Frequency = 1 / (period) = 1 / (0.002 sec) = 500 Hz
Answer:
810.94 m/s
Explanation:
Applying,
v = √(2gR)............. Equation 1
Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth
From the question,
Given: g = 0.636 m/s², R = 517 km = 517000 m
Substitute these values into equation 1
v = √(2×0.636×517000)
v = √(657624)
v = 810.94 m/s
Hence, the escape velocity is 810.94 m/s
Answer:
A. 2.83 m/s
B. 39.55 degrees north of east
Explanation:
The velocity of the ball relative to the ground is the sum vectors of velocity of the ball relative to Mia and the velocity of Mia relative to the ground
If we take north east as positive direction then
Velocity vector of the ball relative to Mia is
Velocity of Mia relative to gorund is
<0, 5.3>
So velocity of the ball relative to the ground is
<1.8, -3.12> + <0, 5.3> = <1.8, 2.18>
Its magnitude is:
m/s
Its direction is
north of east
The answer to your Question is E
Answer:
The cheetahs velocity component due north is 6.99 m/s
The cheetahs velocity component due west is 18.21 m/s
Explanation:
Given;
speed of the cheetah, v = 19.5 m/s
direction of the cheetah's speed, θ = 69°
Find the attached document for explanation