The small spheres that are moving through the circuit are the electric current. Current is the flow or movement of electrons. De
2 answers:
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<h2>Answer: 2T</h2>
According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period of a body (moon, planet, satellite) orbiting a greater body in space with the size
of its orbit.
This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):
(1)
Where:
is the Gravitational Constant
is the mass of planet X
is the radius of the orbit of the satellite around planet X
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
Now, we are asked to find the period when tha mass of the planet is . In order to do this, we have to rewrite equation (2) with this new value:
(3)
Solving:
(4)
On the other hand, if we multiply both sides of equation (2) by 2, we have:
(5)
As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.
Hence, the answer is:
If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>
Answer:
Final temperature is equal to 1291.63°R
Explanation:
given,
p₁ = 100 lb f/in², v₁ = 3.704 ft³/lb, and T₁ = 1000 °R
p₂ = 30 lb f/in² n = 1.4
Δ u = 0.171(T₂ - T₁)
we know for poly tropic process
p vⁿ = constant
p₁ v₁ⁿ = p₂ v₂ⁿ
100 × 3.704¹°⁴ = 30 × v₂¹°⁴
v₂ = 8.753 ft³/lb
work done for poly tropic process
W =
=
= 269.525 lbf/in².ft³
W = Btu/lb
= 49.87 Btu/lb
in the piston cylinder arrangement air is expanding acrobatically
Δ q = Δu + w
Δ u = - w
0.171(T₂ - T₁) = -49.87
0.171(T₁ - T₂) = -49.87
0.171 T₂ = 0.171 × 1000 + 49.87
T₂ = 1291.63 °R
Final temperature is equal to 1291.63°R
Answer:
0.68 s
Explanation:
We are given that
Initial velocity of box=
Final velocity of box=v=11.5 m/s
Distance=d=8.5 m
We have to find the time taken by box to slow by this amount.
We know that
Substitute the values
We know that
Acceleration=
Substitute the values
Hence, the time taken by box to slow by this amount=0.68 s