I):final velocity=initial velocity+acceleration due to gravity*time of travel
Therefore,
v=0+9.8*3.1
=30.38 m/s
If by floor you meant water bed then i think there isn't enough info in the question.
Answer:
A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
