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Sholpan [36]
3 years ago
5

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi

te direction. If the ballis in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is __________
a. 230 N.
b. 400 N.
c. 360 N.
d. 320N,
Physics
1 answer:
trasher [3.6K]3 years ago
3 0

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

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Answer:

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Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

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Now,

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20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

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a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee
Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

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F1 × r1 = k

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A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (
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Answer:

a=9.8 rad/s^{2}

Explanation:

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