Answer:
Total work done in expansion will be 
Explanation:
We have given pressure P = 2.10 atm
We know that 1 atm 
So 2.10 atm 
Volume is increases from 3370 liter to 5.40 liter
So initial volume 
And final volume 
So change in volume 
For isobaric process work done is equal to 
So total work done in expansion will be
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
Answer:
0 m/s
Explanation:
Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.
In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.
As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.
Average velocity = displacement/time
v=0/30
v = 0 m/s
Hence, the correct option is (1).
