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4 years ago

A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Physics

1 answer:

Len [333]4 years ago

7 0

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

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Option (a) $f = \frac{0.5}{1.5} = \frac{1}{3}$

Option (b) uniform sphere

$f = \frac{2}{7}$

C) uniform hollow sphere

$f = \frac{2}{5}$

D) uniform hollow cylinder with inner radius R/2 and outer radius R

$f = \frac{5}{13}$

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$f = \frac{\frac{1}{2}Iw^{2} }{\frac{1}{2}mv^{2} + \frac{1}{2} Iw^{2} }$

$f = \frac{m k^{2}(\frac{v^{2} }{R^{2} }) }{mv^{2} + m k^{2}(\frac{v^{2} }{R^{2} } )}$

$f = \frac{\frac{k^{2} }{R^{2} } }{1 + \frac{k^{2} }{R^{2} } }$

A) uniform Solid cylinder for cylinder we know that

$\frac{k\\ ^{2} }{R^{2} }$ = 0.5

$f = \frac{0.5}{1.5} = \frac{1}{3}$

B) uniform Sphere for sphere we know that

$\frac{k^{2} }{R^{2} } = \frac{2}{5} \\$

$f = \frac{0.4}{1.4} = \frac{2}{7}$

C) uniform hollow sphere for hollow sphere we know that

$\frac{k^{2} }{R^{2} } = \frac{2}{3}$

$f = \frac{\frac{2}{3} }{\frac{5}{3} } = \frac{2}{5}$

D) uniform hollow cylinder with inner radius R/2 and outer radius R for annular cylinder

$\frac{k^{2} }{R^{2} } = \frac{5}{8}$

$f = \frac{\frac{5}{18} }{\frac{13}{8} } = \frac{5}{13}$

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