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Fiesta28 [93]
4 years ago
11

Does the halo of the milky way galaxy include other galaxies?

Physics
2 answers:
Dominik [7]4 years ago
7 0
Yes the halo contains some significantly smaller galaxies
julia-pushkina [17]4 years ago
4 0
Nope.......hope I helppppp
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What is the equation for gravity and what does each part mean?
Andrew [12]
This equation description the force between any two object in the universe in the equation F is the force of the gravity Newton lawe
6 0
4 years ago
A medicine ball has a mass of 6 kg and is thrown with a speed of 4 m/s. What is its kinetic energy?
zhenek [66]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 48 \ J}

Given:

Mass (m) = 6 kg

Speed (v) = 4 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6 \times  {4}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3 \times 16

\sf \implies KE =3 \times 16

\sf \implies KE = 48 \: J

6 0
4 years ago
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on
hodyreva [135]

Solution :

The distance between the starting point and the end point, L_0 = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, L_b = 0.7  \ c

So the distance appeared to Bob due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$

$L_b=10\times \sqrt{1-0.49} \ Ly$

    $=7.1 \ Ly$

Therefore, the time required to finish the race by Bob is

$t_b = \frac{L_b}{V_b}$

  $=\frac{7.1 \ c}{0.7 \ c}$

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, L_c = 0.866 \ c

So the distance appeared to Charlie due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$

$L_b=10\times \sqrt{1-0.75} \ Ly$

    $=5 \ Ly$

The time required to finish the race by Charlie is

$t_b = \frac{L_c}{V_c}$

  $=\frac{5 \ c}{0.866 \ c}$

  = 5.77 year

5 0
3 years ago
A car with mass mc = 1490 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck o
icang [17]

Answer:

v= - 4.507 i - 2.363 j

Explanation:

 Given that

mc= 1490 kg

vc= 9.5 m/s ( - i)

mt=  1650 kg

vt = 6.4 m/s ( -j)

There is any external force so linear momentum will remain conserve.

Lets take final speed is v.

mc .vc + mt . vt = ( mc+mt) v

1490 x 9.5 ( - i) + 1650 x 6.4 ( -j) = ( 1490+1650) v

14,155 ( -i) + 10,560 ( - j) = 3140 v

v= - 4.507 i - 2.363 j

3 0
4 years ago
This is due by 11:59 PM tonight.
svetoff [14.1K]

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

6 0
4 years ago
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