Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
Answer:
Given:
radius of the coil, R = 6 cm = 0.06 m
current in the coil, I = 2.65 A
Magnetic field at the center, B = 
Solution:
To find the number of turns, N, we use the given formula:
Therefore,
N = 22.74 = 23 turns (approx)
