According to galilo in the absence of air resistance fraction then the body mass 60kg and 40kg will reach the earth surface at the same time....
Answer: 0.45
Explanation:
First note that the body that causes the body to move is its moving force (Fm) which is 9.0N
Since the mass of the body is 2.0kg, the weight will be;
W= mg = 2×10
W= 20N
For static body, the frictional force (Ff) acting on the body is equal to the moving force (Fm) since both forces acts along the horizontal on the body.
Ff = Fm = 9.0N
The normal reaction (R) on the body will also be equal to its weight(W) since weight acts downwards and the reaction acts in the opposite direction (upwards).
R = W = 20N
Ff = nR taking 'n' as coefficient of static friction between the drawer and the cabinet.
9.0 = 20n
n = 9/20
n = 0.45
Answer: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
The energy of one massive object has in relation with another massive object due to gravity is called gravity
