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kati45 [8]
3 years ago
6

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area

of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
Physics
1 answer:
LekaFEV [45]3 years ago
8 0
6 cubic feet I’m pretty sure that’s the answer
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The following force diagram represents Newton’s Third Law of Motion:
Julli [10]

Answer:

<u>FALSE.</u>

Explanation:

Newton's third law states that :

  • <em>Every action has equal and opposite reaction</em>
  • <em>That is , the magnitude is the same but the directions are opposite</em>
  • <em>The action reaction forces DONOT operate on the same body.</em>

For example ,

If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>

Thus,

we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>

4 0
3 years ago
please hurry In which state of matter has the LEAST kinetic energy? A) gas B) liquid C) plasma D) solid
anastassius [24]
Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy.
7 0
3 years ago
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When you trace the outline of your palm how do you find its area​
oksian1 [2.3K]

Answer:

Explanation:

if its squares count the squares els messure it i think

6 0
3 years ago
Suppose you ran 53 km in 67 min. With what speed did you run?
melomori [17]

Answer:

47.46 kilometers an hour

Explanation:

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4 0
2 years ago
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A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
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