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Yuliya22 [10]
2 years ago
14

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t

he current through the circuit is _____________ Amps. If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω. The current in the circuit is then __________ Amps. If a third identical lamp is connected in series, the total resistance is now _________Ω. The current through all three lamps in series is now _________ Amps. The current through each individual lamp is __________ Amps. What is the power when a voltage of 120 volts drives a current of 3 amps through a device? What is the current when a 90-W light bulb is connected to 120 V? How much current does a 75-W light bulb draw when connected to 120 V? If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work.
Physics
1 answer:
RSB [31]2 years ago
4 0

Answer:

The answers and workings is in the Explanation section

Explanation:

<em>In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. </em>

According to Ohm’s law V =I*R  

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

<em> If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω</em>.  <em>The current in the circuit is then __________ Amps. </em>

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer =  6Ω  and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer =  9Ω  and 0.67 Amps

<em>The current through each individual lamp is __________ Amps.  </em>

Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

<em>What is the power when a voltage of 120 volts drives a current of 3 amps through a device?  </em>

The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

Answer = 360 Watts

<em>What is the current when a 90-W light bulb is connected to 120 V? </em>

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer =  0.75 Amps

<em>How much current does a 75-W light bulb draw when connected to 120 V?</em>  

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

<em> If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work. </em>

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer =  $4.32

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The question to the above information is;

What is the best use of an atomic model to explain the charge of the particles in Thomson's beams?

Answer;

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation;

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Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula  v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

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f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

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2 years ago
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
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Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0
2 years ago
A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
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Answer:

T=4985.5^{\circ}K

Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

T=\frac{Q}{mc}+T_0

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

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2 years ago
How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

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g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0
1 year ago
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