Answer:
1 N
Explanation:
From coulomb's law,
The force of attraction between two charges is inversely proportional to the square of the distance between the charges.
From the question,
Assuming the charges are the same in both case,
F ∝ /r²....................... Equation 1
Fr² = k
F'r'² = Fr²........................... Equation 2
Where F' = First Force, r'² = First distance, F = second force, r² = second distance.
make F the subject of the equation,
F = F'r'²/r².................... Equation 3
Given: F' = 4 N, r' = 3 m, r = 6 m
Substitute into equation 3
F = 4(3²)/6²
F = 36/36
F = 1 N
Answer:
Less than 18000N
Explanation:
Given
This question will be answered using Newton's third law.
Understanding this law, it implies that reaction force is equal and opposite to the force exerted.
This implies that;
If the force exerted on the ball is 18000N
the force exerted is -18000N
So, the option that answers the question is less than 18000N because -18000N < 18000N
Answer: Option (C) is the correct answer.
Explanation:
Whether an acid is weak or strong it will cause harm if it comes in direct contact with the skin, eyes or mouth etc.
Thus, in order to prevent oneself, it is advised to wear a lab coat while working in laboratory as acids break down fabrics and can cause burns if the acids are strong.
Hence, a lab coat acts as a protective layer on the clothes so that even if acid falls on it then it might not reach the skin immediately and on time prevention can be taken.
Answer:
(a)10.5 rad/s2
(b) 20.9 rev
(c) 47.27 m
Explanation:
As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law
T = mg = 53*9.81 = 519.93 N
Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is
To = TR = 519.93 * 0.36 = 187.17 Nm
This solid cylinder would have a moment of inertia around it's rotating axis of:
(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder
(b) With such constant angular acceleration, the angle it would make after 5s is
Since each revolution equals to 
of angle, we can calculate the number of revolution it makes
(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
