Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.
1 answer:
You might be interested in
Answer:
a) 0.0288 grams
b)
Explanation:
Given that:
A typical human body contains about 3.0 grams of Potassium per kilogram of body mass
The abundance for the three isotopes are:
Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.
a)
Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.
However, the amount of potassium that is present in such person is :
0.012% × 240 grams
= 0.012/100 × 240 grams
= 0.0288 grams
b)
the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:
First the Dose in (Gy) =
=
=
Effective dose (Sv) = RBE × Dose in Gy
Effective dose (Sv) =
Effective dose (Sv) =
Answer:
K.E = 100 J
Final P.E = 100 J
Explanation:
The kinetic energy of any object can be given by the following formula:
where,
K.E = Kinetic Energy
m = mass of ball = 2 kg
v = speed of ball
Initially, v = 10 m/s. Therefore, the initial K.E is given as:
<u>K.E = 100 J</u>
Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:
Initial K.E + Initial P.E = Final K.E + Final P.E
Initial P.E is also zero due to zero height initially.
K.E + 0 = 0 + Final P.E
<u>Final P.E = 100 J</u>