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brilliants [131]
3 years ago
11

TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!

Physics
2 answers:
JulsSmile [24]3 years ago
8 0

Answer:

<em>d = 27.03 m</em>

Explanation:

<u>Constant Speed Motion </u>

An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

It's required to find the distance traveled by someone walking at v=1.7 m/s for t=15.9 s. Substituting in the last equation:

d = 1.7 m/s * 15.9 s

d = 27.03 m

Diano4ka-milaya [45]3 years ago
3 0
The answer is 27.03 I just multiplied the two numbers
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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
3 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
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