Answer:
The answer is choice A.
Explanation:
Assuming you are in a situation with a gravitational field. You can divide the motion of the bullet into two components. One horizontal and the other in the vertical.
"F=Vector Sum Of The Two Forces" Is the answer.
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
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
Answer:
1 * 10^-7 [J]
Explanation:
To solve this problem we must use dimensional analysis.
1 ergos [erg] is equal to 1 * 10^-7 Joules [J]
![1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]](https://tex.z-dn.net/?f=1%5Berg%5D%2A%5Cfrac%7B1%2A10%5E%7B-7%7D%20%7D%7B1%7D%2A%5B%5Cfrac%7BJ%7D%7Berg%7D%20%5D%20%5C%5C%3D%201%2A10%5E%7B-7%7D%5BJ%5D)