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ale4655 [162]
3 years ago
8

Consider a 145 gram baseball being thrown by a pitcher. The ball approaches the batter with a speed of 44 m/s. The batter swings

and sends the ball flying back at a speed of 90 m/s into the outfield.
If the ball was in contact with the bat for a time of 3 ms, what is the average force exerted by the bat during the collision?
Physics
1 answer:
larisa [96]3 years ago
4 0

For the development of this problem it is necessary to apply the concepts related to Force, momentum and time.

By definition we know that momentum can be expressed as

P = F*t

Where

P = Momentum

F = Force

t = Time

At the same time, another way of expressing the momentum is through mass and speed, that is,

P = mv

Where

m = Mass

v = Velocity

Equating both equations

F*t = mv

F = \frac{mv}{t}

Our values are given as

m = 145*10^{-3}Kg

t = 3*10^{-3}s

v = 90-44 = 46m/s \rightarrow Net velocity

Replacing we have then,

F = \frac{mv}{t}

F = \frac{3*10^{-3}*46}{3*10^{-3}}

F = 2223.33N

Therefore the average force exerted by the bat during the collision is 2223.3N

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BartSMP [9]
That type of bending is called "diffraction" of waves.
6 0
3 years ago
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si
Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B)         -  Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m;  c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b  = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0
3 years ago
PLEASE HELP ASAP THANK YOU You're out for run. Your initial velocity is 1.5 m/s. Suddenly a crazy dog starts chasing you and you
Cerrena [4.2K]

Answer:

Your displacement would be

s = 22.5 m

s=1/2(v+u)t

5 0
2 years ago
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Ek=1/2mv^2 = (20)(1/2)(2x2) =40
8 0
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