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3 years ago

The fact that Mercury has no atmosphere is evidence that it _____.

Physics

2 answers:

murzikaleks [220]3 years ago

8 0

Answer:

The correct choice is

"reflects a small percentage of the sunlight that strikes it"

Explanation:

Mercury has no atmosphere and it being very close to sun receives large amount of sunlight. due to absence of atmosphere, much of the sunlight reach the surface of the planet and gets absorbed by the surface making it very hot for life. Only a small percentage of sunlight that strikes the surface gets reflected.

Naddika [18.5K]3 years ago

7 0

We want to know what does the fact that Mercury has no atmosphere tell us. Since Mercury has no atmosphere it cant reflect a lot of sunlight that is hitting its surface. Its constantly being hit by solar wind. So Mercury reflects a small percentage of the sunlight that strikes it.

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago

A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W

maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0

3 years ago

if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

bagirrra123 [75]

Gravitational force between two masses is given by formula

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = 55 kg$

$m_2 = 1234 kg$

$r = 3 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

now from the above equation we will have

$F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}$

$F = 5.03 \times 10^{-7}N$

so above is the gravitational force between car and the person

5 0

3 years ago

A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi

Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0

3 years ago

As the wavelength of an electromagnetic wave ___ the frequency of the wave ________.

adoni [48]

Answer:

As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.

Explanation:

What is the relationship between frequency and wavelength?

Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:

V= fλ

where v= velocity

f= frequency

λ = wavelength

⇒ f = v/λ

also f ∝ 1/λ

For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:

C= fλ

where c= Speed of light

f= frequency

λ = wavelength

⇒ f = v/λ

also f ∝ 1/λ

8 0

3 years ago