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erastova [34]
3 years ago
10

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o

pposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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Jose is batting for the home team when he hits a foul ball that rises straight up over home plate. A fan in the stands notices t
andreyandreev [35.5K]

Answer:

<h3><em>28.01m/s.</em></h3>

Explanation:

Given maximum height reached by the ball as H = 40 metres

Since the ball rises straight up when hit by a ball, then the angle of launch will be perpendicular to the ground and that is 90°.

To determine the upward speed of the ball in meters per second after it got struck by the bat, we will use the formula for calculating the maximum height according to projectile motion;

Maximum Height H = \frac{u^2sin^2\theta}{2g} where;

u is the speed of the ball

\theta is the angle of launch

g is the acceleration due to gravity = 9.81m/s²

Substituting the given parameters into the formula;

40 = \frac{u^2sin^2(90)}{2(9.81)}\\ \\40 = \frac{u^2}{2(9.81)}\\ \\40 = \frac{u^2}{19.62} \\cross\ multiply\\\\u^2 = 40*19.62\\u^2 = 784.8\\u = \sqrt{784.8}\\ u = 28.01 m/s

<em>Hence the upward speed of the ball in meters per second after it got struck by the bat is 28.01m/s.</em>

5 0
3 years ago
An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

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<span>a. A solid will gain kinetic energy and become a liquid.</span>
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