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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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When an electric current flows through a long conductor, each free electron moves
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8 0
3 years ago
Read 2 more answers
If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object
777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0
3 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen
bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

f_{a} =(\frac{v+v_{c} }{v} )f_{s} (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s} (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s

b) Substitute eq. 1 in eq. 2:

f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz

7 0
3 years ago
a certain motor can produce a force of 70 n on a model plane. if the required acceleration is 30 m/s^2 then what is the maximum
8_murik_8 [283]
This is simply F=ma so 70N/30m/s^2 will give you the max mass which would be in kg, and the mass would be 2.333333kg a very light plane I might say
4 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a
zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0
3 years ago
Read 2 more answers
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