To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,
Through the aforementioned formula we will have to
The particulate part of the rest, so the final speed would be
Now from Newton's second law we know that
Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then
Replacing we have that,
Therefore the force that the water exert on the man is 1386.62
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer:
1.8 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)
v = 5.47 m/s
The acceleration down the incline is constant, so given:
Δx = 4.8 m
v₀ = 0 m/s
v = 5.47 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (4.8 m) / (5.47 m/s + 0 m/s)
t = 1.76 s
Rounding to two significant figures, it takes 1.8 seconds.
