While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers awa
y from the original site. Assuming that they were at a height of 137 meters, calculate the horizontal velocity of the spacecraft during touchdown if it lands in a free-fall mode without using retro engines. Consider gravity = 1.63 meters/second2.
1 answer:
Refer to the figure below.
A = original landing site.
B = alternate landing site.
Assume that aerodynamic resistance is negligible.
The craft lands in free-fall mode, therefore its original vertical velocity is zero.
It travels downward by 137 m at gravitational acceleration of g = 1.63 m/s².
The time of travel, t, obeys the equation
(137 m) = (1/2)(1.63 m/s²)*(t s)²
Therefore
t² = (137*2)/1.63 = 168.098 => t = 12.965 s.
The constant horizontal velocity, u, required to travel 4 km (or 4000 m) in time, t, is given by
(u m/s)*(12.965 s) = (4000 m)
u = 4000/12.965 = 308.5 m/s (nearest tenth)
Answer: 308.5 m/s
A = original landing site.
B = alternate landing site.
Assume that aerodynamic resistance is negligible.
The craft lands in free-fall mode, therefore its original vertical velocity is zero.
It travels downward by 137 m at gravitational acceleration of g = 1.63 m/s².
The time of travel, t, obeys the equation
(137 m) = (1/2)(1.63 m/s²)*(t s)²
Therefore
t² = (137*2)/1.63 = 168.098 => t = 12.965 s.
The constant horizontal velocity, u, required to travel 4 km (or 4000 m) in time, t, is given by
(u m/s)*(12.965 s) = (4000 m)
u = 4000/12.965 = 308.5 m/s (nearest tenth)
Answer: 308.5 m/s
You might be interested in
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :
So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
Explanation:
You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:
where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.
by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:
(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.
By doing the same for the electron you obtain:
we can equals these expressions for both proton and electron, because the forces qE are the same:
(a) Greater
The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, :
So we have:
- On wire A, the second-harmonic has frequency of , so the fundamental frequency is:
- On wire B, the third-harmonic has frequency of , so the fundamental frequency is
So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.
(b)
For standing waves on a string, the fundamental frequency is given by the formula:
where
v is the speed at which the waves travel back and forth on the wire
L is the length of the string
(c) Greater speed on wire A
We can solve the formula of the fundamental frequency for v, the speed of the wave:
We know that the two wires have same length L. For wire A, , while for wave B,
, so we can write the ratio between the speeds of the waves in the two wires:
So, the waves travel faster on wire A.