Answer:
the heaviest barbell that could be lifted is 390.6kg
Explanation:
Hello!
To solve this question you must follow the following steps
1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.
2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.
W=mgh
where
W=work=6898.5J
m=mass
g=gravity=9.81m/s^2
h=height=1.8m
now we solve for mass, and use the values.
the heaviest barbell that could be lifted is 390.6kg
Answering the two questions in reverse order:
-- No. I don't need to know how the speed of the person changed before I can answer the question. I can answer it now.
-- The NET work done by the gravitational force is<em> zero</em>.
-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.
-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.
-- The total work done by gravity in one complete revolution is zero.
-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.
Answer:
inelastic collision.
Explanation:
An elastic collision consists in a collision in wich the bodies implied bounce off. Furthermore, the momentum of the bodies is conserved.
In as inelastic collision the momentum is not conserved, and is possible that the bodies implied in the collision stay joint after the collision.
In this case yo have that both bumper car and truck stayed joint after the collision. Hence, this situation is an inelastic collision.
Answer:
Q = 29.4 x 10⁻⁹ C.
Explanation:
Electric field due to a charge Q at distance d is given by coulomb law as follows
Electric field
E = k Q /d²
where for air k which is a constant is 9 x 10⁹
Given E = 32.08 , d = 2.84 m
Putting these values in the relation above, we have
![[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}](https://tex.z-dn.net/?f=%5Btex%5D32.08%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20Q%7D%7B%282.84%29%5E2%7D)
[/tex]
Q = 29.4 x 10⁻⁹ C.
