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3 years ago

Which statement is the best interpretation of the ray diagram shown below?

Physics

2 answers:

andrew11 [14]3 years ago

8 0

Answer:

a concave lens forms a smaller, real image. { a p e x :)) }

Explanation:

hey, friend! this is the answer, but make sure that, next time, you include the diagram to make it easier for us to help you, okay?

Nataly_w [17]3 years ago

4 0

Answer:

Explanation:

hope you fell better my answers thanks god bless keep safe and good luck

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2. Can you rearrange the equation F = kx to get k on one side of the equation

Bumek [7]

k = f/x

x = f/k

( / means divide )

5 0

3 years ago

The pressure of 4.20 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absol

Dmitrij [34]

Answer:

6.30 L

Explanation:

P1 = P, V1 = 4.20 L, T1 = T

P2 = P/3, V2 = ?, T2 = T/2

Where, V2 be the final volume.

Use ideal gas equation

$\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}$

$V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}$

By substituting the values, we get

V2 = 6.30 L

8 0

3 years ago

What is the refractive index of air if light travels through it at 3.0 108 m/s?

olchik [2.2K]

Answer:

n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.

4 0

3 years ago

Where can classic examples of shield volcanoes be found?

Darina [25.2K]

The largest is Mauna Loa on the Big Island of Hawaii; all the volcanoes in the Hawaiian Islands are shield volcanoes. There are also shield volcanoes, for example, in Washington, Oregon, and the Galapagos Islands

4 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago