A balloon contains 0.075 m^3 of
1 answer:
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The picture shows it has a real life something to display conservation of energy with kinetic energy and potential energy.
Five sentences are for potential and kinetic energy. Potential energy is to energy an object when it stores. Kinetic energy is something to motion. When the potential energy is slows down the potential energy it might be increases. As from the object when the speeds up and it is decreases to potential energy.
Kinetic energy is to calculated by KE= mass×velocity²/2 as a fraction.
Potential energy is to calculated by PE= mass×g×height.
And the another picture it has a <span>energy, kinetic energy, mechanical energy, conservation of energy.
</span>
Five sentences are for potential and kinetic energy. Potential energy is to energy an object when it stores. Kinetic energy is something to motion. When the potential energy is slows down the potential energy it might be increases. As from the object when the speeds up and it is decreases to potential energy.
Kinetic energy is to calculated by KE= mass×velocity²/2 as a fraction.
Potential energy is to calculated by PE= mass×g×height.
And the another picture it has a <span>energy, kinetic energy, mechanical energy, conservation of energy.
</span>
Answer:
The quantity of electrons that flows past a given point is 3.0 C.
Explanation:
An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).
i.e I =
It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.
Given that; I = 1.5A and t = 2s, find Q.
Q = It
= 1.5 × 2
= 3.0 C
The quantity of electrons that flows past a given point is 3.0 C.
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
