Answer:
the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
Explanation:
This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law
B- W = 0
B = W
the thrust force is equal to the weight of the liquid that is dislodged
B = ρ g V
we substitute
ρ g V = m g
V = m /ρ_fluid 1
we can write the mass of the pot as a function of its density
ρ_body = m / V_body
m = ρ_body V_body
V_fluid / V_body = ρ_body / ρ _fluid 2
Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
The effect appears the pot as if it had a lower apparent weight
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Answer:
the impedance of the circuit is 25.7 ohms.
Explanation:
It is given that,
Voltage, V = 50 volts
Frequency, f = 60 Hz
Resistance, R = 25 ohms
Capacitive resistance, 
Inductive resistance, 
We need to find the impedance of the circuit. It is given by :
Z = 25.7 ohms
So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
