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artcher [175]
2 years ago
9

In outer space, a piece of rock continues moving at the same velocity for

Physics
1 answer:
Ray Of Light [21]2 years ago
7 0

The absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

<h3>Absence of external force on the outer space</h3>

The outer space is almost an absolute vacuum, because it's nearly empty. There is no matter such as air in the outer space that will provide an external force needed to change the velocity of the piece of rock.

From Newton's first law of motion, an object in a state of rest or uniform motion in a straight line, will continue in that state unless it is acted upon by an external force.

Thus, the absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

Learn more about outer space here: brainly.com/question/24701339

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A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass
ValentinkaMS [17]

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so ,  its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis  after explosion of third part   = mv cosθ

= 10 v cosθ

Momentum along x of first part = -  5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 +  10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°

6 0
3 years ago
How much positive and negative charges is there in a cup of water?
Rina8888 [55]
In a cup of water it is a positive and negative charge of zero
3 0
2 years ago
What acceleration results from exerting a 125N force on a 0.65kg ball?
labwork [276]
first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077
5 0
3 years ago
What is the average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8
Alex

Answer:

The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 \frac{m}{s}.

Explanation:

Velocity ​​is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it. Then, the average velocity relates the change in position to the time taken to effect that change.

velocity=\frac{displacement}{time}

Velocity considers the direction in which an object moves, so it is considered a vector magnitude.

In this case, the displacement is 192 m and the time period is 8 s. Replacing:

velocity=\frac{192 m}{8 s}

Solving:

velocity= 24 \frac{m}{s}

<em><u>The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 </u></em>\frac{m}{s}<em><u>.</u></em>

3 0
2 years ago
The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
2 years ago
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