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2 years ago

Please answer of this question

Physics

1 answer:

Bogdan [553]2 years ago

8 0

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

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A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the groun

zheka24 [161]

Answer:

a) t = 2.64s

b) Vf = -28.7m/s

Explanation:

If the balloon is descending, the velocity is -2.3m/s. So the equation to describe the postion of the falling camera is:

$Y = Vo*t - \frac{g*t^{2}}{2}$

$-41 = -2.3*t - \frac{10*t^{2}}{2}$ Solving for t, we get:

t1 = -3.1s and t2 = 2.64s We discard the negative time and use the positive one.

The velocity of the camera will be:

Vf = Vo - g*t = -2.3 - 10*2.64 = -28.7m/s

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3 years ago

A force of 7.50 N is applied to a spring whose spring constant is .298 N/cm. Find it’s length

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Answer:

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3 years ago

Calculate the efficiency of a machine if the input work is 200J and the output work is 150J.

lidiya [134]

Answer:

75% is the efficency of this maxhine

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3 years ago

What is the three laws of electrostatics?

dangina [55]

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3 years ago

Read 2 more answers

A traffic expert wants to estimate the maximum number of cars that can safely travel on a particular road at a given speed. He a

galben [10]

Answer:

<em>Speed s= 19</em>

<u>Explanation:</u>

Take note of the following parameters:

Speed= s,

Number of cars= N,

Number N of cars that can pass a given spot per minute=

N(s)=88s/16+16(s19)2

The principle of differentiation is here:

We let N(s) = N

N = 86s / (17 + 17((s/19)^2))

17N = 86s /(1 + s²/19²)

17N = 361* 86s /(361 + s²)

17N = 31046s /(361 + s²)

Next step;

Differentiate with respect to s

17N = 31046s /(361 + s²)

Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²

Therefore,

u = 31046s =====> du/ds = u’ = 31046

v = (361 + s²) ====>dv/ds = v’ = 2s

17N = 31046s /(361 + s²)

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

The maximum when dN/ds = 0

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

( 31046(361 + s²) - 31046s(2s) ) = 0

31046 [ (361 + s²) - 2s² ] = 0

31046 (361 - s²) = 0

(361 - s²) = 0

(19 - s)(19 + s) = 0

either s = -19 or s = 19, but s > 0

s = 19

4 0

3 years ago

Please answer of this question​

Please answer of this question