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3 years ago

Describe the path that light waves take when you see your image in a mirror

Physics

1 answer:

ICE Princess25 [194]3 years ago

7 0

Describe the path light waves take when you see your image in the mirror?

<span>light strikes your face and bounces off of it. then, light is reflected off your face and into the mirror. i is hen reflected into your eyes.</span>

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If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

Burka [1]

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

3 0

3 years ago

What is a lack of physical, social, or economic access to safe and nutritious foods called? A) famine B) food security C) food s

Anastaziya [24]

Food insecurity, because you are only lacking safe and nutritious food. It would be famine if you had no food whatsoever, and it would be a food shortage if you were lacking a small amount of ALL food

5 0

4 years ago

Read 2 more answers

Your friends sit in a sled in the snow. If you apply a force of 95 N to them, they have an acceleration of 0.8 m/s2. What is the

ss7ja [257]

F=ma
m=F/a=95/0.8= 118.75kg
your friend is pretty heavy XD

5 0

3 years ago

Read 2 more answers

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

A man weighs 1190 n on earth. what would he weigh on jupiter, where the free-fall acceleration is 25.9 m/s 2 ? the acceleration

Elden [556K]

The answer would be 3,145N. Using W=mg solve for the mass of the man on earth. Once you have the mass you can multiply it by the gravity of Jupiter giving you his weight in Newton’s on Jupiter.

8 0

4 years ago