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olga nikolaevna [1]
3 years ago
6

When a force is applied to an object, what must occur for work to be accomplished on the object

Physics
1 answer:
slava [35]3 years ago
6 0

When a force is applied to an object the object "HAS to be traveling in the same direction as the force." If you put force in a object but it's moving in the opposite direction it would decrease the speed of the object or cause the object to come to a complete stop but if you apply force in a object in the same direction it''s moving in it would increase the speed of that object and cause the object to move faster.

Hope this helps!

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noname [10]
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A lamp that uses 100.0 watts of power requires a voltage of 210.0 volts. What is the resistance of the lamp?
alexandr402 [8]
Resistance = Voltage/Current
Wattage = Voltage * Current

<span> <span> </span> </span> That means the current drawn by the lamp is equal to 100 watts divided by 210 volts.

Resistance =\frac{210}{ \frac{100}{210} }


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What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your
torisob [31]
<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R 
I = MR² = (0.200)(5) = 1.00 kg-m² 
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size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>


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6 0
4 years ago
The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0
monitta
Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2

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4 0
3 years ago
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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
RSB [31]

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

4 0
3 years ago
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