Explanation:
According to the question ,
- mass = 920kg
- time = 1s
- Initial velocity = 16m/s
- Final velocity = 9.5m/s
And we need to find out ,
As we know that Force is the rate of change of momentum ,
F = ∆p/t
F = mv - mu/t
F = m(v - u)/t
F = 920(9.5-16)/1 N
F = 920 * -6.5 N
Force = -5980N
The direction of the force will be opposite to the direction of velocity of the car .
- Distance travelled while breaking ?
Using the 3rd equation of motion , we have ,
2as = v² - u²
2 (v-u)/t *s =(v+u)(v-u)
2(t)(s) = v + u
2 * 1 * s = 16 + 9.5
s = 25.5/2 m
s = 12.75 m
Answer:
B. Resting energy expenditure is the same with basal energy expenditure.
Explanation:
Basal Energy Expenditure can be explained as the energy required to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body system.
On the other hand, Resting Energy Expenditure can be simply explained as the amount of energy expended or burnt when the body is resting.
Hence, in the context of definitions, and relating both definitions, it can be argued that Basal energy expenditure is simply the energy needed to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body, with special emphassy on the body being at rest. Thus, in this context, Basal energy can be looked at through the prism of Resting energy expenditure. Consequently, this two definitions can be used interchangeably, with a special emphassy on perspective.
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
In a stronger gravitational field a given mass will have a larger weight.