Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much
<span>3.2 grams
The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So
72 / 38 = 1.894736842
So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams.
The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power
= 12 g / (e ^ ln(2)*1.8947)
= 12 g / (e ^ 0.693147181 * 1.8947)
= 12 g / (e ^ 1.313305964)
= 12 g / 3.718446464
= 3.227154167 g
So rounded to 2 significant figures gives 3.2 grams.</span>
Answer:
Wavelength = 736.67 nm
Explanation:
Given
Energy of the photon = 2.70 × 10⁻¹⁹ J
Considering:
where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Frequency is:
Frequency = c / Wavelength
So, Formula for energy:
Energy = 2.70 × 10⁻¹⁹ J
c = 3×10⁸ m/s
h = 6.63 x 10⁻³⁴ J.s
Thus, applying in the formula:
Wavelength = 736.67 × 10⁻⁹ m
1 nm = 10⁻⁹ m
So,
<u>Wavelength = 736.67 nm</u>
"These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation. In the Earth's polar regions ocean watergets very cold, forming sea ice." this is what I found...
Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so
Change in entropy 
=3( 3.36035 - 2.88) = 1.44 kJ/kg
