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4 years ago

9

Which describe the image formed by an object on the center of curvature? Check all that apply.

2 answers:

9966 [12]4 years ago

5 0

The image formed by a concave mirror with the object placed at the center of curvature is real inverted and formed at the center of curvature. Using the ray diagram a ray from the top of the object to the mirror through f then reflected parallel to the principal axis,then the ray through the center of curvature reflected through the same point both intersect at a plane through center curvature and perpendicular to the principal axis. The point of intersection forms the top of the image and the center of curvature forms the bottom. Therefore, the correct choices are : real and inverted

schepotkina [342]4 years ago

3 0

The answer is real and inverted. And is not smaller or larger because they both have the same size. :)

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Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by $\frac{1}{2}kx^{2}$ where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, $\frac{1}{2}k(x1)^{2}$

⇒ $\frac{1}{2}k(\frac{w}{k})^{2}$

⇒ $\frac{w^{2}}{2k}$

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, $\frac{1}{2}2k(x2)^{2}$

⇒ $\frac{1}{2}2k(\frac{w}{2k})^{2}$

⇒ $\frac{w^{2}}{4k}$

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is $\frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=$ 2:1

4 0

3 years ago

A cat jumps from 2.5-meter-tall bookshelf to a 1.3-meter-tall countertop. If the cat

UkoKoshka [18]

The change in Potential energy of the cat is 176.4 J.

<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>

ΔP.E = mg(H-h).............. Equation 1

<h3>Where:</h3>

ΔP.E = Change in potential energy
m = mass of the cat
g = acceleration due to gravity
H = First height
h = second height.

From the question,

<h3>Given:</h3>

m = 15 kg
H = 2.5 m
h = 1.3 m
g = 9.8 m/s²

Substitute these values into equation 1

ΔP.E = 15×9.8(2.5-1.3)
ΔP.E = 15×9.8×1.2
ΔP.E = 176.4 J.

Hence, The change in Potential energy of the cat is 176.4 J

Learn more about Potential energy here: brainly.com/question/1242059

5 0

2 years ago

What is the slope of the line?

coldgirl [10]

(30, 5)
(10, 1)

change of y / change of x
= (30 - 10) / (5 - 1)
= 20 /4
= 5

6 0

3 years ago

A student is performing experiments on a particular substance. Which

boyakko [2]

Answer: A.

Explanation:

6 0

3 years ago

adelina 88 [10]

Answer:

I'm pretty sure that the answer is A

5 0

3 years ago