Answer:

Explanation:
speed of motor (N)=1500 rpm
power=4 hp =
=2.9828 KW
service factor(k)= 2.75
now,


torque rating

<h2>
Answer:</h2>
24Ω
<h2>
Explanation:</h2>
When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.
For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;
=
+ 
Solving for Rₓ gives;
=
------------------(i)
From the question;
Let
R₁ = resistance of first resistor = 40Ω
R₂ = resistance of second resistor = 60Ω
Now,
To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;
= 
= 
= 24 Ω
Therefore, the total resistance is 24Ω
Answer:
- The charge on the plates will increase with time
- The potential difference across the capacitor starts with zero and then increases gradually to a maximum value
- The current through the circuit starts high and then drops exponentially
Explanation:
<u>Case : An uncharged capacitor is connected to a resistor and a battery in a closed circuit.</u>
- The charge on the plates will increase with time
applying this equation : Q =
as the value of (t) increases the value of Q increases i.e. charge on the plates
- The potential difference across the capacitor starts with zero and then increases gradually to a maximum value
applying this equation : V = ![V_{0} [ 1 - e^{\frac{-t}{RC} } ]](https://tex.z-dn.net/?f=V_%7B0%7D%20%20%5B%201%20-%20e%5E%7B%5Cfrac%7B-t%7D%7BRC%7D%20%7D%20%5D)
- The current through the circuit starts high and then drops exponentially
current : I = 
Answer:
33.4
Explanation:
Step 1:
\sumMo=0 (moment about the origin)
Fb(15)-Fc(15)=0
Fb=Fc
Step 2:
\sumFx=0
-Fb-Fccos\theta+Ncsin\theta=0
Fc=0.3Nc=Fb
-0.3Nc-0.3Nccos\theta+Ncsin\theta=0
(-0.3-cos\theta+sin\theta)Nc=0----(1)
\sumFy=0
Nccos\theta+Fcsin\theta-Nb=0
Nccos\theta+0.3Ncsin\theta-Nc=0
Nc[cos\theta+0.3sin\theta-1]=0--------(2)
Solving eq (1) and eq (2)
\theta=33.4
Step 3:
As the roller is a two force member
2(90-\phi)+\theta=180
\phi=\theta/2
\phi=Tan(\muN/N)-1
\phi=16.7
\theta=2x16.7=33.4