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Brrunno [24]
3 years ago
14

the distance of the earth to the sun is 1.5x10^11 m and one year is 365days if the distance to the sun increased six time the in

itial value how long it take the earth to orbit the sun?
Physics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

x = 2190[days]

Explanation:

This problem can be solved by means of a simple three rule:

1.5*10^{11}=365[days]\\ 6*(1.5*10^{11})=x\\\\x= \frac{6*(1.5*10^{11})*365[days]}{1.5*10^{11}} \\x=2190[days]

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Please help me on 3,4 and 6. <br> Thanks
rusak2 [61]
3. <span>The second piston will experience the same force as compared with the first. This is because since the </span>pressure is the same everywhere inside the fluid system,<span> the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.


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3 0
4 years ago
An Iowa class warship holds the record for the fastest warship. If the ship accelerates uniformly from rest at 0.15 m/s² for 2 m
Mumz [18]
The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is

a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration

Since it started from rest, v₀ = 0. Using the formula:

0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s
3 0
3 years ago
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
3 years ago
Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen
vovangra [49]

Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>
  • It is defined as the rate of change in velocity.
  • It can also be defined as the rate of change in position in a particular direction.

a =  \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t

Where,

a - acceleration

\Delta v - change in velocity

t - time

Therefore, acceleration can be defined as the rate of change in the velocity of an object.

Learn more about Velocity:

brainly.com/question/2239252

4 0
2 years ago
A man pushes a 10kg box with a constant acceleration of 5m/s2. What force is applied to the box?
drek231 [11]
FORMULA

\boxed {F = m \times a}

F = force
m = mass
a = acceleration

Using the formula

F = 10 \times 5

Multiply

\boxed {\textsf {F = 10 N}}
8 0
3 years ago
Read 2 more answers
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