We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
Answer: 1.3 *10^6 Ω*m
Explanation: In order to explain this problem we have to use the following expression for the resistence:
R=L/(σ*A) where L and A are the length and teh area for the wire, respectively. σ is the conductivity of teh Nichrome.
Then, from mteh OHM law we have V=R*I so R=V/I=2/3.2=0.625 Ω
Finally we have:
σ=L/(R*A)=1.3/(0.625*1.6*10^-6)=1.3*10^6 Ω*m
Answer:
The image distance from right lens is 2.86 cm and image is real.
Explanation:
Given that,
Focal length of left lens = 10 cm
Focal length of right lens = 5 cm
Distance between the lenses d= 15 cm
Object distance = 50 cm
We need to calculate the image distance from left lens
Using formula of lens
Put the value into the formula
We need to calculate the image distance from right lens
The object distance will be
Using formula of lens
Put the value into the formula
The image is real.
Hence, The image distance from right lens is 2.86 cm and image is real.
Angle, θ2 at which the light leaves mirror 2 is 56°
<u>Explanation:</u>
Given-
θ1 = 64°
So, α will also be 64°
According to the figure:
α + β = 90°
So,
β = 90° - α
= 90° - 64°
= 26°
β + γ + 120° = 180°
γ = 180° - 120° - β
γ = 180° - 120° - 26°
γ = 34°
γ + δ = 90°
δ = 90° - γ
δ = 90° - 34°
δ = 56°
According to the law of reflection,
angle of incidence = angle of reflection
θ2 = δ = 56°
Therefore, angle θ2 at which the light leaves mirror 2 is 56°
Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right
