A weightlifter expends 294 w of power lifting a weight to a height of 2 m in 10 seconds. what mass has he lifted?
1 answer:
The power of the weightlifter is equal to the ratio between the work done by it and the time taken:
since we know the power and the time, we can find the work done:
For the work-energy theorem, this work done by the lifter is equal to the gravitational potential energy gained by the mass, which is given by:
where m is the mass and
is the increase in height. Rearranging the formula and using W=U, we can find the mass:
since we know the power and the time, we can find the work done:
For the work-energy theorem, this work done by the lifter is equal to the gravitational potential energy gained by the mass, which is given by:
where m is the mass and
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Answer:
the one with a higher mass
Explanation:
The body with more mass will have the greater kinetic energy of the two.
Kinetic energy is the energy due to the motion of body. It is mathematically expressed as:
K.E = m v²
m is the mass
v is the velocity
Since the velocity of the two bodies are the same, and mass is directly proportional to kinetic energy, the body with more mass will have a higher kinetic energy.
So between mass m1 and mass m2, the one with a greater mass will have a higher kinetic energy
Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =