Answers
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
[tex]v_{f}^{2} = v_{i}^{2}-(2*a*x)\\[/tex]
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
[tex]v_{f} =v_{i} - (a*t)[/tex]
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
[tex]v_{f}= v_{i}+a*t\\[/tex]
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
[tex]v_{f}= v_{i}+a*t\\[/tex]
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
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I guessed and was correct.
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e long answer to the following questions.
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Answers
Answer:
ans: Electromagnetic
Answer:
B on edg.
Explanation:
Carlos gets tired of pushing and instead begins to pull with force Fpull at an angle to the horizontal.
The block slides along the rough horizontal surface at a constant speed. A free-body diagram for the
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Blake: “The velocity of the block is constant, so the net force exerted on the block must be zero.
Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied
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What, if anything, is wrong with this statement? If something is
wrong, identify it and explain how to correct it. If this statement is
correct, explain why.
Answers
Answer:
The wrong items are;
1) The normal for FN equals the weight Fmg
2) The force of friction, Ff, equals the applied force Fpull
The corrected statements are;
1) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ)
Explanation:
The given statement was;
The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull
By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;
The normal force is equal to the weight less the vertical component of the applied force;
That is we have, FN = Fmg - Fpull × sin(θ)
The friction force similarly, is equal to the horizontal component of the applied force;
Ff = Fpull × cos(θ)
The wrong items are therefore as follows;
1) The normal for FN equals the weight Fmg
1 i) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction, Ff, equals the applied force Fpull
2 i) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ).
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Answers
Answer:
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Explanation:
Answer:
answer is B
Explanation:
took the test