Answer:
0.182 m/s
Explanation:
m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s
let the velocity of loaded freight car is v
Use the conservation of momentum
m1 x u1 + m2 x 0 = (m1 + m2) x v
30,000 x 0.85 = (30,000 + 110,000) x v
v = 0.182 m/s
Answer:
Choice a. Approximately
on average.
Explanation:
The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.
Assume that electric charge of size
flowed through a wire cross-section over a period of time
. The average current in that wire would be:
.
For this question:
, whereas
.
Therefore, the average current in this circuit would be:
.
However, the units in the choices are all in
(for Amperes.) One Ampere is equal to one
. It will take some unit conversations to change the unit of
(coulombs-per-minute) to coulombs-per-second.
.
Hence, the most accurate choice here would be choice a.
Answer:


Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/r²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
r: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.
Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.
Known data
q₁ = 63 nC = 63×10⁻⁹ C
q₂ = -47 nC = -47×10⁻⁹ C
k = 8.99*10⁹ N×m²/C²
d₁ = 1.4cm = 1.4×10⁻² m
d₂ = 3.4cm = 3.4×10⁻² m
Calculation of r and β


Problem development
Ep: Total field at point P due to charges q₁ and q₂.

Ep₁ₓ = 0



Calculation of the electric field components at point P


Power is the ratio between energy and time:

In our problem we have E=76 J and t=3.7 s. Therefore, the power is
yup..........................................