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sattari [20]
3 years ago
11

A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop

to the ground.
a. 10sec
b. 8sec
c. 6sec
d. 12sec​
Physics
2 answers:
natali 33 [55]3 years ago
6 0

Answer:

<em>Choice: c. 6sec</em>

Explanation:

<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.

The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:

v_x=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

Where g=10 m/s^2

To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:

\displaystyle t=\sqrt{\frac{2h}{g}}

The marble rolls the edge of the table at a height of h=180 m, thus:

\displaystyle t=\sqrt{\frac{2*180}{10}}

\displaystyle t=\sqrt{36}

t = 6 sec

Choice: c. 6sec

Monica [59]3 years ago
5 0

Answer:

The answer is:

C) 6sec

Hope it helps:D

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A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
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Answer:

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Explanation:

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        \Delta L=\frac{PL}{AE}

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Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

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ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

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6 0
3 years ago
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu
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Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

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                                   qnet = (Iin - Iout)*t

                                   qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

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Where,                        k = 8.99*10^9   ..... Coulomb's constant

                                   qnet = V*r / k

                                   t = 1000*0.1 / (8.99*10^9 * 0.000002)

                                   t = 5.56 ms

                                   

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