Answer:
- The emf of the generator is 6V
- The internal resistance of the generator is 1 Ω
Explanation:
Given;
terminal voltage, V = 5.7 V, when the current, I = 0.3 A
terminal voltage, V = 5.1 V, when the current, I = 0.9 A
The emf of the generator is calculated as;
E = V + Ir
where;
E is the emf of the generator
r is the internal resistance
First case:
E = 5.7 + 0.3r -------- (1)
Second case:
E = 5.1 + 0.9r -------- (2)
Since the emf E, is constant in both equations, we will have the following;
5.1 + 0.9r = 5.7 + 0.3r
collect similar terms together;
0.9r - 0.3r = 5.7 - 5.1
0.6r = 0.6
r = 0.6/0.6
r = 1 Ω
Now, determine the emf of the generator;
E = V + Ir
E = 5.1 + 0.9x1
E = 5.1 + 0.9
E = 6 V
Answer:
by finding melting and boiling points
Explanation:
Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which 
, 
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by 
, the following manipulation confirms the answer to this question.
![\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5C%5C%5Csmall%20F_1%20%26%3D%5Csmall%20%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B4%5CDelta%20t%7D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%5Cbigg%5B%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B%5CDelta%20t%7D%5Cbigg%5D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7DF%5Cend%7Baligned%7D)
Here 
is the force that was applied to the object previously.
#SPJ4
Answer:
(a) the current in the coil is 12.03 A
(b) the maximum magnitude of the torque is 0.0854 N.m
Explanation:
Given;
number of turns, N = 185
radius of the coil, r = 1.6 cm = 0.016 m
magnetic dipole moment, μ = 1.79 A·m²
Part (a) current in the coil
μ = NIA
Where;
I is the current in the coil
A is the of the coil = πr² = π(0.016)² = 0.000804 m²
I = μ / (NA)
I = 1.79 / (185 x 0.000804)
I = 1.79 / 0.14874
I = 12.03 A
Part (b) the maximum magnitude of the torque
τ = μB
Where;
τ is the maximum magnitude of the torque
B is the magnetic field strength = 47.7 mT
τ = 1.79 x 0.0477 = 0.0854 N.m
