Answer:
Energy is transferred from Priya to the box.
Explanation:
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Answer:
Im gonna say it is answer A:) Hope this helps!
Explanation:
Let's break the question into two parts:
1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.
1. Ramp Scenario: In an incline, the only component of cart's weight(
mg) that is in the direction of motion is
. Therefore the effort force in this case must be equal or greater than
.
Now we need to find

.

is the angle between the incline of the ramp and the ground.
Since the height is
5m and the length of the ramp is
8m, 
would be
5/8 or 0.625. Now that you have

, mutiple it with
mg.
=> m*g*

= 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is
125N.
2. Lever Scenario:
Just apply "moment action" in this case, which is:


= ?

= mg = 20 * 10 = 200N

= 10m

= 1m
Plug-in the values in the above equation:

= 200/10=
20NAs 20N << 125N, the best choice is to use lever.
Answer:
the air has to be unstable as well as it needs to be moved upwards.
Explanation:
it needs to be moved upwards and also needs to have unstable air.
Answer:
Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading
E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number
Explanation:
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