An electron is negatively charged.
If they become closer, it is increased, and if the objects become farther away is decreased.
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
Horizontal component of force(Normal reaction) :
Since, box is on the verge of slipping :
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
The answer is b or d u chooose i don’t know which one
Answer:
400 trips
Explanation:
Mechanical energy needed to climb 14 m by a man of 68 kg
= mgh
= 68 x 9.8 x 14
= 9330 J
1 Kg of fat releases 3.77 x 10⁷ J of energy
.45 kg of fat releases 1.6965 x 10⁷ J of energy
22% is converted into mechanical energy
so 22% of 1.6965 x 10⁷ J
= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.
one trip of climbing of 14 m requires 9330 J of mechanical energy
no of such trip possible with given mechanical energy
= 3732.3 x 10³ / 9330
= 400 trips