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Anna007 [38]
2 years ago
9

You leave your house at 7:50

Physics
2 answers:
saw5 [17]2 years ago
6 0

Answer:

Um okay??????????????????

igor_vitrenko [27]2 years ago
3 0

Answer:

LOL what?

Explanation:

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Physical Science Lesson 4 unit 2 questions, I need the answers
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3: Water breaking down into hydrogen and oxygen
4: I think it's Milk but I might be wrong.

I might be wrong about some of them, sorry it's not much.

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Calculate the work done to push a 200-N object 5-meters
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When a cannon fires a cannonball, the cannon will recoil backward because the:
gulaghasi [49]

C) total linear momentum of the ball and cannon is conserved.

Basically it happens that in the beginning before there is a momentum acting on the two bodies, these are a unique system. Here the total momentum of the System is 0. However, when the positive momentum of the cannonball is added, the system will be immediately affected by a negative momentum which will pull back the cannon. Could this be extrapolated as a condition of Newton's third law.

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Just want a ride with​
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8 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
2 years ago
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