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4 years ago

14

In a scale model solar system that used a grapefruit to represent the Sun, how far away should you put another grapefruit to rep

resent Alpha Centauri, the next nearest star?

1 answer:

kotegsom [21]4 years ago

6 0

Answer:

2000 miles

Explanation:

Alpha centaur is the nearest star system to the solar system. It is estimated to be about 2000 miles near the sun

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List at least 5 examples of civil rights

mariarad [96]

Answer:

right to vote, the right to a fair trial, the right to government services, the right to a public education, and the right to use public facilities.

6 0

3 years ago

A car drives 215km east and then 45km north. What is the magnitude of the cars displacement? Round you answer to nearest whole n

kifflom [539]

Answer: $219.65\ \text{km}$

Explanation:

Given

Car drives 215 km east and then 45 km North

Displacement is East direction is

$\Rightarrow \vec{r_1}=215\hat{i}$

Now, the displacement from that to 45 km North is given by

$\Rightarrow \vec{r_{21}}=45\hat{j}$

Net displacement is $\vec{r_2}$

$\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}$

Magnitude of the displacement is

$\Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow \left | r_{2} \right |=219.65\ \text{km}$

5 0

3 years ago

Find the area of rectangle given below

Delicious77 [7]

Answer:

6000 cm<em>²</em>

Explanation:

let width of rectangle = $x$

Using Pythagorean theorem,

$x^{2}$ +<em> 50² = 130²</em>

<em /> $x^{2}$ <em> = 130² - 50² </em>

<em /> $x^{2}$ <em> = 14400</em>

<em /> $x$ <em>= 120 cm</em>

area of rectangle <em>=</em> length x width

<em>=</em> <em>120 x 50</em>

<em>= 6000 cm²</em>

4 0

3 years ago

A person runs with a constant velocity of 1.2 m/s for 30 seconds. how far do they travel?

Nataliya [291]

They will travel 36 m

8 0

4 years ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

4 years ago