence;

S is distance covered

S = 20×10 + 1/2 × 4×(10)^2

= 200 + 200 = 400m

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A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$D^2 = 2 \ d^2$

$D = \sqrt{2 \ d^2}$

$D = \sqrt{2} \ d$

And this is the compression we are looking for

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Tamiku [17]

Answer:

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Explanation:

From the question,

Applying the law of conservation of momentum,

total momentum before collision = Total momentum after collision

mu+Mu' = mv+Mv'........................... Equation 1

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make v the subeject of the equation

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Given: m = 5.00 g = 0.005 kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s

Substitute these values into equation 2

v = [0-(0.12×0.5)]/0.005

v = -0.06/0.005

v = -12 m/s

The negative sign can be ignored since we are looking for the speed, which has only magnitude.

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3 years ago