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3 years ago

11

Why do stars twinkle

1 answer:

vlada-n [284]3 years ago

4 0

As light from a star races through our atmosphere, it bounces and bumps through the different layers, bending the light before you see it. Since the hot and cold layers of air keep moving, the bending of the light changes too, which causes the star's appearance to wobble or twinkle.

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In which situations is heat flow present? Check all that apply.

babunello [35]

A substance feels warm

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4 years ago

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The heat of fusion for water is the amount of energy needed for water to

beks73 [17]

The heat of fusion for water is the amount of energy needed for water to <span>melt. (c)</span>

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3 years ago

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Metamorphic rocks are formed at high temperature and medium to low pressure

storchak [24]

Answer: are u kidding

Explanation:

6 0

3 years ago

An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist

Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0

3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

5 0

3 years ago