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AleksAgata [21]
3 years ago
8

An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Physics
1 answer:
Lynna [10]3 years ago
4 0

Answer:

\boxed{\sf Magnification \ (m) = 2}

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

\boxed{\bold{\sf Magnification  \: (m) = - \frac{Image  \: distance  \: (v)}{Object  \: distance  \: (u)}}}

Substituting values of Image distance(v) & Object distance (u) in the equation:

\sf \implies m =  -  \frac{( - 50)}{25}

-(-50) = 50:

\sf \implies m =  \frac{50}{25}

\sf \implies m =  \frac{2 \times  \cancel{25}}{ \cancel{25}}

\sf \implies m = 2

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The "2" tells us that there are 2 hydrogen atoms in this compound.

Explanation:

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2 years ago
A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field
Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= -  .086 Weber

change in flux

.15 +  .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0
3 years ago
Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp
jekas [21]

Answer:

a=\frac{mBg-mAgSin\theta}{mA+mB}

Explanation:

Given two mass on an incline code mA and mB and an angle of inclination \theta. g. Assume that mA is the weight being pulled up and mB the hanging weight.

-The equations of motion from Newton's Second Law are:

mBg-T=mBa where a is the acceleration.

#Substituting for T (tension) gives:

mBg-mAsin\theta-mAa=mBa

#and solving for a:

a=\frac{mBg-mAgSin\theta}{mA+mB} which is the system's acceleration.

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3 years ago
Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit
marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})

where

E_0 = 13.6 eV is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV

Converting into Joules,

E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J

And now we can find the wavelength of the emitted photon by using the equation

E=\frac{hc}{\lambda}

where h is the Planck constant and c is the speed of light. Solving for \lambda,

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm

And this wavelength corresponds to red light.

5 0
2 years ago
What is a plane mirror? state the characteristics of the image formed by a plane mirror ​
Nadusha1986 [10]
A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!
3 0
2 years ago
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