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3 years ago

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What is the magnitude of the instantaneous velocity (speed) of a freely falling object 25 s after it is released from a position

of rest? Consider the acceleration of gravity to be 10m/s^2. Answer in unit of m/s

1 answer:

Marta_Voda [28]3 years ago

6 0

Answer:

The instantaneous velocity of the object is 250 m/s

Explanation:

Given;

time of the object motion, t = 25 s

acceleration due to gravity, g = 10 m/s²

The instantaneous velocity of the object is given as;

v = u + gt

where;

u is the initial velocity of the object = 0 (since it was released from rest)

Substitute the givens;

v = 0 + (10 x 25)

v = 250 m/s

Therefore, the instantaneous velocity of the object is 250 m/s

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3 years ago

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

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Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

Harman [31]

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

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Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

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