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Hoochie [10]
3 years ago
10

PLEASE HELP!

Physics
2 answers:
Blizzard [7]3 years ago
8 0

Answer : Option A) Upside down and smaller than the flower.

Explanation : Camera lens is usually convex lens which creates real images and the image formed on the human retina is always inverted i.e. upside down and smaller than that of the actual object size. The convex lens converges at the retina screen and forms a clear image of the object.

Nata [24]3 years ago
5 0
Upside-down and smaller than the flower.
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The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag
sukhopar [10]

Answer:

v=2.58\times10^8m/s

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or n=\frac{c}{v}. For our case we want to use v=\frac{c}{n}, which for our values is equal to:

v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the  <em>least</em> number of significant figures):

v=2.58\times10^8m/s

4 0
3 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
madreJ [45]

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

8 0
3 years ago
While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?
timurjin [86]
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x). 
5 0
3 years ago
Which of the following is a category of mechanical wave?
givi [52]

Answer:

a

because the mechanical wave is when it goes over and over again

8 0
3 years ago
Read 2 more answers
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
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