Answer:
<h2>64.4 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
mass = 9.2 kg
acceleration = 7 m/s²
We have
force = 9.2 × 7 = 64.4
We have the final answer as
<h3>64.4 N</h3>
Hope this helps you
Solar energy, wind energy, hydro energy
Answer: 0.62
Explanation:
Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).
Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R
Since W = mg, W = 38.4 ×10
W= 384N =R
Normal reaction = 384N
The horizontal force acting on the body will be the moving force which is 238N
Coefficient of friction = Fm/R
Coefficient of friction = 238/384
Coefficient of friction = 0.62
Therefore, coefficient of kinetic friction between the box and the floor is 0.62
Answer: find the answer in the explanation
Explanation:
When a magnet is placed at the centre of the paper, and the nails are sprinkled on the paper, what will happen to the nails is that, the nails will form a pattern on the paper according to the magnetic field of the bar magnetic pole.
Other phenomena you can observe are:
1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines
2.) The direction of the line of forces
3.) The strength of the magnetic field pole.
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
