Plz give me the answer of qiestion ofno 31

A ball is dropped from the roof of a building. the mass of the ball is 3.0 kg. what is the potential energy of the ball in the i

vitfil [10]

According to given condition there is no height(m) given from roof of building to the ground, there height given 18 m at a point above the ground. So, h=18m , mass=3kg , g=9.8m/s2 P.E=mgh P.E=(3)(9.8)(18) P.E=529J

6 0

3 years ago

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$