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emmasim [6.3K]
2 years ago
14

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

away from the wall. if the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?
Physics
1 answer:
Inga [223]2 years ago
8 0

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

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What is the value of Vo in the
notsponge [240]

Answer: 60m/s

Explanation:

From the diagram:

Θ = 30°

Vertical resolution (y-axis) :

Voy = VoSinΘ

g in the upward direction = negative (-) = - g

Vfinal = 0

Distance (H) traveled along y =

Time taken to reach maximum height :

From v = u + at

0 = usinΘ - gt

gt = usinΘ

t = usinΘ / g

Horizontal resolution:

S = ut + 1/2at^2

Substituting t = usinΘ / g ; Voy = usinΘ

S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2

S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)

S = (u^2sin^2Θ) / 2g

Now if S = maximum height = 45m

Then,

45 = [Vo^2sin^2(30°)] / 2(10)

45 =[ Vo^2 * (0.5)^2] / 20

45 =( Vo^2 * 0.25) / 20

20 * 45 = Vo^2 * 0.25

900 / 0.25 = Vo^2

3600 = Vo^2

Vo = sqrt(3600)

Vo = 60m/s

5 0
3 years ago
A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s
garik1379 [7]

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

4 0
3 years ago
Read 2 more answers
Why does the cyclist have less kinetic energy at position A than at position B?
Semmy [17]
Suppose that the cyclist begins his journey from the rest from the top of a wedge with a slope of a degree above the horizontal.
 At point A (where it starts its journey), the energy is:
 Ea = m * g * h
 In other words, energy is only potential.
 At point B (located at the bottom of the wedge), the energy is:
 Eb = (1/2) * (m) * (v ^ 2)
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 For energy conservation we have:
 Ea = Eb
 That is, we have that all potential energy is transformed into kinetic energy.
 Which means that the cyclist has less kinetic energy at point A because that's where he has more potential energy.
 answer:
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6 0
3 years ago
Identify the following as
bazaltina [42]
3. Kinetic energy
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6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
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6 0
3 years ago
What is the impulse of a 3kg object accelerating from rest to 12m/s?
ale4655 [162]
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).

The answer is 36 kg m/s
6 0
3 years ago
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