A projectile fired upward from the Earth's surface will usually slow down, come momentarily to rest, and return to Earth. For a certain initial speed, however it will move upward forever, with its speed gradually decreasing to zero just as its distance from Earth approaches infinity. The initial speed for this case is called escape velocity. You can find the escape velocity v for the Earth or any other planet from which a projectile might be launched using conservation of energy. The projectile of mass m leaves the surface of the body of mass M and radius R with a kinetic energy Ki = mv²/2 and potential energy Ui = -GMm/R. When the projectile reaches infinity, it has zero potential energy and zero kinetic energy since we are seeking the minimum speed for escape. Thus Uf = 0 and Kf = 0. And from conservation of energy,
Ki + Ui = Kf + Uf
mv²/2 -GMm/R = 0
∴ v = √(2GM/R)
This is the expression for escape velocity.
Answer:
16.6 °C
Explanation:
From the question given above, the following data were obtained:
Temperature at upper fixed point (Tᵤ) = 100 °C
Resistance at upper fixed point (Rᵤ) = 75 Ω
Temperature at lower fixed point (Tₗ) = 0 °C
Resistance at lower fixed point (Rₗ) = 63.00Ω
Resistance at room temperature (R) = 64.992 Ω
Room temperature (T) =?
T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ
T – 0 / 100 – 0 = 64.992 – 63 / 75 – 63
T / 100 = 1.992 / 12
Cross multiply
T × 12 = 100 × 1.992
T × 12 = 199.2
Divide both side by 12
T = 199.2 / 12
T = 16.6 °C
Thus, the room temperature is 16.6 °C
If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)
Answer:
I think it's C I'm so sorry if I'm wrong.
Explanation:
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
