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hram777 [196]
2 years ago
6

3. A 3.0 × 10−3 kg copper penny drops a distance of 50.0 m to the ground. If 65 percent of the initial potential energy goes int

o increasing the internal energy of the penny, determine the magnitude of that increase.
Physics
1 answer:
Ganezh [65]2 years ago
5 0

Answer:

The magnitude of the increase is 0.9555J

Explanation:

Potential Energy (PE) = mgh

mass (m) = 3×10^-3kg, acceleration due to gravity (g) = 9.8m/s^2, distance (h) = 50m

PE = 3×10^-3 × 9.8 × 50 = 1.47J

Increase in internal energy = 65% of 1.47J = 0.65 × 1.47J = 0.9555J

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A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o
Vlada [557]

Answer:

The temperature is   T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

Explanation:

From the question we are told that

      The height on a cool spring day is H_O

      The temperature on a cool spring day is  T_O

      The difference in height between a cool spring day and a summer day  is     h

     The coefficient of static friction is \alpha _{steel}

The mathematical relation for the linear expansion of the steel buiding is represented as

               h  =  H_o \alpha_{steel}  [T-T_O]

Where T is the temperature of the steel during summer

Now making T the subject we have

                T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

5 0
2 years ago
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
2 years ago
PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north
Serggg [28]

I think this is right I hope this is right for you

7 0
2 years ago
Read 2 more answers
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
11111nata11111 [884]

Answer:

\sigma=0.014\ C/m^2

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2

So, the surface charge density on the sphere is 0.014\ C/m^2.

7 0
2 years ago
Can light cause the rubber to become solid? Why or why not? Does it matter what type of light she shines on the rubber?
ELEN [110]

Answer:

Yes, ultraviolet light can turn a rubber into solid due to prolong exposure.

Explanation:

A rubber is a material with an elastic property, causing it to be deform by an external force but takes its shape when the force is removed. Light is an electromagnetic wave which causes the sensation of vision. It transfers energy to a medium during propagation through the medium.

Generally, most light do not cause hardness of a rubber. But an ultraviolet light can cause rubber to become solid over a period of time. This is possible if there is a prolong exposure of the rubber, and because of the evaporation of volatiles in the polymer material. Ultraviolet light are known to cause a rubber to become solid.

8 0
2 years ago
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