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scoundrel [369]
3 years ago
13

If you have nearsighted eyes, you need lenses that _____ light.

Physics
2 answers:
WINSTONCH [101]3 years ago
4 0

Answer:

B. focus

Explanation:

  • myopia "nearsighted", a common type of refractive error where close objects appear clearly, but distant objects appear blurry
Zielflug [23.3K]3 years ago
3 0
D spread out because you can see farther then
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20 POINTS!!!
bogdanovich [222]

a) Average velocity for her run north: 2.68 m/s north

b) The average velocity for the whole trip is zero

c) The average speed for the whole trip is 1.95 m/s

Explanation:

a)

In this part of the problem we only want to consider the part when Micha is running north.

The average velocity is given by:

v=\frac{d}{t}

where

d is the displacement (a vector connecting the initial position to the final position of motion)

t is the time

For the part running north, we have

d=2 mil \cdot 1609 = 3218 m north is the displacement

t = 20 min \cdot 60 = 1200 s is the time interval

Substituting,

v=\frac{3218}{1200}=2.68 m/s north (we have to specify also the direction, since velocity is a vector)

b)

As we said, average velocity is given by

v=\frac{d}{t}

where

d is the displacement

t is the time

If we consider the whole trip, however, the displacement is zero:

d = 0

Because Micha returns home, so the initial position corresponds to the final position of motion. Therefore, the average velocity is also zero:

v = 0

c)

The average speed is given by

s=\frac{d}{t}

where

d is the  distance covered (the total length of the path, regardless of the direction)

t is the time interval

Since MIcha ran 2 miles north and then 2 miles back, the total distance is

d=2 mi + 2mi = 4 mi \cdot 1609 = 6436 m

And the time taken is

t=20 min + 15 min + 20 min = 55 min \cdot 60 = 3300 s

So, the average speed is

s=\frac{6436}{3300}=1.95 m/s

And since speed is a scalar, there is no need to specify a direction.

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

5 0
3 years ago
A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?
eduard

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of  motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

4 0
3 years ago
Properties of convex mirror
Ahat [919]

Answer:

The image result of an object reflected by a convex mirror is typically virtual, upright, and smaller. Discover how moving the object farther away from the mirror's surface affects the size of the virtual image formed behind the mirror

Explanation:

6 0
3 years ago
What is Earth described as?
Ymorist [56]

Answer:

Earth, our home, is the third planet from the sun. It's described as the only planet known to have an atmosphere containing free oxygen, oceans of water on its surface and, of course, life. Earth is also the fifth largest of the planets in the solar system. Hope this helped.

Explanation:

7 0
4 years ago
A particle has rest mass 10-30 kg and travels at 0.5 c with respect to the Laboratory. Find its inertial mass, its momentum, its
astra-53 [7]

Explanation:

It is given that,

The rest mass of the particle, m_o=10^{-30}\ kg

Speed of the particle, v = 0.5c

Firstly calculating the relativistic factor. The formula is as follows :

\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}

\gamma=\dfrac{1}{\sqrt{1-\dfrac{(0.5c)^2}{c^2}}}

\gamma=1.032

1. Inertial mass is given by :

m=\gamma \times m_o

m=1.032 \times 10^{-30}

2. The momentum is given by :

p=\gamma\times mv

p=1.032\times 1.032 \times 10^{-30}\times 0.5c

p=1.59\times 10^{-22}\ kg-m/s

3. Kinetic energy of the particle is given by :

E_k=(\gamma-1)mc^2      

E_k=(1.032-1)\times 1.032 \times 10^{-30}\times (3\times 10^8)^2  

E_k=2.97\times 10^{-15}\ J

4. The total energy of the particle is :

E=\gamma mc^2

E=1.032\times 1.032 \times 10^{-30}\times (3\times 10^8)^2

E=9.58\times 10^{-14}\ J

Hence, this is the required solution.

7 0
4 years ago
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