Answer:
120,000J
Corrected question;
In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?
Explanation:
Given;
Total energy Et = 500,000 J
Wasted Energy Ew = 380,000J
The amount useful energy is the amount of energy that is available for supply.
This can be derived by subtracting the wasted energy from the total energy.
Useful energy = Total Energy - wasted energy
Eu = Et - Ew
Substituting the given values;
Eu = 500,000J - 380,000
Eu = 120,000 J
The amount of useful energy produced in one hour is 120,000 J
Answer:
The answer is
<h2>28 kg</h2>
Explanation:
The mass of an object given it's momentum and velocity / speed can be found by using the formula
where
m is the mass
p is the momentum
v is the speed or velocity
From the question
p = 280 kg/ms
v = 10 m/s
The mass of the object is
We have the final answer as
<h3>28 kg</h3>
Hope this helps you
Answer:
The maximum current, in amperes, that a conductor can carry continuously under the conditions of abuse without exceeding its temperature rating.
<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>
