Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as


= 1.87 A
Answer:
a) τmax = 586.78 P.S.I.
b) σmax = 15942.23 P.S.I
Explanation:
D = 3.81 in
d = 3.24 in
P = 930 lb
L = 3.7 ft = 44.4 in
a) The maximum horizontal shear stress can be obtained as follows
τ = V*Q / (t*I)
where
V = P = 930 lb
Q = (2/3)*(R³- r³) = (1/12)*(D³- d³) = (1/12)*((3.81 in)³- (3.24 in)³)
⇒ Q = 1.7745 in³
t = D - d = 3.81 in - 3.24 in = 0.57 in
I = (π/64)*(D⁴-d⁴) = (π/64)*((3.81 in)⁴- (3.24 in)⁴) = 4.9341 in⁴
then
τ = (930 lb)*(1.7745 in³) / (0.57 in*4.9341 in⁴)
⇒ τmax = 586.78 P.S.I.
b) We can apply the following equation in order to get the maximum tension bending stress in the pipe
σmax = Mmax *y / I
where
Mmax = P*L = 930 lb*44.4 in = 41292 lb-in
y = D/2 = 3.81 in /2 = 1.905 in
I = 4.9341 in⁴
then
σmax = (41292 lb-in)*(1.905 in) / (4.9341 in⁴) = 15942.23 P.S.I
Answer:
A. 95N
B. ?
C.225N
D.?
Explanation:
For Mars and Pluto im not so sure about those too but for A and C I am positive those are correct.
Sorry I could not help you all the way, please dont be mad ;(
The potential energy of the car when it let go is 20,000 J.
The speed of the car at the bottom of the ramp is 20 m/s.
The given parameters;
- <em>mass of the car, m = 100 kg</em>
- <em>height of the car, h = 20 m</em>
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The potential energy of the car is calculated as follows;
P.E = mgh
P.E = 100 x 10 x 20
P.E = 20,000 J
The speed of the car at the bottom of the ramp is calculated as follows;

Learn more here:brainly.com/question/18597080
Answer:
Normal force of 10,000N
Explanation:
From the question, the weight the car exerts on the pavement is 10,000N.
The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.
Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.