Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer:
Explanation:
we know angular velocity in terms of moment of inertia and angular speed
ω .... (1)
moment of inertia of rod rotating about its center of length b
........ .(2)
using v = ωr
where w is angular velocity
and r is radius of rod which is equal to b
so we get 2v = ωb
ω = 2v/b ................. (3)
here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v
put second and third equation in ist equation
×
so final answer will be 
Answer:
a) Velocity = 4.2m/s
b) Acceleration = 2.94m/s^2
c) Force exerted on the floor= 1401.4×10^3N
Explanation:
a) Velocity,V=sqrt(2×9.8×0.900)
V= 4.2m/s
b) Vf2= V^2+2ay2
a= 4.2^2 - 0/2×3
a= 17.64/6= 2.94m/s^2
c) Newton's 2nd law indicates:
Fnet= F - mg=ma
F= m(g+a)
F=110(9.8+2.94)
F=110×12.94
F= 1401.4N
Answer:
90.77%
its capacity utilization rate for the month is 90.77%
Explanation:
The capacity utilisation rate can be expressed mathematically as;
Capacity utilisation rate = capacity used/Best operating level × 100%
Given;
Total Number of production time = 205hours
Production output/capacity used = 21400 units
Best operation rate = 115units/hour
Best operation output for the month of July( at best operation level )
=115units/hour × 205 hours = 23575 units
Capacity utilisation rate = 21400/23575 × 100%
= 90.77%
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u />
<u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u />![-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]](https://tex.z-dn.net/?f=-k%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20x%7D%5D_%7Bx%3DL%7D%3Dh%5BT%28L%2Ct%29-T_%7B%5Cinfty%7D%5D)
<u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
