All categories

3 years ago

The location where an earthquake begins

Physics

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

Explanation:

the location where an earthquake begin is called

You might be interested in

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

=》

${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

$v = 2 \sqrt{7} \: \: ms {}^{ - 1}$

=》

$5.29 \: \: ms {}^{ - 1}$

7 0

3 years ago

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the

dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

$S=\frac{1}{2}gt^2$

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s$

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

$v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s$