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The correct answer is option A. 1/4 times
This is because distance and force are inversely proportional
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To solve this problem, we will calculate the entropy for both cases, remembering that the concept of entropy is the relationship between the heat released / gained and the temperature. After calculating the entropy in the sun and on the earth we will find the difference between the two. So that,
Entropy at Sun
Replacing,
The entropy at Earth,
The values are,
Replacing at the equation,
Then the total change in entropy will be,
Therefore the entropy change is 
Answer:
There are different ways to investigate density. In this required practical activity, it is important to:
record the mass accurately
measure and observe the mass and the volume of the different objects
use appropriate apparatus and methods to measure volume and mass and use that to investigate density
Explanation:
Answer:
82.7 m
Explanation:
u= 22m/s
a= 2.4 m/s^2.
t= 3.2 secs
Therefore the distance travelled can be calculated as follows
S= ut + 1/2at^2
= 22 × 3.2 + 1/2 × 2.4 × 3.2^2
= 70.4 + 1/2×24.58
= 70.4 + 12.29
= 82.7 m
Hence the distance travelled by the truck is 82.7 m
Answer:
q = -2 m and q = -0.5 m
Explanation:
For this exercise we must use the equation of the optical constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image, respectively
Let's start with the far vision point, in this case the power of the lens is
P = -0.5D
power is defined as the inverse of the focal length in meter
f = 1 / D
f = -1 / 0.5
f = - 2m
the object for the far vision point is at infinity p = infinity
1 / f = 1 / p + i / q
1 / q = 1 / f - 1 / p
1 / q = -1/2 - 1 / ∞
q = -2 m
The sign indicates that the image is on the same side as the object
Now let's lock the near view point
D = +2.00 D
f = 1 / D
f = 0.5m
the near mink point is p = 25 cm = 0.25 m
1 / f = 1 / p + 1 / q
1 / q = 1 / f - 1 / p
1 / q = 1 / 0.5 - 1 / 0.25
1 / q = -2
q = -0.5 m
the sign indicates that the image is on the same side as the object in front of the lens
