The scale would need 10 aluminum cubes on one side. Figure out how many paper clips would be needed on the other side to balance this. You have to use more than one aluminum cube because you need to have enough cubes so that you get a whole number mass. 10 cubes gives you a total mass of 27 g for the aluminum.
Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=>
0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
Answer:
Conservation of Energy: the total energy of the system is constant. Conservation of Momentum: the mass times the velocity of the center of mass is constant. Conservation of Angular Momentum: The total angular momentum of the system is constant.
Explanation:
none
First we need to convert the mm to inches to make our computation
easier.
1mm = 0.0393701
32mm * 0.0393701 = 1.25 in
Solution:
C = 1/2d = ½ (1.25) = 0.625 in^4
Tension: tension = Te/J = 2T/ piC^3
= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi
Bending:
I = pi/4 * c^4 = 119.842 x 10^-3 in^4
M = (5)(600) = 3600 lb in
G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2
psi = -18.775ksi
Gx = -18.775 ksi
Gy = 0
Txy = 6.519 ksi
G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi
1.
G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi
G2 = Gave - R = -9/387 - 11.429 = -20.8
Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 =
-1.3889
ϴp = -27.1 degrees and 62.9 degrees
2.
Tmax = R = 11.43 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2
+ (6.519)^2 = 11.429 ksi
<span> </span>
