Answer:
The maximum load that this person is able to lift is 34.3 N
Explanation:
Applying the balancing torque, the expression is equal:
F₁L₁ = F₂L₂
Where
g = 9.8 m/s² = gravity
L₁ = 0.8 m
F₂ = 527 N
L₂ = 6 - 0.8 = 5.2 m
Replacing and clearing the mass m:
The maximum load that this person is able to lift is:
F = m * g = 3.5 * 9.8 = 34.3 N
Answer:
4.2 x 10⁷N
Explanation:
Given parameters:
Charge on ball:
q₁ = 3C
q₂ = 14C
Distance between balls = 9000m
Unknown:
Force acting on the two balls
Solution:
The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;
F =
where k = 9 x 10⁹Nm²/C²
q is the charges
r is the distance
Input the variables and solve;
F = = 4.2 x 10⁷N
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
Answer:
Amplitude will be equal to 0.091 m
Explanation:
Given mass of the slits = 41 gram = 0.041 kg
Frequency f = 1.65 Hz
So angular frequency
Angular frequency is equal to
Squaring both side
k = 4.40 N/m
For vertical osculation
A = 0.091 m
So amplitude will be equal to 0.0391 m