Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
Hello,
<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:
p=mv
Where,
p = momentum
m = mass
v = </span>velocity
The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:
p=2.0 x 10^3 *15 or 2000(15)
p=30000
Thus, the cars momentum is 30000 kg m/s
Faith xoxo
The scientists describe them as D. Elliptical
<span>Product can be defined as a thing or person that is the result of an action or process</span>
Answer:

Explanation:
Given that
B(y, t) = k y ³t²
To find the total flux over the loop we have to integrate over the loop

Given that loop is square,so

B(y, t) = k y ³t²


We know that emf given as


So
