Answer:
Force of static friction between the two surfaces
Explanation:
When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.
This force is given by the relation
Where,
μ - coefficient of static friction
η - normal force acting on the body
When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.
So, in order to move the body, the applied force should be greater than the force of static friction.
The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.
Look at the formula for the gravitational force:
F = G m₁ m₂ / R² .
If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:
(6.1 N) / (2.5 kg) = (F) / (1 kg)
Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)
Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .
Answer:
Work- Activity involving mental or physical effort done in order to achieve a purpose or result.
Explanation:
This was the first definition from the dictionary on google :)
Answer:
The value is the temperature of the air inside the tire 
340.54 K
% of the original mass of air in the tire should be released 99.706 %
Explanation:
Initial gauge pressure = 2.7 atm
Absolute pressure at inlet 
= 2.7 + 1 = 3.7 atm
Absolute pressure at outlet 
= 3.2 + 1 = 4.2 atm
Temperature at inlet 
= 300 K
(a) Volume of the system is constant so pressure is directly proportional to the temperature.
340.54 K
This is the value is the temperature of the air inside the tire
(b). Since volume of the tyre is constant & pressure reaches the original value.
From ideal gas equation P V = m R T
Since P , V & R is constant. So
m T = constant
value of the original mass of air in the tire should be released is 
⇒ -0.99706
% of the original mass of air in the tire should be released 99.706 %.
