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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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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?
Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, q=6.4\times 10^{-6}\ C

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV

So, the required electric potential is 192 kV.

3 0
2 years ago
An eastbound car travels a displacement of 80.0 Km in 45 minutes. What is the velocity
alexgriva [62]

Explanation:

d= 80km = 8000m

t = 45 min = 45/60 h

= 0.75 h

V= ?

we know that,

V = d /t

or,V= 80 km / 0.75 h

  1. or, V= 106.67 km/hr

or,V= 106.67×1000m / 3600 s

2. or, V= 29.63 m/s

4 0
2 years ago
If a train is traveling at 85 mph I need to travel 17,000 feet how many hours will it take? Break it down to show the math
Ahat [919]
1 hour for the train to travel 17,000 feet
6 0
3 years ago
Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro
LUCKY_DIMON [66]

Let \mathbf r_A denote the position vector of the ball hit by player A. Then this vector has components

\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}

where g=9.8\,\dfrac{\mathrm m}{\mathrm s^2} is the magnitude of the acceleration due to gravity. Use the vertical component r_{Ay} to find the time at which ball A reaches the ground:

1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s

The horizontal position of the ball after 0.49 seconds is

\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector \mathbf r_B of the ball hit by player B has

\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}

Again, we solve for the time it takes the ball to reach the ground:

1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s

After this time, we expect a horizontal displacement of 12 meters, so that v_0 satisfies

v_0(0.57\,\mathrm s)=12\,\mathrm m

\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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