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3 years ago

What factors in Earth's relationship to the Sun make life on Earth possible?

2 answers:

5 0

The answer is B.

The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life.

lawyer [7]3 years ago

5 0

C because without the gravitational pull from the sun we would be flying throughout space right now.

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Help please<br> It’s kinda urgent

user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

$(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]$

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0

2 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

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#LearnwithBrainly

8 0

4 years ago

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years