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maw [93]
2 years ago
10

Height of the image formed by a lens is-2cm.0what is the nature of the image?

Physics
1 answer:
MA_775_DIABLO [31]2 years ago
3 0

Answer:

As the height is negative , image is inverted. This means it is inverted and real

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A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular
larisa [96]

Answer:

0.38°

Explanation:

\theta = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

m\lambda=dsin\theta

For air

m\lambda_a=dsin\theta_a

For liquid

m\lambda_l=dsin\theta_l

Dividing the two equations

\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}

Wavelength ratio = n

n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}

The angular separation is 0.38°

5 0
3 years ago
Read 2 more answers
Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6
Lorico [155]

Answer:

k=784 N/m

\Delta x=8,8 cm

Explanation:

Usando la ley de Hook tenemos:

F=k\Delta x

Solving it for k we have:

k=\frac{F}{\Delta x}

k=\frac{39,2}{0,05}

k=784 N/m

Usando la misma ecuación y sabiendo k tenemos:

\Delta x=\frac{F}{k}

\Delta x=\frac{68,6}{784}

\Delta x=8,8 cm

Espero esto te ayude!

6 0
2 years ago
CAN SOMEONE DO THIS TO ME ITS OK IF YOUR DRAWING IS NOT THAT COOL OR GOOD AND DONT GET A PHOTO FROM SOCIAL MEDIA
Svet_ta [14]

bc i cant draw good sorry but hope it helps!

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7 0
3 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
Pulmonary circulation involves blood flow to and from the heart and the ____?
olga nikolaevna [1]
I believe the answer would be lungs
3 0
3 years ago
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