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4 years ago

A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second

2 answers:

hjlf4 years ago

6 0

The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.

Charra [1.4K]4 years ago

4 0

On Apex the answer is 1.96J

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A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i

evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

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Vsevolod [243]

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Explain what happens to the pitch of a cell phone ring when the wavelength of a sound wave increases

max2010maxim [7]

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3 years ago

Read 2 more answers

Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill

laila [671]

Answer:

a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d) , c) m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

the maximum for m = 1 is at the angle

θ = sin⁻¹ λ / d

the second maximum m = 2

θ = sin⁻¹ ( λ / 2d)

the third maximum m = 3

θ = sin⁻¹ ( λ / 3d)

the fourth maximum m = 4

θ = sin⁻¹ ( λ / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Ф = π d sin θ /λ

x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference d sin θ = m λ

first diffraction minimum a sin θ = λ

we divide the two expressions

d / a = m

In our case

3a / a = m

m = 3

order three is no longer visible

7 0

3 years ago