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Murljashka [212]
2 years ago
12

A small candle is 34 cm from a concave mirror having a radius of curvature of 22 cm. Where will the image of the candle be locat

ed? Follow the sign conventions.
Physics
1 answer:
xxTIMURxx [149]2 years ago
5 0

Answer:

16.26 cm in front of the mirror

Explanation:

Using,

1/f = 1/u+1/v....................... Equation 1

Where f = focal focal length of the concave mirror, u = object distance, v = image distance.

make v the subject of the equation

v = fu/(u-f)................... Equation 2

Note: The focal length of a concave mirror is positive

Using the real- is- positive convention

Given: f = 22/2 = 11 cm, u = 34 cm.

Substitute into equation 2

v = (34×11)/(34-11)

v = 374/23

v = 16.26 cm.

The image will be formed 16.26 cm in front of the mirror.

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PLEASEE HELPP
topjm [15]

Explanation:

u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s

by the formula, S=ut+½at².

S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation

S = 1427.6+362.404

S=1790.004m

hope this helps you.

4 0
2 years ago
Which type of mountain is created when rock layers are pushed up by forces inside the Earth?
skad [1K]

Correct answer choice is :


B) Upwarped


Explanation:


An upwarped mountain is a mountain consisting of a large area of the Earth's coat that has led smoothly upward without much visible deformation and normally including sedimentary, igneous, and metamorphic rocks. Sedimentary rocks are set down in layers called beds or layers. A bed is described as a layer of rock that has a similar lithology and character. Beds form by the removal of layers of sand on top of each other.

7 0
3 years ago
Read 2 more answers
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
2 years ago
A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0
2 years ago
Read 2 more answers
A tennis ball is tossed upward with a speed of 3.0\,\dfrac{\text m}{\text s}3.0 s m ​ 3, point, 0, start fraction, start text, m
MrRa [10]

Answer:

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

Explanation:

Hi there!

The equation for the velocity of an object thrown upward is the following:

v = v0 + g · t

Where:

v = velocity of the ball.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:

v = v0 + g · t

v = 3.0 m/s - 9.8 m/s² · 0.40 s

v = -0.92 m/s

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

8 0
3 years ago
Read 2 more answers
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