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Answer:
F = 41,954 N
Explanation:
given,
mass of bucket = 580 Kg
length of the cable = 20 m
velocity = 40 m/s
angle made = 38.0°
T cos 38° = m g..............(1)
T sin 38^0 = \dfrac{mv^2}{l} + F......(2)
dividing equation (2) by (1)



F = -46400 + 4445.36
F = -41,954 N
hence, the force is acting in the opposite direction as assumed.
F = 41,954 N
Answer:
If you push horizontally with a small force, static friction establishes an equal and opposite force that keeps the book at rest. As you push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the book moves.
Explanation:
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
Answer:
Net force exerted on the radio is 27.5 Newton.
Given:
Mass = 5.5 kg
Acceleration = 5 
To find:
Force exerted on the radio = ?
Formula used:
F = ma
Where F = net force
m = mass
a = acceleration
Solution:
According to Newton's second law of motion,
F = ma
Where F = net force
m = mass
a = acceleration
F = 5.5 × 5
F = 27.5 Newton
Hence, Net force exerted on the radio is 27.5 Newton.