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horsena [70]
3 years ago
15

A human-operated spaceship reaches the moon in 3 days. The moon is about 386,400 km from Earth. Mars, our closest planetary neig

hbor, is, at its closest, about 140 times farther away from us than the moon if Mars stays in place. Assuming a very simple model of the solar system, about how long would it take that same spaceship to reach Mars?(1 point) about 22.5 years about 1 year about 2.5 years about 7.5 years
Physics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

About 1 year.

Explanation:

Since the spacecraft travels the 386,400 kilometers of distance between the Earth and the Moon in 3 days, and the planet Mars is 140 times further from the Earth than the Moon, to obtain the time it will take for the same spacecraft to reach Mars must perform the following calculation:

140 x 3 = 420 days

If a year has 365 days, the travel time that said spacecraft would take to reach Mars would be just under 1 year and 2 months.

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Difference between 3.15 m and 2.0 m with the correct number of significant figures?
pentagon [3]

Answer:

1.2

Explanation:

took the test

3 0
3 years ago
What is the purpose of a free body diagram
Amiraneli [1.4K]
<span>A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.


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4 0
3 years ago
Read 2 more answers
A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

4 0
3 years ago
An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how
Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

v=u+at

a is the acceleration, here, a=\mu g

0=u+\mu gt

t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

8 0
3 years ago
A 15kg ball accelerates at a rate of 3m/s/s. What force was required?
hoa [83]

Answer:

<h2>45 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 3

We have the final answer as

<h3>45 N</h3>

Hope this helps you

3 0
3 years ago
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