Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
Answer:
The answer is "36 grams".
Explanation:
In this question, the weight of the ball is not mentioned but is the weight of the cookies is declared, which is equal to 36 grams, and all the cookies are squeezes into the ball and after that, it calculates the overall weight so, let assume that ball weight is =0 and then the overall weight is:
Answer:
Explanation:
Given that
The speed of the airplane ,v= 142 m/s
The speed of the air ,u = 30 m/s
Lets take angle make by airplane from east direction towards north direction is θ .
Now by using diagram ,we can say that
Now by putting the values in the above equation we get
Therefore the angle will be 12.19° .
Answer:
Net Force = 10N
Acceleration = 2m/s^2
Explanation:
calculate the net force and the acceleration on the block
Net force on the block F = mass * acceleration
Net force acting in the positive direction = 4N + 6N = 10N
Mass = 5kg
According to newton's second law;
a = F/m
a = 10N/5
a = 2m/s^2
hence the acceleration on the block is 2m/s^2
Answer:
<u><em>Rate of dissolving compounds:</em></u>
If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.
<u><em>Boiling Point of Compounds:</em></u>
If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.
