For the first one 320
second 
1200W
Data
R = 12 Ω ∆V = 120V I =? P =?
Solution:
According to Ohm’s law,
∆V = I R
I = ∆V / R  
= 120 / 12  
= 10 A
Power P = I ∆V  
= 10 x 120  
= 1200 W
Third
∆V = 120 V P = 60 W I =? R =?
Use the formula, P = I ∆V
I = P / ∆V = 60 / 120 = 0.5 A
∆V = I R
R = ∆V / I = 120 / 0.5 = 240 Ω
 
        
             
        
        
        
Impulse is the integral of a force, F.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!
        
             
        
        
        
Answer:
The change of the volume of the device during this cooling is 
Explanation:
Given that, 
Mass of oxygen = 10 g
Pressure = 20 kPa
Initial temperature = 110°C
Final temperature = 0°C
We need to calculate the change of the volume of the device during this cooling
Using formula of change volume


Put the value into the formula



Hence, The change of the volume of the device during this cooling is 
 
        
             
        
        
        
and closing
.
The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.  
The mitral valve is located between the left atrium and the left ventricle. It closes the left atrium to collect oxygenated blood from the lungs and opens to pass it on to the left ventricle.
The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.
The aortic valve is located between the aorta and the left ventricle. It closes the left ventricle and opens to the aorta to pass on the oxygen-rich blood to the body.
The pulmonary valve is located between the pulmonary artery and the right ventricle. It closes off the right ventricle and opens to pass on unoxygenated blood to the lungs.