9.3 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 70kg
Mass 2 = 2000kg
distance = 1m
Unknown:
force between them =
Solution:
The force between the two masses will be a gravitational force of attraction.
F = 
G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²
r is the distance between the two masses
Substituting the parameters:
F = 
= 9.3 x 10⁻⁶N
Learn more:
Universal gravitation constant
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Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
Answer:
If a body is electrically neutral it means it has no net charge.
Explanation:
Because it has the same number if protons as it does electron , which are opposite charges that offset eachother.
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
