Explanation:
Given that,
Weight of the engine used to lift a beam, W = 9800 N
Distance, d = 145 m
Work done by the engine to lift the beam is given by :
W = F d

Let W' is the work must be done to lift it 290 m. It is given by :

Hence, this is the required solution.
Answer:
A) reduced air pressure on the ball.
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:
Explanation:
1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is
100kg/h*0.5h = 50kg

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity
3. If we assume that the force of the boat before the raining is

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts
And if we take the net force as

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.
I hope this is useful for you
regards
We will apply the concept of period in a pendulum, defined as the product between 2
by the square root of the length over gravity, this is mathematically

Here,
T = Period
L = Length
g = Acceleration due to gravity
For the period to be 1 second, then we must look for the necessary length for such a requirement so




The meter's length would be slight less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g, because the time has been defined to be exactly 1s.
Therefore the correct answer is C.