*Fig is Attached with answer*
Answer:
(a) Weight = Normal Force = 196.2 N
(b) Normal force table on 20 kg box = 294.3 N
Normal force 20 kg box on 10 kg box = 98.1 N
Explanation:
(a) Mass = m = 20 kg g = 9.81 m/s²
Weight = w = mg
w = 20 × 9.81
w = 196.2 N
As the box rests on the table so, normal force (NF) must be equal to the weight of the box.
NF = w = 196.2 N
(b)
m₁ = 20 kg m₂ = 10 kg
total mass = M = 30 kg
Total Weight = W = Mg
= 30 × 9.81
= 294.3 N
As both the boxes rest on the table so, normal force (NF) must be equal to the total weight of the boxes.
NF = W = 294.3 N
Weight of 10 kg box = 10 × 9.81 = 98.1 N
As the 10 kg box is placed on the top of 20 kg box, So 20 kg box must exert a normal force that is equal to the weight of 10 kg box.
Normal Force = Weight of 10 kg box = 98.1 N
Answer:
40000 N/m²
Explanation:
Applying,
P = F/A................... Equation 1
Where P = Pressure, F = Force, A = Area.
From the question,
The force(F) exerted by the person's foot is thesame as it's weight.
F = W = mg............ Equation 2
Where m = mass of the person, g = acceleration due to gravity.
Substitute equation 2 into equation 1
P = mg/A................ Equation 3
Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²
Substitute these values into equation 3
P = (60×10)/0.015
P = 600/0.015
P = 40000 N/m²
Answer:
Friction of the road on the motorcycle in the opposite direction
Explanation:
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