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zimovet [89]
3 years ago
15

The Milky Way is our Galax. Where are we located in Milky Way?​

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

Earth is located in one of the spiral arms of the Milky Way (called the Orion Arm) which lies about two-thirds of the way out from the center of the Galaxy.

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A center-seeking force related to acceleration is _______ force.
zvonat [6]
<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>
5 0
3 years ago
The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
Crazy boy [7]

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

4 0
2 years ago
WILL GIVE BRAINLY!!!
Gnoma [55]
The answer is apogee my dude
7 0
3 years ago
A wooden block is sitting on an inclined plane near the bottom. The student gave the block a flick and it moved up the inclined
sladkih [1.3K]

Answer:

The block didn't slide due to balancing of gravitational force with friction force

Explanation:

When the block was given a flick the force provided an acceleration to it and it moved up the inclined plane. when the block reached top it was expected that it would slide back but it didn't this happened because of the frictional force acting on the bottom the block which was balancing the gravitational force component along the plane and this prevented sliding back of the block.

static friction  was balancing mg*sin(theta)

fs = mg*sin(theta)

6 0
3 years ago
A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a
dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

A_1v_1 = 18 A_2v_2

now we know that

A_1 = 1 cm^2

A_2 = 0.4 cm^2

now from above equation

1 cm^2 v_1 = 18(0.400 cm^2)v_2

\frac{v_2}{v_1} = \frac{1}{18\times 0.4}

\frac{v_2}{v_1} = 0.14

so velocity will reduce by factor 0.14

3 0
3 years ago
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