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4 years ago

X-rays travel in vacuum at a higher speed than the visible light. (A) True (B) False

Physics

1 answer:

AleksandrR [38]4 years ago

3 0

Answer:

(B) False

Explanation:

The visible light and the x-rays are part of the electromagnetic spectrum, all in the spectrum travel in the vacuum at the same speed, at the speed of light (c = 299,792,458 m/s ). If the visible light ant the x-rays are traveling inside a medium, their velocity will change due to the properties of the medium but in the vacuum, their speed is the same.

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Io experiences tidal heating primarily because __________.

Inga [223]

Answer:

the orbit of Io around Jupiter causes the tidal heating.

Explanation:

Io's elliptical orbit causes the tidal force to vary as it orbits Jupiter.

Because Jupiter is a very massive planet, it has a very large gravitational pull on its moons, so the side of Io facing the planet is experiencing a greater gravitational pull than the opposite side. This causes a distortion in the shape of the Io, and a friction that causes the tidal heating.

This friction generates heat currents, which is what makes Io the place with the most volcanic activity in the solar system.

7 0

3 years ago

In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

= (10970)²/(2x220)

= 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

7 0

3 years ago

Describe how to determine you maximum heart rate and why heart rate important to monitor while exercise ?

PtichkaEL [24]

You can look it up on google but heres pt1!

4 0

3 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

4 years ago

Is this float sing or neutrally buoyant

Triss [41]

1=float in liquid

2=sink in liquid

3=float in liqid

4=neutrally byoyant

5=sink in liquid

6=neutrally byoyant

8 0

3 years ago