To answer this problem, we must remember that momentum is
conserved. Therefore,
Initial momentum = Final momentum
In this case we use angular momentum which it is defined
as:
Momentum =0.5 m ω^2
So finding for final angular velocity, ωf:
mi * ωi^2
= mf * ωf^2
120 kg * (0.5 rev / s)^2 = (120 kg + 22 kg) * ωf^2
ωf^2
= 0.2113
ω<span>f = 0.46 rev/s</span>
Answer:
v=15.24 m/s
Explanation:
Given that
Mass ,m= 1600 kg
radius ,r= 28 m
Coefficient of friction ,μ = 0.83
The radial force on the car when it takes turn
The friction force on the car
Fr= μ m g
The condition for motion without losing traction
F= Fr
v²=μ r g
Now by putting the values
( take g=10m/s²)
v=15.24 m/s
The speed of the car will be 15.24 m/s
Answer:
it does use it at doppler
Explanation:
My dad told me the hes a cop so yw
Answer:
L = 0.475 m = 475 mm = 18.7 inches
Explanation:
A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb ) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in).
E = 207 GPa = 207*10⁹ Pa
D = 10.2 mm = 0.0102 m
P = 8900 N
ΔL = 0.25 mm = 2.5*10⁻⁴ m
L = ?
We can use the Equation of the Hooke's Law
ΔL = P*L / (A*E) ⇒ L = ΔL*A*E / P
⇒ L = (2.5*10⁻⁴ m)*(π*(0.0102 m)²*0.25)*(207*10⁹ Pa) / (8900 N)
⇒ L = 0.475 m = 475 mm = 18.7 inches
Answer:
The magnetic field exerts net force and net torque on the loop.
