<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
Answer:
The strength of the magnetic field that the line produces is 
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where 
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is .
<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
Answer:
μ = 0.33
Equal to 3.2 m/s²
Explanation:
Draw a free body diagram of the block. There are three forces:
Normal force N pushing up.
Weight force mg pulling down.
Friction force Nμ pushing opposite the direction of motion.
Sum of forces in the y direction.
∑F = ma
N − mg = 0
N = mg
Sum of forces in the x direction.
∑F = ma
Nμ = ma
Substitute.
mgμ = ma
μ = a/g
μ = (3.2 m/s²) / (9.8 m/s²)
μ = 0.33
As found earlier, the acceleration is a = gμ. Since g and μ are constant, a is also constant, so it does not change with velocity.
Well, it goes 60 miles in one hour......so set up a ratio.....
60 miles/5 miles = 1 hr/x.....you'll get 60x = 5....then 5/60 would be 0.083
