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2 years ago

Please answer faast!!!

Physics

1 answer:

Drupady [299]2 years ago

7 0

Answer:

Any medium or material of high refractive density e.g Water which refracts the light rays 1&2 away from their normal

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a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin

creativ13 [48]

If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

(2.5 x 1 newton) = 2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

(2.5 x 120 newtons) = 300 newtons.

4 0

3 years ago

A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc

Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

1/f = 1/di + /1do

then the image distance

di = fd_o / d_o - f

= (10)(40) / 40-10

= 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

di = 10×20/ 20-10

= 20

Hence, the image must be real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 = ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0

3 years ago

Read 2 more answers

Write an equation that expresses the first law of thermodynamics in terms of heat and work.

OleMash [197]

Answer:

Heat Input = Work Output (at 100% efficiency)

ΔQ = ΔW

(you cannot get something for nothing)

3 0

1 year ago

because lighting heats up the surrounding air slowly, it realationship to thunder is currently unknown true or false

NeTakaya

Answer:

False

Explanation:

8 0

2 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$