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Nesterboy [21]
3 years ago
13

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

is the primary voltage? (in this case, the voltages and currents are inversely proportional.)
Physics
1 answer:
Lynna [10]3 years ago
7 0

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

V_1 i_1 = V_2 i_2

here we know that

V_2 = 140 volts

i_2 = 3.5 A

i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1

V_1 = 49 volts

So primary coil voltage is 49 Volts

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A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantit
Bess [88]

Answer:

C. 85%

Explanation:

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A. 15%

B. 30%

C. 85%

D. 100%

work done by the system will be

W=PdV

p=pressure

dV=change in volume

3tam will be changed to N/m^2

3*1.01*10^5

W=3.03*10^5*(1.5-1)

convert 0.5L to m^3

5*10^-4

W=3.03*10^5*5*10^-4

W=152J

therefore

to find the percentage used

152/1000*100

15%

100%-15%

85% uf the fuel's energy was lost to friction and heat

6 0
2 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
2 years ago
When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each pla
hodyreva [135]

Answer:

42.9 μF

Explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF

5 0
2 years ago
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h

Next, you calculate the difference between both times t1 and t2:

\Delta t=t_1-t_2=2.30h-1.875h=0.425h

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0
2 years ago
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1
Free_Kalibri [48]

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

7 0
3 years ago
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