Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
Kinetic energy E = m * v^2
<span>Since the acceleration of both books will be -neglecting air resistance - the same, the kinetic energy will be directly proportional to the mass of the book.</span>
Answer: 3.4s
Explanation:
There are three stages in the motion of the ball, so you have to calculate the times for every stage.
1) Ball dropping from 9.5m: free fall
d = Vo + gt² / 2
Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²
⇒ t = √ (1.94 s²) = 1.39s
2) Ball rising 5.7m (vertical rise)
i) Determine the initial speed:
Vf² = Vo² - 2gd
Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²
⇒ Vo = 10.6 m/s
ii) time rising
Vf = Vo - gt
Vf = 0 ⇒ Vo = gt ⇒
t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s
3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)
i) d = 5.7m - 1.20m = 4.5m
ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²
⇒ t = √ (0.92 s²) = 0.96s
4) Total time
t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from

to

(the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates
1) The heat needed for process 1) is

where

is the water mass

is the water specific heat

is the variation of temperature of the water
If we plug the numbers into the equation, we find

2) The heat needed for process 2) is

where

is the water mass

is the latent heat of evaporation of water
If we plug the numbers into the equation, we find

So, the total heat needed for the whole process is