The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>
The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;
The magnification of the lens is found as;
Hence, the image distance and magnification of The image will be 30 cm and 3.
To learn more about the focal length refer;
brainly.com/question/16188698
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Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement<span> is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
</span>To calculate displacement<span>, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your </span>displacement<span> is 0. In physics, </span>displacement<span> is represented by Δs.
For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.
https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php
Hope I helped.
</span>
Answer:
θ = 36.2º
Explanation:
When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law
I = I₀ cos² tea
The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization
I₁ = I₀ / 2
I₁ = 845/2
I₁ = 422.5 W / m²
In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of
I = 275 W / m
²
Cos² θ = I / I₁
Cos θ = √ I / I₁
Cos θ = √ (275 / 422.5)
Cos θ = 0.80678
θ = cos⁻¹ 0.80678
θ = 36.2º
This is the angle between the two polarizers
Answer: B
Explanation: look at the chart, easy
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
