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topjm [15]
3 years ago
10

An object is placed 11.0 cm in front of a concave mirror whose focal length is 24.0 cm. The object is 2.60 cm tall. What is the

height of the image? (cm)
Physics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

Height of the image is 4.79 cm.

Explanation:

Object distance, u = -11 cm

Focal length of the mirror, f = -24 cm

Height of the object, h = 2.6 cm

We need to find the height of the image. Firstly, using the mirror's formula as :

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

v is the image distance

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{(-24)}-\dfrac{1}{(-11)}

v = 20.30 cm

The magnification of the image is given by :

m=\dfrac{-v}{u}=\dfrac{h'}{h}, h' is the height of the image

\dfrac{-v}{u}\times h={h'}

\dfrac{-20.30}{-11}\times 2.6={h'}

h' = 4.79 cm

So, the height of the image is 4.79 cm. Hence, this is the required solution.

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Answer:

<h2>5850 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

<h3>5850 N</h3>

Hope this helps you

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Which would BEST describes what occurs when a ball is thrown against a wall? A) The ball will not bounce off the wall. B) The ba
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Answer:

D) The ball exerts a force on the wall and the wall exerts a force back.

Explanation:

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"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"

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When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

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c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

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