Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:
Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same
Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:
As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:
b) If the temperatures changes, we can still reuse the ideal gas equation above:
Answer:
v = 2.21 m/s
Explanation:
The foreman had released the box from rest at a height of 0.25 m above the ground.
We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :
So, its velocity at the bottom of the ramp is 2.21 m/s.
The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:
5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
Answer:C
Explanation:
Mass energy of hydrogen fusing into helium
