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ad-work [718]
3 years ago
9

How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?

Physics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

W = 47040  J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040  J

So, the required work done is 47040  J.

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Which of the following statements must be true?
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A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the
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A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
A small lead ball, attached to a 1.70-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
Otrada [13]

Answer:

H = 54.37

Explanation:

given,

lead ball attached at = 1.70 m

rate of revolution = 3 revolution/sec

height above the ground = 2 m

velocity = \dfrac{distance}{time}

circumference of the circle = 2 π r

                                             = 2 x π x 1.7

                                             = 10.68 m

velocity = \dfrac{3 \times 10.68}{1}

v = 32.04 m/s

using conservation of energy

\dfrac{1}{2}mv^2 + mgh = mgH

\dfrac{1}{2}v^2 + gh = gH

\dfrac{1}{2}32.04^2 + 9.8\times 2 = 9.8\times H

532.88 = 9.8\times H

H = 54.37

the maximum height reached by the ball is equal to H = 54.37

4 0
3 years ago
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