Answer:
The density of the fluid is 1100 kg/m³.
Explanation:
Given that,
Height = 5.00 cm
Pressure at top =594 Pa
Pressure at bottom = 1133 Pa
We need to calculate the change in pressure
Using formula of change in pressure
Where, 
= Pressure at bottom
= Pressure at top
put the value into the formula
Using formula of pressure for density
Where, 
= density
P = pressure
h = height
Put the value in to the formula
Hence, The density of the fluid is 1100 kg/m³.
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
Answer:
option D.
Explanation:
The correct answer is option D.
When an object is in equilibrium torque calculated at any point will be equal to zero.
An object is said to be in equilibrium net moment acting on the body should be equal to zero.
If the net moment on the object is not equal to zero then the object will rotate it will not be stable.
Explanation:
Force Description
1. 
It is also known as the weight of an object. It is the force that is exerted on an object due to its mass
2. 
It is force which is exerted by a push or a pull on an object. It is also known as applied force.
3. 
It is known as resistive force. It opposes the motion of an object.
4. 
It is the force which is at a right angle to the surface or perpendicular to the surface.
