All categories

2 years ago

Please help me with this question

Engineering

1 answer:

kap26 [50]2 years ago

3 0

Answer:

The current through each lamp is 0.273 Amperes

Power dissipated in each lamp is 0.082W

Explanation:

Battery v = 1.5 V

Each lamp has resistance, r = 1.1 Ohms

The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms

The current through each lamp, I = v/R = 1.5 / 5.5 = 0.273 Amperes

Power dissipated in each lamp = I² * r = 0.273² * 1.1 = 0.082W

You might be interested in

A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?

Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0

1 year ago

How to find the voltage(B Aab) in series parallel circuit?

Sindrei [870]

Answer:

Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

(1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

(V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0

2 years ago

The power supply converts the wall outlet AC power into DC power. T or F

Alika [10]

Answer:

True

Explanation:

All computer parts require DC power to operate, and wall outlets provide AC Power.

7 0

2 years ago

A cylindrical space capsule lands in the ocean. This capsule is 2.44 m long, 1.10 m in diameter, and weighted at one end so that

sukhopar [10]

Answer:

The correct answer is "15456.8 N".

Explanation:

According to the question,

The inside volume will be:

= $3.14\times (\frac{1.1}{2} )^2\times (2.44-0.82)$

= $3.14\times \frac{1.21}{4}\times 1.62$

= $3.14\times 03025\times 1.62$

= $1.538757 \ m^3$

hence,

The buoyant force will be:

= $V\times Pw\times g$

= $1.538757\times 1025\times 9.8$

= $15456.8 \ N$

7 0

2 years ago

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile

Ber [7]

Answer:

A.) Time = 17.13 s

B.) Distance = 32 m

C.) V = 11.18 m/ s

D.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

7 0

3 years ago

Please help me with this question​

Please help me with this question