<u>We are given:</u>
Mass of Neptune = 1.03 * 10²⁶ kg
Distance from the center of Neptune (r) = 2.27 * 10⁷
now, computing the value of the acceleration due to gravity (g)
<u>Finding g:</u>
We know the formula:
g = G(mass of planet) / (r)²
g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]
g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)
which can be rewritten as:
g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15
g = (6.87 * 10¹⁵⁻¹⁴) / 5.15
g = (6.87/5.15) * 10
g = 1.34 * 10
g = 13.4 m/s² <em>(approx)</em>
Answer:
The speed is 0.97 c.
Explanation:
Given that,
Dilated time t'= 50.0 years
Rest time t = 13.0 years
We need to calculate the speed
Using formula of time dilation

Where, t' = change in time
t = rest time
v = velocity
c = speed of light
Put the value into the formula




Hence, The speed is 0.97 c.
The distance from the Earth's center to the point outside the Earth is 55800 Km
<h3>How to determine the distance from the surface of the Earth</h3>
- Acceleration due to gravity of Earth = 9.8 m/s²
- Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
- Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
- Mass of the Earth (M) = 5.97×10²⁴ Kg
- Distance from the surface of the Earth (r) =?
g = GM / r²
Cross multiply
GM = gr²
Divide both sides by g
r² = GM / g
Take the square root of both sides
r = √(GM / g)
r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]
r = 4.94×10⁷ m
Divide by 1000 to express in Km
r = 4.94×10⁷ / 1000
r = 4.94×10⁴ Km
<h3>How to determine the distance from the center of the Earth</h3>
- Distance from the surface of the Earth (r) = 4.94×10⁴ Km
- Radius of the Earth (R) = 6400 Km
- Distance from the centre of the Earth =?
Distance from the centre of the Earth = R + r
Distance from the centre of the Earth = 6400 + 4.94×10⁴
Distance from the centre of the Earth = 55800 Km
Learn more about gravitational force:
brainly.com/question/21500344
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Answer:
Temperature of the hot reservoir is 1540K
Explanation:
![E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}](https://tex.z-dn.net/?f=E%3D%201-%20%5Cfrac%7BT_%7Bc%7D%7D%5Btex%5D%7BT_h%7D%3D308%2B%7BT_c%7D%5C%5C%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EEfficiency%20of%20a%20carnot%20engine%20is%20given%20by%20the%20above%3C%2Fp%3E%3Cp%3ETc%3Dtemperature%20of%20the%20cold%20reservoir%3C%2Fp%3E%3Cp%3ETh%3D%20temperature%20of%20the%20hot%20reservoir%3C%2Fp%3E%3Cp%3EK%3D273%2B%2035%20%20%28convert%20%2035%C2%B0C%20to%20kelvin%29%3C%2Fp%3E%3Cp%3EK%3D308k%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%7BT_h%7D%3D%7BT_c%7D%2B308-----------------------%28equation%20%201%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E20%25%3D1-%7BT_c%7D%2F%7BT_h%7D)
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K