In 2005, the space probe Deep Impact launched a 370 kg projectile into Comet Temple 1. Observing the collision helped scientists
1 answer:
You might be interested in
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
Answer:
The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.
Explanation:
From Physics we get that expansion of the rod portion is found by this formula:
(Eq. 1)
Where:
- Expansion of the rod portion, measured in meters.
- Linear coefficient of expansion for titanium, measured in
.
- Initial length of the rod portion, measured in meters.
- Initial temperature of the rod portion, measured in Celsius.
- Final temperature of the rod portion, measured in Celsius.
If we know that ,
,
and
, the expansion experimented by the rod portion is:
The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.