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lions [1.4K]
2 years ago
14

I dont understand pojectile motion can someone help with this

Physics
1 answer:
katrin2010 [14]2 years ago
3 0

When thrown from the balloon, the sandbag has a constant horizontal/forward velocity. As the text puts it, "the sandbag continues forward as if it had not been dropped." What this means is that the sandbag has no acceleration in the horizontal plane of motion, so it moves in this direction at a constant speed. So for any time t, the horizontal velocity is given by

v_x(t) = 12\dfrac{\rm m}{\rm s}

In the vertical plane of motion, the sandbag has no vertical velocity until it's released and allowed to fall. As it's falling, however, the sandbag is being pulled down by gravity and is accelerating with magnitude g=9.8\frac{\rm m}{\mathrm s^2}, so its vertical/downward velocity at time t is given by

v_y(t) = -gt

So, after 3 seconds have passed, the sandbag has horizontal velocity v_x = 12\frac{\rm m}{\rm s} and vertical velocity v_y = -g(3\,\mathrm s) = -29.4\frac{\rm m}{\rm s}. If the question is specifically asking about speed you would just ignore the negative signs.

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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

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Plugging in our numbers (and assuming that

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10

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m

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2

⋅

(

10

19

m

)

V

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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

×

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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

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n

V

Using the volume of a sphere

V

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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
2 years ago
Read 2 more answers
Question 2 The gravitational force between two objects with identical masses that are 10 m apart, is 2.67 x10-10 N. To the neare
xxMikexx [17]

Answer  888990,0 kg

Explanation:

3 0
3 years ago
Which value is equivalent to 7.2 kilograms?
Elis [28]
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3 0
3 years ago
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Marta_Voda [28]

Answer:

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Explanation:

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5 0
2 years ago
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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

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