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erastova [34]
3 years ago
10

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o

pposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
3 years ago
A piece of wood on top of an ocean wave stays in the same location, only moving up and down as the wave passes. It is energy tha
Ierofanga [76]

Answer:

Hey

Yes, this is true.

As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.

8 0
3 years ago
In February 1955, a paratrooper fell 365 m from an airplane without being able to open his chute but happened to land in snow, s
DaniilM [7]

Answer:

a. i=4760 kg*m/s

b. D_U= 1.11 m

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

K_E= (1/2)*85kg*(56m/s)^2

K_E=133280 J

D_U= \frac{133280J}{120000N} = 1.11 m

b)

The momentum is equal to the impulse on him from the snow so:

p=m*v

p=85kg*56m/s

i=4760 kg*m/s

7 0
3 years ago
In the middle of a thunderstorm, a lightning bolt flashes. It takes Roberto 5 seconds to
Gemiola [76]

Answer:

 v = 344.1 m / s    

 d = 1720.5 m

Explanation:

For this problem we must calculate the speed of sound in air at 22ºC

           v = 331 RA (1+ T / 273)

we calculate

           v = 331 RA (1 + 22/273)

           v = 344.1 m / s

the speed of the wave is constant,

           v = d / t

           d = v t

we calculate

           d = 344.1   5

           d = 1720.5 m

5 0
3 years ago
When light passes from water into air, the light bends away from the normal. Question 4 options: True False
kykrilka [37]

Answer:

True

Explanation:

8 0
3 years ago
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