All categories

2 years ago

What are northern lights? a lights on mars b lights on the sun c lights on venus

1 answer:

8 0

The northern lights are shafts or curtains of colorful light that occasionally appear in the night sky. They are one of the numerous astronomical phenomena known as polar lights (aurora Polaris).This phenomenon may be observed in mars.

Earth's magnetic field directs electrons and protons from the sun to the poles, where they excite atmospheric gas molecules and cause them to glow, resulting in the aurora borealis and aurora australis, two nocturnal light displays. You might refer to it as the aurora Universalis on Mars. This is because Mars does not direct the energetic particles from the sun to its poles since it lacks an internal magnetic field. Today, researchers utilizing the MAVEN (Mars Atmosphere and Volatile Evolution) spacecraft find evidence for an aurora that may potentially cover the whole nightside of the planet. Venus lacks a magnetic field, thus it would not experience the same kind of nighttime aurora that we do.

To know more about aurora borealis go here:-

brainly.com/question/12757223

#SPJ4

You might be interested in

A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the

Elan Coil [88]

Answer:

The horse is going at 12.72 m/s speed.

Explanation:

The initial speed of the horse (u) = 3 m/s

The acceleration of the horse (a)= 5 m/ $s^{2}$

The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m

Applying the second equation of motion to find out the time,

$s=ut+\frac{1}{2}at^{2}$

$15.3=3t+2.5t^{2}$

$2.5t^{2}+3t-15.3=0$

Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.

Now applying first equation of motion, to find out the final velocity,

$v=u+at$

$v=3+1.945*5$

$v=3+9.72$

v=12.72 m/s

The horse travels at a speed of 12.72 m/s after covering the given distance.

7 0

3 years ago

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

goldenfox [79]

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval $\Delta t$

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction

7 0

3 years ago

A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters

vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

$a=4 m/s^2$ is the acceleration

s = 3 m is the distance covered

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s$

3 0

3 years ago

Explain at least 3 primary and secondary online sources of data<br>

Slav-nsk [51]

Answer And Explanation:

Primary sources of online data could be blog posts, twitter posts, and vlogs because the person producing forms of work is giving first-person account of his experience.

Secondary sources online data could be articles, news, and documentaries because people have gathered several primary sources to build their case that they are making.

Hope this is clear, and good luck with studies :)

7 0

3 years ago

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

8 0

3 years ago