Answer:
Speed of the satellite V = 6.991 × 10³ m/s
Explanation:
Given:
Force F = 3,000N
Mass of satellite m = 500 kg
Mass of earth M = 5.97 × 10²⁴
Gravitational force G = 6.67 × 10⁻¹¹
Find:
Speed of the satellite.
Computation:
Radius r = √[GMm / F]
Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)
Radius r = 8.146 × 10⁶ m
Speed of the satellite V = √rF / m
Speed of the satellite V = √(8.146 × 10⁶)(3,000) / 500
Speed of the satellite V = 6.991 × 10³ m/s
Answer:
Options d and e
Explanation:
The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.
We can get the length of the pendulums likely to oscillate with the formula;
where g=9.8m/s
ω= 2rad/s to 4rad/sec
when ω= 2rad/sec
L = 2.45m
when ω= 4rad/sec
L = 9.8/16
L=0.6125m
L is between 0.6125m and 2.45m.
This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.
Have a great day ahead
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
Can you please translate to English?
Answer:
p = mv
m = p/v = 125000/22 = 5682 kg
Explanation:
Direct application of the momentum equation
p = mv
where,
p: momentum
m: mass
v: object velocity
steps:
-------
1) check for units consistency ( SI or Imperial)
2) separate the variable you are looking for.
3) DONE! :DD
