Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
F=mg
if m doubled, F would double as well
Answer:
955.36 seconds ≈ 16 minutes
Explanation:
Power(P) is the rate of doing work(W)
That is, P = W/t, where t is the time.
multipying both sides with 't' and dividing with 'P', we get: t=W/P
Here, W = 5.35 x 10^10 J and P = 5.6 x 10^7 W ( 1 W = 1 J/s).
Therefore , on dividing W with P, we get 955.36 seconds.
