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Alborosie
4 years ago
6

how would the velocity of the book change if the applied force were equal to the sliding friction force

Physics
1 answer:
Inessa [10]4 years ago
3 0
Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is  <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>
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3 years ago
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g100num [7]

Answer:

<u>Question 2</u>

<u>Part (a)</u>

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<u>Part (b)</u>

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<u>Question 3</u>

This circuit is in parallel.

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4 0
2 years ago
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

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         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

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to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

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we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
3 years ago
If a vibrating string is made shorter (i.e., by holding your finger on it), what effect does
Ghella [55]
  <span>Pitch and frequency are more or less the same thing - high pitch = high frequency. 
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4 0
3 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

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Now, in our case

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E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
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