Apple hits the surface with speed 16.2 m/s
The angle made by the apple velocity with normal to the incline surface is given as 20 degree
now the component of velocity which is parallel to the surface and perpendicular to the surface is given as


so here we have


<em>so its velocity along the incline plane will be 5.5 m/s</em>
Answer:
v = 1.32 10² m
Explanation:
In this case we are going to use the universal gravitation equation and Newton's second law
F = G m M / r²
F = m a
In this case the acceleration is centripetal
a = v² / r
The force is given by the gravitational force
G m M / r² = m v² / r
G M/r = v²
Let's calculate the mass of the planet
M = v² r / G
M = (1.75 10⁴)² 5.00 10⁶ / 6.67 10⁻¹¹
M = 2.30 10²¹ kg
With this die we clear the equation to find the orbit of the second satellite
v = √ G M / r
v = √ (6.67 10⁻¹¹ 2.30 10²¹ / 8.75 10⁶)
v = 1.32 10² m
Answer:
4.2 J
Explanation:
Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K
From specific heat capacity,
Q = cmΔt.............................. Equation 1
Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.
Given: m = 1 g = 0.001 kg, Δt = 1 °C
Constant : c = 4200 J/kg.°C
Substitute into equation 1
Q = 0.001×4200(1)
Q = 4.2 J.
Hence the energy absorbed or lost = 4.2 J
Answer:
(A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Explanation:
Given that,
Voltage = 8.2 V
Inductor = 38 mH
Resistance = 150 Ω
Time t = 0.110 ms
The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance
We need to calculate the current
Using formula of current

Put the value into the formula



(B). We need to calculate the store energy in the inductor
Using formula of energy

Put the value into the formula


{tex]E=7.04\ \mu J[/tex]
Hence, (A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.