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Contact [7]
4 years ago
6

An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T

he three devices are in parallel when plugged into the same socket).
A) What current is drawn by each device?
B) Will this combination blow the 15-A fuse?
Engineering
1 answer:
Viktor [21]4 years ago
5 0

Answer:

A)

Current drawn by toaster = 15.33 A

Current drawn by electric frying pan = 11.83 A

Current drawn by lamp = 0.58 A

B)

The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

Explanation:

There are three devices plugged into the same outlet.

Toaster = 1840 W

Electric frying pan = 1420 W

Lamp = 70 W

Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.

A) What current is drawn by each device?

The current flowing through the device is given by

I = P/V

Where P is the power and V is the voltage.

Current drawn by toaster:

I = 1840/120

I = 15.33 A

Current drawn by electric frying pan:

I = 1420/120

I = 11.83 A

Current drawn by lamp:

I = 70/120

I = 0.58 A

B) Will this combination blow the 15-A fuse?

The total current drawn by all three devices is

Total current = 15.33 + 11.83 + 0.58

Total current = 27.74 A

Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

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Which of the following statements is not necessarily true for a well-arranged floor plan?
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A well arranged floor plan is one that optimises the given floor area. Floor plans are useful to help design furniture layout, wiring systems, and much more. Option B is the answer, since it does not meet the standard of a plan

<h3>What is a Floor Plan?</h3>

A floor plan is a scaled diagram of a room or building viewed from above. The floor plan may depict an entire building, one floor of a building, or a single room. It may also include measurements, furniture, appliances, or anything else necessary to the purpose of the plan.

<h3>Other properties of a floor plan are:</h3>
  1. Maximize the property
  2. Utilize space effectively
  3. Accessibility
  4. Flexibility
  5. Functionality
  6. Maximize the use of light
  7. Attention to size
  8. Fitting to your lifestyle

Learn more:

brainly.com/question/25057316

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3 years ago
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It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore
Dmitry [639]

Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

C. 9.1125 x 10^14 kg of CO2

Explanation:

A. Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Initial area of the rainforest = 20 x 10^12 m^2

Therefore to calculate exponential rate in 1/day,

Rate of rainforest destruction/ initial area of rainforest

= 139.2 x 10^6/20 x 10^12

= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

(40 * 365 )days + 10 days (leap years)

= 14610 days

Area of rainforest in 1975 = 24.5 x 10^12m^2

Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

= 2.034 x 10^12 m^2

Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

= 24.5 x 10^12 - 2.034 x 10^12

= 22.466 x 10^12 m^2

C. How much CO2 will be removed in 2025

Recall: Photosynthesis is the process of plants taking in CO2 and water to give glucose and O2.

So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

6 0
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.Block A of mass 30 kg is resting on Block B of 15 kg and both blocks are connected via a
Gala2k [10]

The tension in the cable and the <em>net</em> acceleration of the block B are 291.814 newtons and 0.112 meters per square second, respectively.

<h3>How to determine the tension and the acceleration of block B</h3>

Based of the representation seen in the image, an <em>external</em> force moves the block A <em>downwards</em> and the block B moves <em>upwards</em> and with an net acceleration of same magnitude due to the cable. By Newton's laws we construct the free body diagrams for each mass and that can be seen in the image seen below.

The <em>motion</em> equations are described below:

Mass A

250 N - 0.4 · Na - T + (30 kg) · (9.807 m / s²) · sin 30° = (30 kg) · a      (1)

Na - (30 kg) · (9.807 m / s²) · cos 30° = 0     (2)

Mass B

- T + 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30° = - (15 kg) · a     (3)

Nb - Na - (15 kg) · (9.807 m / s²) · cos 30° = 0     (4)

By (2):

Na = (30 kg) · (9.807 m / s²) · cos 30°

Na = 254.793 N

By (4):

Nb = 254.793 N + (15 kg) · (9.807 m / s²) · cos 30°

Nb = 382.190 N

Then, we have the following system of <em>linear</em> equations:

T + (30 kg) · a = 250 N - 0.4 · Na + (30 kg) · (9.807 m / s²) · sin 30°

T + 30 · a = 295.188

T - (15 kg) · a = 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30°

T - 15 · a = 290.127

The solution of the <em>linear</em> system is T = 291.814 N and a = 0.112 m / s².

To learn more on Newton's laws: brainly.com/question/27573481

#SPJ1

6 0
2 years ago
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