Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure,
= 2.95 atm
Atmospheric pressure,
= 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;

where;
P₁ = 
P₂ = 
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;

where;
is the density of seawater = 1030 kg/m³

Therefore, the water is flowing at the rate of 28.04 m/s.
Answer:
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Explanation:
Total heat content of the fat = heat content of water +heat content of the lipids
Let it be Q
the Q= (mcΔT)_lipids + (mcΔT)_water
total mass of fat M= 0.63 Kg
Q= heat supplied = 100 W in 5 minutes
ΔT= 20°C
c_lipid= 1700J/(kgoC)
c_water= 4200J/(kgoC)
then,

solving the above equation we get
m= 0.46 kg
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Answer:
The force is pull or push acting on the body which tends to change its state of rest or of motion is called force.
There are two types of force:
1.Contact force
2. Non-Contact
Answer:
Final velocity v = 8.944 m/sec
Explanation:
We have given distance S = 40 meters
Time t = 10 sec
As it starts from rest so initial velocity u = 0
From second equation of motion 


Now from first equation of motion
, here v is final velocity, u is initial velocity, a is acceleration and t is time
So 
Answer:
The work done by this force can be found via the following formula

Explanation:
Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.
