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faltersainse [42]
3 years ago
5

A baseball has a mass of 0.145 kg and approaches a bat at 40.0 m/s. After it is hit, the ball leaves the bat at 50.0 m/s directl

y back. Find the impulse of the bat on the ball.
7.25 kg*m/s
13.1 kg*m/s
5.80 kg*m/s
1.45 kg*m/s
Physics
1 answer:
topjm [15]3 years ago
3 0

Impulse is defined as change in momentum of object

so here we can say

Impulse = final momentum - initial momentum

I = P_f - P_i

I = m(v_f - v_i)

here we know that

m = 0.145 kg

v_f = 50 m/s

v_i = - 40 m/s

now we will have

Impulse = 0.145 kg(50 m/s + 40 m/s)

Impulse = 13.05 kg m/s

so impulse of the bat on ball is approx 13.1 kg m/s

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A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that
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Answer:

Part(a): The value of the spring constant is 3.11 \times 10^{2}~Kg~s^{-2}.

Part(b): The work done by the variable force that stretches the collagen is 1.5 \times 10^{-6}~J.

Explanation:

Part(a):

If 'k' be the force constant and if due the application of a force 'F' on the collagen '\Delta l' be it's increase in length, then from Hook's law

F = k~\Delta l....................................................................(I)

Also, Young's modulus of a material is given by

Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)

where 'A' is the area of the material and 'l' is the length.

Comparing equation (I) and (II) we can write

&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation (III), we have

k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}

Part(b):

We know the amount of work done (W) on the collagen is stored as a potential energy (U) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)

Substituting all the values, we can write

W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J

3 0
3 years ago
A person is lifting a bag of groceries with a uniform force of 75.0 N upward for 2.50 s. They raised the bag 1.50 m to place on
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Answer:

45 W

Explanation:

First of all, we need to find the work done by the person to lift the bag. The work done is given by

W=Fd

where

F = 75.0 N is the force applied

d = 1.50 m is the displacement of the bag

Substituting numbers into the formula,

W=(75.0 N)(1.5 m)=112.5 J

Now we can find the power used, which is given by

P=\frac{W}{t}

where

t = 2.50 s is the time taken

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P=\frac{112.5 J}{2.5 s}=45 W

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Two parallel conductors are carrying currents in the same direction. The currents are non-zero and not necessarily equal. The ma
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Answer:

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