Forces between objects will decrease as the objects are moved farther away from each other.
In this case the force is decreased by 4 times because the objects are moved 4 times away from each other.
hope this helps :)
Answer:
(c) no different than on a low-pressure day.
Explanation:
The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.
Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.
<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>
Answer:

Explanation:
Each time the work done to raise a given mass is

here we know that


now we have


since it is just 25% of actual energy consumed as we know its efficiency is 25%
so we have total energy consumed in this way


now if it took N number of times so burn the fat of a pizza then


Answer:
no
actually no one weres the scarf in sunny ʕ•ε•ʔ
Explanation:
The colours we see are the wavelengths that are reflected or transmitted. For example, a red shirt looks red because the dye molecules in the fabric have absorbed the wavelengths of light from
Concept: According to Ohm's Law, the flow of electric current through a conductor is directly proportional to the potential difference across it, provided physical conditions (like temperature, pressure, volume etc.) remains same.
v = ir
or, r = v / i
Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.
voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).
In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.
In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.