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Maksim231197 [3]
3 years ago
15

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 light bulb to a higher resistance, but she wanted to keep the current the same. What will Venus need to do?
A) Increase the voltage by adding a bigger battery


B) Decrease the voltage by adding a bigger battery


C) Keep the same exact setup as she originally made


D) It is impossible to change the resistance and keep the current the same.
Physics
2 answers:
miskamm [114]3 years ago
7 0

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

Venus decided to change the light bulb to a higher resistance, but she wanted to keep the current the same.

In order to keep the same current flowing through a greater resistance, Venus will need to increase the voltage by changing to <em>a battery with greater voltage</em>. (A)

Dmitriy789 [7]3 years ago
4 0
<h2>Answer:</h2>

<u>A) Increase the voltage by adding a bigger battery </u>

<h2>Explanation:</h2>

According to Ohm's law

V = IR

where V is voltage, I is current and R is the resistance. If we write the equation for resistance we would get

R= V / I

Here we can see that Voltage is directly proportional to Resistance so in order to keep the balance if we increase the resistance then we must increase the voltage to keep the current constant.


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3 years ago
A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015
vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

N=\dfrac{1}{e}N_0

N_0 = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s

Using exponential decay formula

N=N_0e^{\dfrac{-bt}{m}}

Final amplitude = Initial times decay

\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s

The time taken is 33.33 seconds

7 0
3 years ago
A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit
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Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)=1.31 m/s^2

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

5.47=0+1.31\times t

t=\frac{5.47}{1.31}=4.17 s

Distance traveled during this time is

s=ut+\frac{1}{2}at^2

s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m

So remaining distance left to travel with constant velocity=153.57 m

thus time =\frac{distance}{speed}

t_2=\frac{153.57}{5.47}=28.07 s

Total time=28.07+4.17=32.24 s

6 0
3 years ago
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yKpoI14uk [10]

bonded pairs of electrons, lone pairs of electrons.

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3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
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Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
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