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3 years ago

Is number six correct? If so why???

Physics

1 answer:

Yuki888 [10]3 years ago

3 0

In question #5, the one where you have to label the three pictures,
all of your labels are correct !

Look at the 3rd picture, where the light is coming down slanted,
hitting some kind of surface, and bouncing away slanted.
You labelled that picture "reflection", and that's correct.

Now, look at the two angles marked in the picture .... the angle
between the arriving light and the surface, and the angle between
the departing light and the surface.

There is a statement in Science called the "Law of Reflection".
It simply says that those two angles are EQUAL.
I don't think we have to get into WHY right now. I think that
at this point, your class is just trying to teach you what the law says.
If you go on and study more physics in High School, then you'll learn
WHY the angles are equal.

If the light ray arrives and hits the mirror at a 40° angle,
then it'll leave the mirror at a 40° angle. It's that simple.

The two angles are the ones marked equal in the picture
where you wrote 'reflection'.

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9 The diagram shows a uniform beam PQ. The length of the beam is 3.0 m and its weight is 50 N. The beam is supported on a pivot

tankabanditka [31]

equilibrium = 1/1

50 N/x = 1/1

x = 1/1 × 50 N

x = 50 N (B)

#LearnWithEXO

6 0

3 years ago

Explain why your hands feel cool if they get wet?

Furkat [3]

Answer:

How hot or cold you feel depends on the rate at which your body is losing heat to the environment

Water on your skin acts pretty much like sweat. Water is more thermally conductive than air; therefore, the skin loses its heat to it much faster than it would to air.

Also, because water evaporates, it carries heat away from the skin and this increases the rate at which the skin loses its heat. The faster heat loss from the skin to water is what makes us feel cold when we are wet. But of course, the temperature of water has to be lower than the skin for this to occur, which is usually the case.

8 0

3 years ago

A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an angle of 2.5° below horizontal. The catcher

rodikova [14]

Answer:

The initial velocity of the pitch is approximately 36.5 m/s

Explanation:

The given parameters of the thrown fastball are;

The height at which the pitcher throws the fastball, h₁ = 2.65 m

The angle direction in which the ball is thrown, θ = 2.5° below the horizontal

The height above the ground the catcher catches the ball, h₂ = 1.02 m

The distance between the pitcher's mound and the home plate = 18.5 m

Let 'u' represent the initial velocity of the pitch

From h = $u_y$ ·t + 1/2·g·t², we have;

$u_y$ = The vertical velocity = u·sin(θ) = u·sin(2.5°)

h = 2.65 m - 1.02 m = 1.63 m

uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m

∴ t = 18.5 m/(u·cos(2.5°))

∴ h = $u_y$ ·t + 1/2·g·t² = (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²

1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²

t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087

t = √(0.25691469087) ≈ 0.50686752763

t ≈ 0.50686752763 seconds

u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s

The initial velocity of the pitch = u ≈ 36.5 m/s.

3 0

3 years ago

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

8 0

3 years ago

Interstellar space (far from any stars) contains atomic hydrogen (H) with a density of 1 atom/cm3 and at a temperature of about

Mandarinka [93]

Explanation:

Given that,

Number density $n= 1\ atom/cm^{3} =10^{6}\ atom/m^3$

Temperature = 2.7 K

(a). We need to calculate the pressure in interstellar space

Using ideal gas equation

$PV=nRT$

$P=\dfrac{nRT}{V}$

$P=\dfrac{10^{6}\times8.314\times2.7}{6.023\times10^{23}}$

$P=3.727\times10^{-17}\ Pa$

$P=36.78\times10^{-23}\ atm$

The pressure in interstellar space is $36.78\times10^{-23}\ atm$

(b). We need to calculate the root-mean square speed of the atom

Using formula of rms

$v_{rms}=\sqrt{\dfrac{3RT}{Nm}}$

Put the value into the formula

$v_{rms}=\sqrt{\dfrac{3\times8.314\times2.7}{1.007\times10^{-3}}}$

$v_{rms}=258.6\ m/s$

The root-mean square speed of the atom is 258.6 m/s.

(c). We need to calculate the kinetic energy

Average kinetic energy of atom

$E=\dfrac{3}{2}kT$

Where, k = Boltzmann constant

Put the value into the formula

$E=\dfrac{3}{2}\times1.38\times10^{-23}\times2.7$

$E=5.58\times10^{-23}\ J$

The kinetic energy stored in 1 km³ of space is $5.58\times10^{-23}\ J$ .

Hence, This is the required solution.

7 0

4 years ago