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2 years ago

When the force acting on the body equal to acceleration?

Physics

1 answer:

topjm [15]2 years ago

5 0

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

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(10%) Problem 10: A 7.25-kg bowling ball moving at 9.85 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle

siniylev [52]

Answer

given,

mass of bowling ball = 7.25 Kg

moving speed of the bowling ball = 9.85 m/s

mass of bowling in = 0.875 Kg

scattered at an angle = θ = 21.5°

speed after the collision = 10.5 m/s

angle of the bowling ball

$tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}$

$tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}$

$tan \theta_1 = \dfrac{-[3.3672]}{62.86}$

$tan \theta_1 = 0.0536$

$\theta_1 =-3.066^0$

b) magnitude of final velocity

$v = \dfrac{-m_2v_2sin\theta_2}{m_1 sin\theta_1}$

$v = \dfrac{-0.875 \times 10.5 sin21.5^0}{7.25 sin(-3.066^0)}$

v = 8.68 m/s

5 0

3 years ago

Which type of force could be used to pull up on a metal paper clip without

solmaris [256]

Answer:

Magnetic force

Explanation:

Because a paper clip is type of a metal and magnets and paper clip get pulled to each other.

5 0

3 years ago

How much force is required to accelerate a 12 kg mass at 5 m/s 2

Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0

2 years ago

A race car moves along a circular track at a speed of 0.512 m/s. If the car's centripetal acceleration is 15.4 m/s2, what is the

tiny-mole [99]

Answer:

The radius is $r = 0.0170 \ m$

Explanation:

From the question we are told that

The speed at which the race car moves is $v = 0.512 \ m/s$

The centripetal acceleration is $a _r = 15.4 \ m/s$

Generally the centripetal acceleration is mathematically represented as

$a_r = \frac{v^2 }{r}$

=> $15.4 = \frac{0.512^2 }{ r}$

=> $r = 0.0170 \ m$

6 0

3 years ago

A ball with a mass of 275 g is dropped from rest, hits the floor, and re-bounds upward. If the ball hits the floor with a speed

disa [49]

Answer:

a) $\Delta p = 1.350\,\frac{kg\cdot m}{s}$ , b) $\Delta p' = -0.454\,\frac{kg\cdot m}{s}$ , c) D. The magnitud of the change in the ball's momentum.

Explanation:

a) The magnitude of the change in the ball's momentum is:

$\Delta p = (0.275\,kg)\cdot \left[\left(1.63\,\frac{m}{s} \right)-\left(-3.28\,\frac{m}{s} \right)\right]$

$\Delta p = 1.350\,\frac{kg\cdot m}{s}$

b) The change in the magnitude of the ball's momentum:

$\Delta p' = (0.275\,kg)\cdot \left[(1.63\,\frac{m}{s} )-(3.28\,\frac{m}{s} ) \right]$

$\Delta p' = -0.454\,\frac{kg\cdot m}{s}$

c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.

3 0

3 years ago