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marysya [2.9K]
3 years ago
15

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow- dryer is 11 Amps, and th

at of the vacuum cleaner is 4.0 Amps. Determine the power consumed by (a) the blow-dryer and (b) the vacuum cleaner, (c) Determine the ratio of the energy used by the blow-dryer in 15 minutes to the energy used by the vacuum cleaner in one-half hour.
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

(a) 1320 W

(b) 480 W

(c) E':E ≈ 11:2

Explanation:

(a) Applying,

P' = VI'................. Equation 1

Where P' = Power of the blow-dryer, V = Voltage, I = current rating of the blow-dryer.

From the question,

Given: V = 120 V, I' = 11 A

Substitute these values into equation 1

P = (120×11)

P = 1320 W

(b) Similarly,

P = VI................... Equation 2

Where P = Power of the vacuum cleaner. I = current rating of the vacuum cleaner.

Also Given: I = 4 A,

Therefore

P = 4(120)

P = 480 W

(c)

E' = P'/t'............. Equation 3

E = P/t................ Equation 4

Where E' = Energy of the blow-dryer, t' = time of use of the blow-dryer, E = Energy of the vacuum cleaner, t = time of use of the vacuum cleaner

From the question,

Given: t' = 15 minutes = (15×60) = 900 seconds, t = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 3 and 4

E' = 1320/900

E' = 1.47 J,

E = 480/1800

E = 0.267

Therefore,

E':E = 1.47:0.267

E':E ≈ 11:2

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Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.
Tcecarenko [31]
The energy of a photon is given by
E=hf
where
h=6.6 \cdot 10^{-34} Js is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is 
f=5.49 \cdot 10^{14}Hz
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J
8 0
3 years ago
You input 75 J of work with a wedge. If the wedge does 65 J of useful work, what if the efficiency of the wedge ?
zavuch27 [327]

Answer:

Efficiency of wedge is the ratio of "work done by the machine to the work supplied to the machine".

           Efficiency (η) = Work done by machine ÷ Work supplied

                                  = 65 ÷ 75

                                 = 0.86%

<em>Efficiency of wedge is 86%</em>

5 0
3 years ago
Help me with this please
pochemuha

Answer:

1. 31.5

3. 3.5

Explanation:

4 0
3 years ago
ASAP WILL GIVE BRAINLIEST!!!
Oksana_A [137]

Answer:

<em>The change in momentum of the car is 3575 Kg.m/s</em>

Explanation:

<u>Impulse and Momentum</u>

The impulse (J) experienced by the object equals the change in momentum of the object (Δp).

The formula that represents the above statement is:

J = Δp

The impulse is calculated as

J = F.t

Where F is the applied force and t is the time.

The car hits a wall with a force of F=6500 N and stops in 0.55 s. Thus, the impulse is:

J = 6500 * 0.55

J = 3575 Kg.m/s

The change in momentum of the car is:

\Delta p= J = 3575\ Kg.m/s

The change in momentum of the car is 3575 Kg.m/s

5 0
3 years ago
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