Answer:
F = 479.21 N
Explanation:
given,
initial velocity = 0 m/s
final velocity = 16.7 m/s
time taken = 20.7 s
combined mass of the boat and trailer = 594 kg
tension in the hitch = ?
using equation of motion
v = u + a t
16.7 = 0 + a × 20.7
a = 0.807 m/s²
Force = mass × acceleration
F = 594 × 0.807
F = 479.21 N
Hence, the tension in the hitch that connects the trailer to the car is F = 479.21 N
Answer:
Explanation:
<u>Elastic Potential Energy
</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:
Substituting:
Calculating:
<h3>Answer</h3>
m/s^2 (meter per sec square)
Explanation:
acc = change in velocity/time
= distance/time
----------------
time
= m/s
------
s
=m/s^2
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
<span> (26 m/s)(1 rotation/0.62π m) ≈ 13.35 rotations/s that will do pig that'll do</span>
