Answer:0.302
Explanation:
Given
mass of crate m=27 kg
Force required to set crate in motion is 80 N
Once the crate is set in motion 56 N is require to move it with constant velocity
i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force
thus
where 
is the coefficient of static friction and N is Normal reaction
<span>Charge of the glass bead Q = 8.0 x 10^-9 C
Distance d = 2.0 cm = 0.02 m
Coulombs constant K = 8.99 x 10^9 Nm^2/C^2
Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2
E = 71.92 / 0.0004 = 17.98 x 10^4
The electric field is 1.8 x 10^5 N/C</span>
Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
