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MissTica
2 years ago
7

The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an obj

ect during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t1 = 0.5 s. The force as a function of time is given by F(t) = ať+ b, where a = –28 N/s2 and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po 0) plus its change in momentum due to the applied impulse: pf = po+SET+ F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for Pf?
a) tl tl Pf= 0++)16
b) 0+*+*+b)
c) 0+++bt) 0++) ti
Physics
1 answer:
aliina [53]2 years ago
4 0

Answer:

Follows are the solution to this question:

Explanation:

By checking the value in which we have calculated by performing its differentiation of \frac{a}{3}t^3+bt, the correct form of its integer value is calculating  with regard to t, that also provides as expected at^2+b = F(t).

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There are some missing data in the text of the problem. I've found them online:
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Solution:
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F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
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b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
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2 years ago
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Case A :

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x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

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(0.5) k x² + mgh = (0.5) m v²

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Case B :

B .60 kg 35 N/m .9 m

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k = spring constant of the spring = 35 N/m

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x = compression of spring = 0.25 m

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Case C :

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Using conservation of energy between Top of hill and Bottom of hill

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Case D :

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k = spring constant of the spring = 32 N/m

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x = compression of spring = 0.25 m

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hence closest is in case C at 5.1 m/s




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