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Gwar [14]
4 years ago
7

Of the planets with atmospheres, which is the warmest? a. Venusb. Earthc. Marsd. Jupiter

Physics
1 answer:
tangare [24]4 years ago
7 0

The Atmosphere in Jupiter is full of gases that move at high speeds in giant eddies. Its atmosphere consists mostly of gases such as hydrogen that generate a temperature fluctuation of around 128K.

On Earth, due to the protection of the Ozone Layer and the presence of Nitrogen and Oxygen, the temperature fluctuates by an average of 300K.

In the case of Mars, its atmosphere is thin, mostly composed of Carbon Dioxide and Diatomic Nitrogen, which allow a temperature oscillation of 210K.

In contrast, the atmosphere of Venus is thick and is composed of carbon dioxide that does not allow the sun's rays to escape, generating an extreme 'greenhouse effect' with temperatures ranging from 737K,

Correct Answer is A.

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What is the period of a wave with a frequency of 0.75 Hz?
Wewaii [24]

Answer:

Period = 1.33 seconds

Explanation:

Period = 1/0.75

8 0
3 years ago
What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?
mixer [17]

Hi there!

Recall Newton's Second Law:

\large\boxed{\Sigma F = ma}

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We must begin by solving for the acceleration using the following:

a = Δv/t

In this instance:

Δv = 3 m/s

t = 2.5 sec

a = 3/2.5 = 1.2 m/s²

Now, plug this value along with the mass into the equation for net force:

\Sigma F = 5(1.2) = \boxed{6 N}}

6 0
3 years ago
Read 2 more answers
Three currents are passing through a surface bounded by a closed path. The currents have different values and directions. Accord
GaryK [48]

Answer:

See Explanation Below

Explanation:

This question is incomplete.

I'll answer this question on general terms. You'll get your result if you apply the steps I'll highlight below.

To start with; what's Ampere law;

It states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

The simple translation of this law is that, the sum of current in the close loop gives us the desired result.

Rephrasing your question;

Three currents, (I1 = +3A, I2 = +4A and I3 = -5A) are passing through a surface bounded by a closed path. The currents have different values and directions. According to Ampere’s law, what is the value of I on the right side of this equation?

First, we take note of the signs in front of the given currents.

The negative sign in front of I3 means that; it is moving in opposite direction of I1 and I2.

To calculate the value of I.

The value of I is the sum of the three currents:

i.e. 3A + 4A - 5A

I = 2 A

7 0
3 years ago
if a car engine does 600,000 J of work over a 500m distance and the mass of the car is 250Kg then what is the final velocity of
drek231 [11]

Given that

Work = 600,000 J ,

distance(S) = 500 m ,

mass (m) = 250 Kg ,

Determine the velocity of car (v) = ?

                 We know that,

                                Work = Force × distance

                               => Force = Work ÷ distance

                                              = 600,000 ÷ 500

                                              = 500 N .

                   Also Force F =  m.a  ; from Newtons II law

                                      500 = 250 × a  

                                             a = 2 m/s.

<em>Final Velocity from the given  formula </em>

                                     V² = u² + 2.a.s

                                         = 0 + 2 × 2 × 500

                                         = \sqrt{2000}

                                    <em>   v = 44.7 m/s</em>

8 0
3 years ago
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric
gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

KE = \fract{1}{2}mv^2

PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

m = 910 Kg

r_1 = 1200 + 6371 km = 7571km

r_2 = 6371 km,

Replacing we have,

\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})

v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})

v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})

v = 4456 m/s

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0
4 years ago
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