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azamat
2 years ago
10

Interactive Solution 22.61 offers one approach to problems such as this one. The secondary coil of a step-up transformer provide

s the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 45:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 2.0 x 10-3 A. Find the power consumed by the air filter.
Physics
1 answer:
vichka [17]2 years ago
8 0

Answer:

The power consumed by air filter is 10.8W

Explanation:

Voltage V=120 V

Current I=2.0×10⁻³A

Turn ration of transformer=45:1

To find

The power consumed by the air filter

Solution

The power used by filter is

P=I_{p}V_{p}\\Solving \\\frac{I_{s}}{I_{p}}=\frac{N_{p}}{N_{s}}\\  shows\\I_{p}=I_{s}(\frac{N_{s}}{N_{p}})

Substituting this result for power

P=I_{p}V_{p}\\P=I_{s}(\frac{N_{s}}{N_{p}})V_{p}\\P=(2.0*10^{-3}A)(\frac{45}{1} )(120V)\\P=10.8W

The power consumed by air filter is 10.8W

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An astronaut floating in space throws a wrench forward with the force of 10 N.
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Answer:

10N

Explanation:

1. Every Action has an equal and opposite reAction.

2. If 10N of force is acted upon an wrench, then the wrench will react with an equal amount of force, but in the opposite direction.

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Manuel is holding a 5kg box. How much force is the box exerting on him?In what direction
soldi70 [24.7K]
The force the box is exerting on Manuel is the weight of the box, downward:
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3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
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