Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
The velocity of tennis racket after collision is 14.96m/s
<u>Explanation:</u>
Given-
Mass, m = 0.311kg
u1 = 30.3m/s
m2 = 0.057kg
u2 = 19.2m/s
Since m2 is moving in opposite direction, u2 = -19.2m/s
Velocity of m1 after collision = ?
Let the velocity of m1 after collision be v
After collision the momentum is conserved.
Therefore,
m1u1 - m2u2 = m1v1 + m2v2
![v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2](https://tex.z-dn.net/?f=v1%20%3D%20%28%5Cfrac%7Bm1-m2%7D%7Bm1%2Bm2%7D%29u1%20%2B%20%28%5Cfrac%7B2m2%7D%7Bm1%2Bm2%7D%29u2)
![v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96](https://tex.z-dn.net/?f=v1%20%3D%20%28%5Cfrac%7B0.311-0.057%7D%7B0.311%2B0.057%7D%2930.3%20%2B%20%28%5Cfrac%7B2%20X%200.057%7D%7B0.311%20%2B%200.057%7D%29%20X-19.2%5C%5C%5C%5Cv1%20%3D%20%28%5Cfrac%7B0.254%7D%7B0.368%7D%20%2930.3%20%2B%20%28%5Cfrac%7B0.114%7D%7B0.368%7D%29%20X%20-19.2%5C%5C%20%5C%5Cv1%20%3D%2020.91%20-%205.95%5C%5C%5C%5Cv1%20%3D%2014.96)
Therefore, the velocity of tennis racket after collision is 14.96m/s
The answer is B tell me if I am wrong.
2040
15.4+2.2/2 until it equals 2.2
( divide by 3)
680(years)*3 devisions = 2040
Answer:
F = 0.00156[N]
Explanation:
We can solve this problem by using Newton's proposed universal gravitation law.
![F=G*\frac{m_{1} *m_{2} }{r^{2} } \\](https://tex.z-dn.net/?f=F%3DG%2A%5Cfrac%7Bm_%7B1%7D%20%2Am_%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%5C%5C)
Where:
F = gravitational force between the moon and Ellen; units [Newtos] or [N]
G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1= Ellen's mass [kg]
m2= Moon's mass [kg]
r = distance from the moon to the earth [meters] or [m].
Data:
G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1 = 47 [kg]
m2 = 7.35 * 10^22 [kg]
r = 3.84 * 10^8 [m]
![F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]](https://tex.z-dn.net/?f=F%3D6.67%2A10%5E%7B-11%7D%20%2A%20%5Cfrac%7B47%2A7.35%2A10%5E%7B22%7D%20%7D%7B%283.84%2A10%5E8%29%5E%7B2%7D%20%7D%5C%5C%20F%3D%200.00156%20%5BN%5D)
This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.