Ashot-putter puts a shot (weight=71.1n) that leaves

Ответ:

rakanmadi87

11.01.2022

Physics

Explanation:

It is given that,

weight of the shot- putter, W = 71.1 kg

Initial position, $x_i=1.52\ m$

Final position, $x_f=2.13\ m$

To find,

Work done and the change in potential energy.

Solution,

(a) Let W is the work done by the gravitational force when the shot has risen to a height of 2.13 m above the ground. It is given by :

$W_f=F\times x_f$

$W_f=71.1\times 2.13$

$W_f=151.44\ J$

(b) We know that the potential energy is equal to the work done by an object such that,

$\Delta P=P_f-P_i$

$\Delta P=W_f-W_i$

$\Delta P=151.44-71.1\times 1.52$

$\Delta P=43.36\ J$

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Ответ:

karenjunior

10.03.2020

Physics

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = $6.37\times 10^{6}\ m$

g = Acceleration due to gravity = 9.81 m/s²

$\omega$ = Angular speed

Force at equator would be

$F_e=m(g-\omega^2R)$

Force at pole

F_p=mg

From the question

$F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}$

Time period is given by

$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

The rotational period of the planet is 7160.2812 s or 1.988 hours

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