It gets warmer because the forming ice gives heat up
Answer: That is to say, a satellite is an object upon which the only force is gravity. Once launched into orbit, the only force governing the motion of a satellite is the force of gravity. Newton was the first to theorize that a projectile launched with sufficient speed would actually orbit the earth.
Explanation: Because Earth-orbiting objects follow elliptical paths around Earth and not a straight line, forces cannot, by definition, be balanced. Force is directional. It is a push or a pull in a particular direction.
Answer:
atmosphere
Explanation:
majority of nitrogen found in atmosphere,it's about 78% and present in form of gas
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
Answer:
The correct answer is "
".
Explanation:
The given problem seems to be incomplete. Please find the attachment of the complete query.
According to the question,
At point A, we have
⇒ 
or,
⇒ ![E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ]](https://tex.z-dn.net/?f=E_x%20%3D%209%5Ctimes%2010%5E9%5Ctimes%20%5B%5Cfrac%7B6%5Ctimes%2010%5E%7B-9%7D%7D%7B15%5E2%7D%5Ctimes%20%5Cfrac%7B9%7D%7B15%7D-%5Cfrac%7B8%5Ctimes%2010%5E%7B-9%7D%7D%7B20%5E2%7D%5Ctimes%20%5Cfrac%7B16%7D%7B20%7D%20%5D)
and,
⇒ 
or,
⇒ ![E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ]](https://tex.z-dn.net/?f=E_y%20%3D%209%5Ctimes%2010%5E9%5Ctimes%20%5B%5Cfrac%7B6%5Ctimes%2010%5E%7B-9%7D%5Ctimes%2012%7D%7B15%5E2%5Ctimes%2015%7D%2B%20%5Cfrac%7B8%5Ctimes%2010%5E%7B-9%7D%5Ctimes%2012%7D%7B20%5E2%5Ctimes%2020%7D%20%5D)
