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Kryger [21]
3 years ago
10

While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c

hange the battery to a higher voltage, but she kept the same light bulb. What will happen to the current?
Physics
2 answers:
MakcuM [25]3 years ago
7 0
When Venus put in a different battery with higher voltage ... no matter
what other components were in the circuit ... the voltage across the
light bulb, and the current through it, both had to increase, and the
light bulb had to shine brighter than before.  
krok68 [10]3 years ago
5 0

The correct answer to the question is : Current increases.

EXPLANATION:

Before going to answer this question, first we have to understand Ohm's law.

From Ohm's law, we know that current flowing through a conductor is directly proportional to the potential difference maintained across the two ends of the conductor at constant temperature and pressure.

But, mathematically it is written as -

                  V ∝ I

               ⇒ V = IR

Here,  V is the potential across the two ends of the conductor.

           I is the current flowing through the conductor.

          R is the resistance of the wire.

As per the question, the voltage is increased.

Hence, the current flowing through the wire is increased.


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A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat
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Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

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v₀ = launching velocity = ?

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Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

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θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

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