All categories

3 years ago

If a 2 kg ball traveling at 10 m/s hits a wall and stops in 0.03 seconds, then how much force will the ball experience?

1 answer:

hjlf3 years ago

8 0

Newton's second law is:
F=m*a,

where a=dv/dt, so

F=m*(dv/dt)

Rearranging gives:
F*dt=m*dv.

Basic integration gives:
F*t=m(vf-v0),
where vf and v0 are the final and initial velocities of the object respectively.

In your case vf=0, because the ball stops completely, and v0=10m/s.

Rearranging the last expression gives F=(m(vf-vo))/t.

Plug in numbers to find F=(2*10)/0.03=666.6 N

You might be interested in

Hydrogen has only one electron in its only (and outer) electron shell. If a hydrogen atom were to absorb a small amount of energ

Xelga [282]

Answer:

D. move up to another shell that would form

Explanation:

An atom has protons, neutrons and electrons. Protons and neutrons are present in the nucleus and electrons orbit the nucleus in fixed shells. An electron can jump to higher shell when it gains energy and lower one when it loses energy. Thus, when single electron in hydrogen atom is given a small amount of energy, it would jump to another higher shell.

8 0

2 years ago

A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m

katovenus [111]

Answer:

217.43298 m/s

Explanation:

$m_1$ = Mass of bullet = 19 g

$m_2$ = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

$\theta$ = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s$

As the momentum is conserved

$m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s$

The speed of the bullet is 217.43298 m/s

5 0

3 years ago

A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5

masya89 [10]

Answer:

39 m/s

Explanation:

7 0

2 years ago

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

Anni [7]

Answer:

$m=18000kg$

Explanation:

From the question we are told that:

Crane Length $l=20m$

Crane Mass $m_a=3000kg$

Arm extension at lifting end $l_l=15m$

Arm extension at counter weight end $l_c=5m$

Load $M_l=5000kg$

Generally the equation for Torque Balance is mathematically given by

$T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0$

$mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0$

$m=18000kg$

7 0

2 years ago

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is