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3 years ago

What property of light waves does the michelson-morley interferometer directly demonstrate?

1 answer:

8 0

The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.

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You toss a racquetball directly upward and then catch it at the same height you released it 1.82 s later. assume air resistance

Anton [14]

Neglecting air resistance, the acceleration of the ball is
the acceleration of gravity ... 9.8 m/s² downward.

It doesn't matter what you toss, what it's mass is, what it weighs,
what color it is, how much it cost, what its shape or size is, how
fast you toss it, in what direction, or how long it's in the air.

Its horizontal acceleration is zero and its vertical acceleration
is 9.8 m/s² downward, from the moment it leaves your hand
until the moment somebody catches it or it hits the ground.

5 0

3 years ago

a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

3 years ago

A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the

Andre45 [30]

The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
$p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s$
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is $v_f$ . The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
$p=12960 kg m/s =(1200 kg+m)v_f$
and so the final speed is
$v_f = \frac{12960 kg m/s}{1200 kg +m}$

5 0

3 years ago

Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.

BlackZzzverrR [31]

The easiest way I know to explain it is this:

-- Take a flashlight and a ball into a dark room.

-- Turn on the flashlight and point it at the ball.

-- Half of the ball is lighted up by the flashlight, and the other half is dark.

-- There is no way you can turn or twist the ball to make more or less
than 50% of it lighted up and more or less than 50% of it dark.

<em>Everything</em> in the solar system ... as long as it's shaped like a ball ... is
half illuminated by the sun and half dark.

7 0

3 years ago

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mixas84 [53]

Answer:

nice!!! but did you know that geico can save you 15% or more on car insurance!

Explanation:

8 0

3 years ago

Read 2 more answers