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MAXImum [283]
2 years ago
14

A tennis ball traveling horizontally at 22.0 m/s suddenly hits a vertical brick wall and bounces back with a horizontal velocity

of 18.0 m/s
Physics
1 answer:
Shalnov [3]2 years ago
3 0
This problem has three parts.

I enclose a pdf file with the three body diagrams requested.

Here is an explanation of them.

Part A: Make a free-body diagram of this ball just before it hits the wall.

The only force acting on the ball is the pull of the Earth, this is its weight; so the diagram is a vertical vector downwards.

<span>
Part B: Make a free-body diagram of this ball just after it has bounced free of the wall.
</span>

<span>Again, the only force acting on the ball is the pull of the Earth, its weight, and the free-body diagram is identical to that of the part A.
</span>

<span>
Part C: Make a free-body diagram of this ball while it is in contact with the wall.</span>

When the ball is in contact with the wall, two forces act over it: the reaction of the wall, which is represented as a horizontal vector toward the left, and the gravity (weight), which is represented as a vertical vector downwards.
Download pdf
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What name is given to the wall of water that makes landfall just ahead of a hurricane?
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What is the answer?
Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE    =   KE

mgh = 1/2 mv^2     divide both sides of the equation by 'm'

gh    = 1/2 v^2             multiply both sides by 2

2 gh = v^2               take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

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2 years ago
Which condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge
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8 0
3 years ago
Read 2 more answers
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca
Aleks04 [339]

(a) Let v be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

a_{\rm rad} = \dfrac{v^2}R

where R is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

F = (1.500\,\mathrm{kg}) a_{\rm rad}

so that

(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N

Solve for v :

v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}

(b) The net force equation in part (a) leads us to the relation

F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}

so that v is directly proportional to the square root of R. As the radius R increases, the maximum linear speed v will also increase, so the cord is less likely to break if we keep up the same speed.

6 0
1 year ago
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
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