A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
1 answer:
Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
You might be interested in
Answer:
The work done is 360 J.
Explanation:
Given that,
Mass = 50 kg
Distance =3 m
We need to calculate the work done
The work done is equal to the product of force and displacement.
Using formula of work done
Where, F = force
D = distance
θ = Angle between force and displacement
Put the value into the formula
Hence, The work done is 360 J.
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of