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3 years ago

PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST

Physics

1 answer:

n200080 [17]3 years ago

6 0

Answer:

all of those are pisitions

Explanation:

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1. A 3,000 N force is applied to a car which

Marizza181 [45]

Answer:

The answer is c

Explanation:

3 0

3 years ago

Dois automóveis A e B percorreram uma trajetória retilínea conforme as equações horárias Sa = 30 + 20t e Sb = 90 – 10t, sendo a

Ray Of Light [21]

Answer:

a yeHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Explanation:

4 0

3 years ago

Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is t

uysha [10]

<u>I have assumed a weight of 120 N on Earth.</u>

Answer:

<em>The object weighs 20 N on the moon</em>

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

$g_m=1/6 g_e$

Dividing by ge:

$g_m/g_e = 1/6$

An object of weight We=120 N on planet Earth has a mass of:

$m = 120 / g_e$

Multiplying by gm:

$m.g_m=120 g_m/g_e$

Substituting the ratio of accelerations of gravity:

$m.g_m=120 * 1/6$

Since m.gm is the weight on the Moon Wm:

$W_m=20~N$

The object weighs 20 N on the moon

5 0

3 years ago

Ne W2

levacccp [35]

Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

5 0

4 years ago

What’s at the bottom of a black hole?

Anettt [7]

Answer:

Black holes do not have a bottom.

Explanation:

Hope this helps :)

8 0

3 years ago