Normal reaction force on the block while it is at rest on the inclined plane is given as
here we know that
m = 46 kg
now we will have
now the limiting friction or maximum value of static friction on the block will be given as
Above value is the maximum value of force at which block will not slide
Now the weight of the block which is parallel to inclined plane is given as
here we know that
Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.
So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>
Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
The maximum magnetic field is calculated using the following intensity formula;
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N
Answer:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s
Its like a shoter way or writing "to the power" I think.
