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Black_prince [1.1K]
3 years ago
15

You replace a 200 W lamp with a more efficient 55 W LED lamp. If you leave your lights on 24 hours a day, how much energy are yo

u saving each day by replacing this bulb?
A.21,225,000 J
B.17,280,000 J
C.4,752,000 J
D.12,528,000 J
Physics
2 answers:
umka2103 [35]3 years ago
6 0

Answer:

its d

Explanation:

mezya [45]3 years ago
4 0
P_1=200W\\P_2=55W\\t=24h=86400s\\\\Find:\\\Delta W=?\\\\Solution:\\\\\Delta W=W_1-W_2\\\\W=Pt\\\\\Delta W=P_1t-P_2t=t(P_1-P_2)\\\\\Delta W=86400s(200W-55W)=12,528,000J\\\\Answer\;\;D
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Why does the area around the equator stay about the same temperature year round?
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Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.
3 0
3 years ago
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest
shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0
4 years ago
Charging an object without touching it is called?
Airida [17]
<span>Charging by induction</span>
5 0
4 years ago
Will mark brainlest helpp!!!!!!​
Luda [366]

Answer:

I don't know if it is correct or not.

6 0
3 years ago
Read 2 more answers
A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and
stira [4]

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with  constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

y(t) = h + v_{0y} t - \frac{1}{2}gt^2

where y is the vertical position at time t, h is the initial height, and v_{0y} is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, v_{0y}=0. So the equation becomes

y(t) = h - \frac{1}{2}gt^2

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

t=\sqrt{\frac{2h}{g}}

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

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