Answer:
at x=0
=W*L/L
and
=W-
Explanation:
Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.
moments is the product of force and the perpendicular distance in line of the action of the given force
from the principle of moments which states that the sum of clockwise moments ,must be equal to the sum of anticlockwise moments.
also, sum of upward forces must be equal to sum of downward forces
Going by the aforementioned,
taking moments about
W*(l-x)=*(L)
=W*(L-x)/(L)..............1
at x=0
=W*L/L
also
+ =W
=W-
Centripetal force of satellite = gravitational force between earth and the satellite
(m•v^2)/r = (G•Me•m)/r^2
v^2 = (G•Me)/r
v^2 = 6.67x10^-11 x 5.598x10^24 / 7.5x10^7
v = 2231 m/s
so answer is B
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.
Hi there!
This collision is an example of an inelastic collision since kinetic energy is lost from the collision.
We can represent this using the conservation of momentum formula:
m1v1 + m2v1 = m1vf + m2vf
Where:
m1 = blue ball
m2 = green ball
We know that the final velocity of the blue ball is 0, so:
m1v1 + m2v1 = m2vf
Rearrange to solve for the speed of the green ball:
(m1v1 + m2v1)/m2 = vf
Plug in given values:
((0.15 · 3) + (0.15 · 2)) / 0.15 = 5 m/s
