Answer:
125.83672 seconds
Explanation:
P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)
m = Mass of professor = 103 kg
g = Acceleration due to gravity = 9.8 m/s²
h = Height of professor = 93 m
Work done would be equal to the potential energy
Power is given by
The time taken by the horse to pull the professor is 125.83672 seconds
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
Answer:
3.7kg
Explanation:
The following data were obtained from the question:
Volume = 3.7L
Mass =?
Next, we shall convert 3.7L to m³.
This is illustrated below:
1000L = 1m³
Therefore, 3.7L = 3.7/1000 = 0.0037m³
Now, we can obtain the mass of the water as shown below:
Density of water = 1000kg/m³
Volume of water = 0.0037m³
Mass of water =..?
Density = Mass /volume
1000kg/m³ = Mass /0.0037m³
Cross multiply
Mass = 1000Kg/m³ × 0.0037m³
Mass = 3.7Kg
Therefore, the mass of the water is 3.7Kg.
