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Fantom [35]
4 years ago
11

You travel in a circle, whose circumference is 8kilometers, at an average speed of kilometer/hour. If you stop at the same point

you started from, what is your average velocity
Physics
1 answer:
dimaraw [331]4 years ago
3 0

You skipped over a number in the question, and you didn't tell me what my average speed is. Lucky for you, my average speed has NO EFFECT on the answer to the question.

When you calculate velocity, you only use the straight-line distance between the start-point and the end-point. It doesn't matter what route the thing took to get there, or how much ground it actually covered.

If I travel in a circle and stop at the same point I started from, then the size of the circle doesn't matter, and neither does my speed. The distance between my start-point and my end-point is zero, and my average velocity is zero.

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It is the thermal energy transferred from a hot object to a cold object.
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The following statement listed above is correct.

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4 years ago
Using a simple machine, a student is able to lift a 500N weight by applying only 100N. What is the mechanical advantage of the s
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Explain the Kinetic Theory of Matter<br> and use an example from everyday life
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3 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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