Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
Answer:
It is (1/5)th as much.
Explanation:
If we apply the equation
F = G*m*M / r²
where
m = mass of a man
M₀ = mass of the planet Driff
M = mass of the Earth
r₀ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
F₀ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
g₀ = the gravitational acceleration on the surface of the planet Driff
we have
F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²
⇒ F₀ = G*m*M / (5*r²) = (1/5)*F
If
F₀ = (1/5)*F
then
W₀ = (1/5)*W ⇒ m*g₀ = (1/5)*m*g ⇒ g₀ = (1/5)*g
It is (1/5)th as much.
Gravitational potential energy can be described as m*g*h (mass times gravity times height).
Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.
After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.
To find how much energy was added, we subtract final energy from initial energy:
147 J - 44.1 J = 102.9 Joules.
Mechanical and electrical
Answer:
Hydroelectricity
Explanation:
Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.
