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kondaur [170]
3 years ago
5

Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres

, and that there are 5.0 cm3 in a teaspoon, what is the thickness (in cm) of a later of oil?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

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A current of 0.2 A flows through a conductor for 5minutes. How much charge would have passed through the conductor?​
Montano1993 [528]

As we know,

electric \:  \: current =  \dfrac{charge}{time}

so, let's solve for charge (q) :

time = 5 minutes = 5 × 60 seconds = 300 seconds.

  • 0.2 =  \dfrac{q}{300}

  • q = 300 \times 0.2

  • q = 60

hence, the charge = 60 coulombs (C)

4 0
3 years ago
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

7 0
3 years ago
Is 250 000 miles from the earth to the moon" is a qualitative<br> Observation<br> TRUE<br> Or false
lys-0071 [83]

Answer:

True

Explanation:

3 0
3 years ago
The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.
loris [4]
<u>Answer</u>

 48 Volts  

<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
            Ns ⇒number of turns in the seconndary coil
            Vp ⇒ primary voltage
             Vs ⇒secondary voltage

Np/Ns = Vp/Vs

10/4 = 120/Vp

Vp = (120 × 4)/10

      = 480/10
      = 48 Volts  

 

5 0
3 years ago
Read 2 more answers
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