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3 years ago

15

You exert a force of 75 newtons on a rock. You push and you push, but you can’t budge it. You are exhausted! How much work did y

ou perform?

1 answer:

dalvyx [7]3 years ago

4 0

Answer: Work W = 0

Explanation: Work W = F·s. Because rock does not move, s = 0 and

work done is zero.

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Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

Friction force equation

Sedaia [141]

It’s FF= μ•Fn
Ff stands for friction force
The weird symbol is your coefficient of friction which has no units
Fn stands for normal force

4 0

3 years ago

Suppose the roller coaster in Fig. 6-38 (h1 = 33 m, h2 = 13 m, h3 = 25) passes point A with a speed of 2.10 m/s. If the average

Kamila [148]

Answer:

22.67 m/s

Explanation:

See in the picture.

7 0

3 years ago

A constant force of 3.2 N to the right acts on a 18.2 kg mass for 0.82 s. (a) Find the final velocity of the mass if it is initi

zheka24 [161]

Explanation:

The given data is as follows.

F = 3.2 N, m = 18.2 kg,

t = 0.82 sec

(a) Formula for impulse is as follows.

I = Ft = $\Delta P$

Ft = $m(v_{f} - v_{i})$

or, $v_{f} = \frac{Ft}{m} + v_{i}$

Putting the given values into the above formula as follows.

$v_{f} = \frac{Ft}{m} + v_{i}$

= $\frac{3.2 \times 0.82}{18.2} + 0$

= 0.144 m/s

Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.

(b) When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.

Ft = $m(v_{f} - v_{i})$

or, $v_{f} = \frac{Ft}{m} + v_{i}$

= $\frac{3.2 \times 0.82 sec}{18.2} - 1.85$

= -1.705 m/s

Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.

4 0

3 years ago

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1000-kg car travelling down on a steepest road, which has inclined angle of 18

eduard

Answer:

1

Explanation:

it was thinking about how much and then the other than you can get it would be able to ask her to be able and then the way you have been sent from your browser

8 0

3 years ago