Answer:
(orbital speed of the satellite) V₀ = 3.818 km
Time (t) = 4.5 × 10⁴s
Explanation:
Given that:
The radius of the Earth is 6.37 × 10⁶ m; &
the acceleration of gravity at the satellite’s altitude is 0.532655 m/s
We can calculate the orbital speed of the satellite by using the formula:
Orbital Speed (V₀) = √(r × g)
radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m
= (2.1 × 10⁷ + 6.37 × 10⁶) m
= 27370000
= 2.737 × 10⁷m
Orbital Speed (V₀) = √(r × g)
Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )
= 3818.215
= 3.818 × 10³
= 3.818 Km
To find the time it takes to complete one orbit around the Earth; we use the formula:
Time (t) = 2 π × 
= 2 × 3.14 × 
= 45019.28
= 4.5 × 10 ⁴ s
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-diameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. If the total irreversible head loss of the system is determined to be 1.5 m, determine the initial velocity of the water from the tank. Disregard the effect of the kinetic energy correction factors.
Ferromagnetic, paramagnetic, and diamagnetic
Answer: N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
Explanation:
See attached for a sketch.
From the attachment.
.
N = normal reaction force on block
W = weight of the block
theta = angle of the inclined plane to the horizontal
From the sketch, we can see that
N is equal in magnitude but opposite direction to Wy
N = Wy
And
Wy = Wcos(theta)
Wx = Wsin(theta)
Then,
N = Wy = Wcos(theta)
But W = mass × acceleration due to gravity = mg
N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
