Energy Conservation Theory,




<h3>What is law of energy conservation?</h3>
The principle of energy conservation states that energy is neither created nor destroyed. It may change from one sort to another. Just like the mass conservation rule, the legitimacy of the preservation of energy depends on experimental perceptions; hence, it is an experimental law. The law of preservation of energy, too known as the primary law of thermodynamics
To learn more about Energy Conservation Theory, visit;
brainly.com/question/8004680
#SPJ4
Answer:
W=315 x 10⁵ J
Explanation:
Given that
F= 2.5 x 10⁵ N
d= 90 m
K.E.=5.4 x 10⁷ J
We know that work done by all force is equal to the change in kinetic energy
Lets take work done by catapult is W
W + F.d= K.E.
W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J
W= (540 - 25 x 9) 10⁵ J
W=315 x 10⁵ J
Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.
Answer:
they cross over one another between charge.
Explanation:
Given that,
Terminal voltage = 3.200 V
Internal resistance 
(a). We need to calculate the current
Using rule of loop


Where, E = emf
R = resistance
r = internal resistance
Put the value into the formula


(b). We need to calculate the terminal voltage
Using formula of terminal voltage

Where, V = terminal voltage
I = current
r = internal resistance
Put the value into the formula


(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf


Hence, This is the required solution.