If "0.3 minute" is correct, then it's 9,543,272 Joules.
If it's supposed to say "0.3 SECOND", then the KE is 2,651 Joules.
The basic principles that apply to circuits is that electrons must receive energy from a source, and electrons transfer energy to perform some useful function.
<h3 /><h3>What is circuit?</h3>
Individual electronic components, like resistors, transistors, are connected by metallic wires or traces by which the electric current can flow to form a circuit design.
The basic principles that apply to circuits will be;
1. Electrons must receive energy from a source.
2. Electrons transfer energy to perform some useful function.
Hence, option 1 and 2 are correct.
To learn more about the circuit, refer to the link;
brainly.com/question/21505732
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Answer:
3WGGRGWERGRG
Explanation:
GERAGAETGAERR GHERUERHGRRGF;SBDF;JKSRDMFNSDFLDGGD;GDVF
Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Answer:
13.1 m/s
Explanation:
Given that a baseball is tossed up into the air at an initial velocity 18 m/s. The height of the baseball at time t in seconds is given by h(t) = 18t−4.9t 2 (in meters).
a) What is the average velocity for [1,1.5]?
To calculate the velocity travelled by the ball, differentiate the function.
dh/dt = 18 - 9.8t
Substitute t for 1 in the above Differential function
dh/dt = 18 - 9.8 (1)
But dh/dt = velocity
V = 18 - 9.8
V = 8.2 m/s
Average velocity = ( U + V ) / 2
Average velocity = (18 + 8.2)/2
Average velocity = 26.2/2
Average velocity = 13.1 m/s
