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Leona [35]
3 years ago
10

A baseball player applies a force of 200N onto a baseball. Determine the time that baseball player exerts on the baseball if an

impulse of 50 Ns is exerted on the baseball.
Physics
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

Time, t = 0.25 seconds  

Explanation:

Given that,

Force applied by a baseball player, F = 200 N

We need to find the time that baseball player exerts on the baseball if an impulse of 50 Ns is exerted on the baseball. The impulse exerted on the baseball is given by :

J=F\times t

t is time taken

t=\dfrac{J}{F}\\\\t=\dfrac{50\ N-s}{200\ N}\\\\t=0.25\ s

So, the time for which the baseball player exerts on the baseball is 0.25 seconds.

jolli1 [7]3 years ago
3 0

Answer:A football player kicks a ball with a force of 50N. Find the impulse on the ball if his foot stays in contact with the football for ... A hockey player applies an average force of 80N to a 0.25kg hockey puck for a ... She uses a hammer to drive the nail into the wall. ... A 0.5kg baseball experiences a 10N force for a duration of 0.1s.

Explanation:

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A loud sound is produced in the downtown section of a city. Which of the following is least likely to occur with the sound wave
JulsSmile [24]

Answer:

A. The sound wave will reflect off Buildings and automobiles.

Explanation:

This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.

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3 years ago
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Answer:

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Explanation:

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5 0
2 years ago
Why do clothes often cling together after tumbling in a clothes dryer?
nadya68 [22]

Answer:

Explained

Explanation:

The electrons get rubbed off some items onto others. This causes an excess or electron on one item and deficiency of electron on the other. Therefore an electrostatic force of attraction is produced and hence clothes often cling together after tumbling in a clothes dryer

8 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
3 years ago
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Answer:

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Explanation:

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