Electric potential difference is also known as

Which of the following is most critical for our study of outter space

nordsb [41]

In my opinion, the answer would be d: a universal language. D would technically be more important because there are so many foreign scientific companies and to get the information we need to study the universe, we should communicate and give eachother what we have of our resources as we work for it.

6 0

3 years ago

A ball with mass 0.11 kg is thrown upward with initial velocity 10 m/s from the roof of a building 20 m high. Neglect air resist

inessss [21]

Answer:

a) $H=25.1020\ m$

b) $t=3.2838\ s$

Explanation:

Given:

mass of the ball thrown up, $m=0.11\ kg$

initial velocity of the ball thrown up, $u=10\ m.s^{-1}$

height above the ground from where the ball is thrown up, $h=20\ m$

a)

Maximum height attained by the ball above the roof level can be given by the equation of motion.

As,

$v^2=u^2-2g.h'$

where:

$v=$ final velocity at the top height of the upward motion $=0\ m.s^{-1}$

$g=$ acceleration due to gravity

$h'=$ height of the ball above the roof

Now,

$0^2=10^2-2\times 9.8\times h'$

$h'=5.10\ m$

Therefore total height above the ground:

$H=h+h'$

$H=20+5.1020$

$H=25.1020\ m$

b)

Now we find the time taken in raching the height $h'$ :

$v=u-gt'$

$v=$ final velocity at the top of the motion $=0\ m.s^{-1}$

So,

$0=10-9.8\times t'$

$t'=1.0204\ s$

Now the time taken in coming down to the ground from the top height:

$H=u'.t_d+\frac{1}{2} g.t_d^2$

where:

$u'=$ is the initial velocity of the ball in course of coming down to ground from the top $=0\ m.s^{-1}$

Here the direction acceleration due to gravity is same as that of motion so we are taking them positively.

$25.1020=0+0.5\times 9.8\times t_d^2$

$t_d=2.2634\ s$

Therefore the total time taken in by the ball to hit the ground after it begins its motion:

$t=t'+t_d$

$t=1.0204+2.2634$

$t=3.2838\ s$

4 0

3 years ago

A bus with a vertical wind shield moves horizontally

Katena32 [7]

Answer:

(3) 53°

Explanation:

We want to measure the angle that the rains form with the vertical wind shield, so we have to measure the angle relative to the vertical. This means that we can write the following equation

$tan \theta = \frac{v_x}{v_y}$