I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door )...
The door forms a right triangles that satisfies
We also have
so if you happen to know the height of the door, you can solve for and .
is fixed, so
We can solve for the angular velocity :
At the point when and ft/s, we get
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
<h3>
How to find the initial speed of the rock as it left the astronaut's hand?</h3>
- We have given with the following values,
- We have the expression for the initial velocity as,
- Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,
- Now, the velocity will become,
<h3>How to find the speed of the satellite?</h3>
- As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Learn more about the equations of planetary motion here:
brainly.com/question/28105768
#SPJ4
Answer
given,
diameter of segment 1, d₁ = 1 cm
diameter of segment 2, d₂ = 2 cm
diameter of segment 3 , d₃ = 0.5 cm
speed of liquid in first segment, v₁ = 8 m/s
a) using continuity equation
A₁ v₁ = A₂ v₂
v₂ = 2 m/s
speed in the second segment = 2 m/s
b) again, using continuity equation
A₂ v₂ = A₃ v₃
v₂ = 32 m/s
speed in the second segment = 32 m/s
c) Flow rate in the pipe
Q = A₁ V₁
Q = 6.28 x 10⁻⁴ m³/s
Q = 62.8 L/s
discharge in the pipe is equal to 62.8 L/s
Answer:
884 m
Explanation:
The motion of the baseball is a projectile motion: a uniform motion along the horizontal direction and a uniformly accelerated motion along the vertical direction.
The horizontal range of a projectile (the horizontal distance covered by the object before it reaches the original vertical height) is given by
where
u is the initial speed
is the angle of launch
g = 9.8 m/s^2 is the acceleration of gravity
For the ball in this problem,
u = 100 m/s
Substituting into the equation, we find the range of the ball:
so, the ball will cover 884 m before reaching the original vertical level.
Answer:
speed are different at different places
Explanation:
because it's top speed is 50km/h,so it's initial speed may be less. when it covers some distance it's speed changes again. then we have given a mean speed .mean speed means sum of all speed divided by sum of total time.so the mean speed and final speed differs from each other