When a satellite is revolving into the orbit around a planet then we can say
net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have


now we can say that kinetic energy of satellite is given as


also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it
so we will have

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite
Answer:
It would take
time for the capacitor to discharge from
to
.
It would take
time for the capacitor to discharge from
to
.
Note that
, and that
.
Explanation:
In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is
, and the initial charge of the capacitor be
. Then at time
, the charge stored in the capacitor would be:
.
<h3>a)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
<h3>b)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
Answer:
26945.6 ft⋅lbf
Explanation:
Volume of Right Circular Cone = pi*(radius^2)*(height/3)
Pi*(4)*(5/3) = 20.94 ft^3
Density = Mass / Volume
Mass = Density*Volume
Mass = (40)*(20.94)
Mass = 837.6 lb
Work = Force*Height
Force = Mass*Acceleration
Acceleration will be gravitational acceleration
Work = (837.6)*(32.17)*(1)
Work = 26945.6 ft⋅lbf
Answer:
<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

- v is image distance
- u is object distance, u is 10 cm
- f is focal length, f is 5 cm

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>
• Let's derive this formula from the lens formula:

» Multiply throughout by fv

• But we know that, v/u is M

- v is image distance, v is 10 cm
- f is focal length, f is 5 cm
- M is magnification.

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>
- Image is magnified
- Image is erect or upright
- Image is inverted
- Image distance is identical to object distance.
Answer:
1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%
Explanation:
For the first part, which is speed just before the crash, we can use energy conservation
Initial. Highest point
Em₀ = U = mg y
Final. Low point just before the crash
Emf = K = ½ m v²
Em₀ = Emf
m g y = ½ m v²
v = √ 2 g y
Let's calculate
v = √ (2 9.8 0.05)
v = 0.99 m / s
1) the moment before the crash is
p₀ = m v
p₀ = 0.221 0.99
p₀ = 0.219 kg m / s
After the collision, the car's speed is zero, so its moment is zero.
p = 0
2) change of momentum
Δp = p - p₀
Δp = 0- 0.219
Δp = -0.219 kg m / s
3) the reason is
Δp / p = 1
In percentage form it is 100%