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Temka [501]
3 years ago
7

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

andescent bulb or a 24-W CFLbulb.All threebulbsproduce 1600 lumens of light. What is the current through each bulbwhen placed in a circuitwith a 110-Vsource
Physics
1 answer:
Yanka [14]3 years ago
5 0

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

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Any projectile always has a constant vertical acceleration of _______.
Mashcka [7]

Answer:

The answer to your question is: C. -9.81 m/s²

Explanation:

A. 9.81 m/s²  acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.

B. 0 m/s²  This option is incorrect because acceleration is 0 for a linear motion without acceleration.

C. -9.81 m/s²  If a projectile goes to the sky, then the acceleration will be negative.

D. It is not constant. Acceleration is constant.

8 0
3 years ago
Please help with vectors (will give BRAINLIEST answer)
zhenek [66]

just analyze it in this way:

20cos30*=10( radical 3 )

20sin30*=10

7 0
3 years ago
when you place your feet near the fireplace and they become warm what type of energy conversion occurs
JulsSmile [24]
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "Radiant to thermal" is the type of energy conversion that occurs when <span>you place your feet near the fireplace and they become warm. I hope the answer has come to your great help.</span>
8 0
3 years ago
Read 2 more answers
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0
3 years ago
The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what di
ss7ja [257]

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 =  .05 cm ( away from film )

3 0
3 years ago
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