1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Answer:
(a) 61.25 N
(b) 6.25 kg
(c) 6.25 Kg
Explanation:
Weight on moon = 10 N
Acceleration due to gravity on moon = 1.6 m/s^2
Acceleration due to gravity on earth = 9.8 m/s^2
Let m be the mass of the package.
(a) Weight on earth = mass x acceleration due to gravity on earth
Weight on earth = 6.25 x 9.8 = 61.25 N
(b) Weight on moon = mass x acceleration due to gravity on moon
10 = m x 1.6
m = 6.25 kg
(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.
Both are constants used in the definition of Forces (gravitational and electric,respectively)
Since those constants are proportional to the magnitude of the forces:
Having a small gravitational constant explains why there is no apparent force of attraction with objects of considerable low mass (they would need to have great value of mass for the equation to give an apreciable force)
Electrical interactions are usually strong, and thus require an appropiate constant to depict the phenomenon. We deal in this case with charges really small, but the forces are in different order of magnitude.
Answer:
frictonal force due to the surface of irregularities
Hello!
This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).
We can write out the conservation of momentum formula:
m1vi + m2vi = m1vf + m2vf
Let:
m1 = mass of puck
m2 = mass of the goalkeeper
We know that the initial velocity of the goalkeeper is 0, so:
m1vi + m2(0) = m1vf + m2vf
m1vi = m1vf + m2vf
The final velocities will be the same, so:
m1vi = (m1 + m2)vf
Plug in the given values:
(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s
Using the equation for momentum:
p = mv
The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of
p = (120)(0.0533) = 6.396 kgm/s.
