When a bus is moving (especially with a high speed) on the road suddenly stops or suddenly changes its direction, the luggage on
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Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given
M₀ = 9 Kg (Initial mass)
me = 0.225 Kg/s (Rate of fuel consumption)
ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)
v = ? if t = 20 s
y = ?
We use the equation
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
then we have
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
v = 1176.23 m/s
then we apply the formula
y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt
⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt
⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)
For t = 20 s we have
y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)
⇒ y = 741192.997 m = 741.19 km
The graphs are shown in the pics.
Answer:
P = 10 Watts
Explanation:
Given that,
The voltage of a mobile charger, V = 5 V
Current, I = 2 A
We need to find the power of the charger. The formula for the electric power is given by:
P = VI
So,
So, the power of the mobile charge is 10 watts.