Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) = 
First, let us find the area of the circular rod.
Area is given as:
Now, the resistance of the material is given by the formula:
Express this in terms of 'L'. This gives,
Now, plug in the given values and solve for length 'L'. This gives,
Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.
Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
Pitch is related to frequency
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
