Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Answer:0.27*10^{6} (m)
Explanation:
with a wave we know that speed= wavelenght* frecuency so
wavelenght= 3*10^{7} / 109 = 0.27*10^{6} (m)
Answer:
t = 1.62 s
Explanation:
given,
mass of the block m₁ = 16.5 Kg
m₂ = 8 Kg
angle of inclination = 60°
μs = 0.400 and μk = 0.300
time to slide 2 m = ?
a) let a is the acceleration of the block m₁ downward.
Net force acting on m₂,
F₂ = T - m₂ g
m₂a = T - m₂ g
.......(1)
net force acting on m₁
F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T
m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T
.........(2)
from equations 1 and 2
T = 90.61 N
from equation (1)
.......(1)
a = 1.52 m/s²
let t is the time taken
Apply,
d = ut + 0.5 a t²
2 = 0 + 0.5 x 1.52 x t²
t = 1.62 s
The force is proportional to the product of the charges. So if either one increases, the force between them increases.
