Answer:
0.24 kg used up
Explanation:
He has a mass of 67 kg
The gravitational constant is 9.81 m/s^2
The distance upward is 3500 m
W = m*g*h
W = 67 * 9.81 * 3500
Work = 2,300,445 Joules
Work = 2300 kj
work = 2.30 * 10^6 joules in scientific notation.
Part B
He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.
4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.
The total amount in a kg of fat = 3.8 * 10^7 joules
x kg of fat is needed to provide 9.20.*10^6 joules
1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)
9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x
9.20 * 10 ^6 / 3.8 * 10^7 = x
x = 0.24 kg of fat are needed
The answer is A because when you relocate something you’re moving it to a different place from where you had it at first
Answer:
The difference in the phase angle between any two waves at any given position along the waves.
Explanation:
Phase difference describes the difference in the phase angle between any two waves at any given position along the waves. The equation of wave is given by :
Wave 1, 
Wave 2, 
Where
A is the amplitude of wave
is the angular frequency
is the phase difference between two waves
If
is the path difference, it is given by :

Hence, the correct option is (a) "the difference in the phase angle between any two waves at any given position along the waves".
I have no idea
sorry
good luck thooooo!!
Answer: The frequency heard will be f = 275.675Hz
Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.
The formula to calculate the Doppler effect is as follows
f = (
) · f₀
f is the observed frequency;
c is the speed of sound;
Vs is velocity of the source;
f₀ is the emitted frequency of source;
Substituting and calculating,
f =
· 300
f = 275.675 Hz
Thus, the frequency heard by the police officer is 275.675Hz.