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2 years ago

What kind of object are the light rays interacting with in the model below?

Physics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

Concave lens

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Pls heplp 70 points!!!!!

Rasek [7]

Answer

the answer is d for sure

5 0

3 years ago

Explain why a cow that touches an electric fence experiences a mild shock

Minchanka [31]

The voltage exists between the fence and the ground. The cow is grounded. The cow is touching the ground, completing the circuit of electricity. <span>When the cow comes into contact with the fence, it becomes an electric ground which sends an electric current into the cow, through the cow, and into the ground. The pain experienced from the shock is due to the current that flows through the cow.</span>

8 0

3 years ago

Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen

stich3 [128]

Answer:

The distance is $D = 2.6 \ m$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m$

The distance between the slit is $d = 0.29 \ mm = 0.29 *10^{-3} \ m$

The between the first and second dark fringes is $y = 4.2 \ mm = 4.2 *10^{-3} \ m$

Generally fringe width is mathematically represented as

$y = \frac{\lambda * D }{d}$

Where D is the distance of the slit to the screen

Hence

$D = \frac{y * d}{\lambda }$

substituting values

$D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }$

$D = 2.6 \ m$

7 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$