Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :

Where
is the density of air, 
So,

Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.
Answer:
B) Power is the rate at which work is done