Answer:
if there is no friction in a simple machine, work output and work input are found equal in that machine
Explanation:
Answer:
450 kJ
Explanation:
Q = mCΔT
where Q is heat (energy),
m is mass,
C is specific heat capacity,
and ΔT is the temperature change.
Q = (1.2 kg) (4180 J/kg/°C) (100°C − 10°C)
Q = 451,440 J
Q ≈ 450 kJ
Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:
-4.8 m / s²
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
F = m a
Where the bold letters indicate vectors, F is the force, m the masses and the acceleration
The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system
-W = m a
Whera
is the trampoline force
Body weight is
W = mg
We substitute
- mg = ma
a =
Let's calculate
a = 
a = -4.8 m / s²
The negative sign indicates that the acceleration is directed downward.
In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is
-4.8 m / s²
Learn more here: brainly.com/question/19860811
Answer:
The weight of the body, W = 793.8 m/s²
Explanation:
Given,
The volume of the body, v = 45,000 cm³
The density of the body, ρ = 1.8 g/cm³
The mass of the body is given by the formula,
m = ρ x v
= 1.8 g/cm³ x 45,000 cm³
= 81,000 g
Hence, the mass of the body is m = 81 kg
The weight of the body,
W = m x g
= 81 kg x 9.8 m/s²
= 793.8 m/s²
Hence, the weight of the body, W = 793.8 m/s²