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3 years ago

A tennis ball rolls off the edge of a table. the table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Physics

1 answer:

k0ka [10]3 years ago

3 0

Answer:

v = 0.363 m/s

Explanation:

Given that,

The table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Here, u = 0 (at rest) for initial vertical velocity as it rolls off the edge of a table.

Let t is the time to fall from the vertical height. So,

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 0.55}{9.8}} \\\\t=0.33\ s$

It can be assumed to find the initial horizontal velocity of the tennis ball. It can be given by :

$v_x=\dfrac{x}{t}\\\\v_x=\dfrac{0.12}{0.33}\\\\v_x=0.363\ m/s$

Hence, the initial horizontal velocity is 0.363 m/s.

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A spring with spring constant 13.1 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

$T = 2\pi \sqrt{\frac{m}{K}}$

$m = \frac{T^{2}K}{4\pi ^{2}}$

$m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}$

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

v = 0.691 m/s

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3 years ago

In what way is a screw similar to an inclined plane?

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Explanation:

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A particle travels 15 times around a 10-cm radius circle in 42 seconds. what is the average speed (in m/s) of the particle?

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3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

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Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

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So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

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Answer:

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