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stepladder [879]
2 years ago
6

The center of our galaxy lies in the direction of the constellation of:.

Physics
1 answer:
Sphinxa [80]2 years ago
8 0

Answer:

It lies in the direction of the constellation of Sagittarius (The Archer), and is marked with a red circle in the image. This map shows most of the stars visible to the unaided eye under good conditions.

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What is the difference between speed and velocity?
Wittaler [7]
It’s C) because we know that velocity is a vector quantity with direction and speed is only quantity. For example, velocity could be 5 m/s to the east but speed would only be 5 m/s
7 0
3 years ago
A(n) is a line in which waves that are moving together are all in the same phase.
Ksivusya [100]
The line in which waves that are moving together are all in the same phase is referred to as the wave front. 
For example, in a series of waves, all of the crests that are in a single line make up a single wave front. This is the case for every set of particles at the same specific point in a series of waves.
5 0
3 years ago
A 6.0-kg object moving 5.0 m/s collides with and sticks to a 2.0-kg object. after the collision the composite object is moving 2
vredina [299]
  <span>m1v1 + m2v2 = (m1+m2)*(-vf) 
6*5 + 2*v2 = (6 + 2)*(-2) 
v2 = -16/2 = -8 m/s i</span>
5 0
4 years ago
One billiard ball strikes another billiard ball, and they move away from each other
xxTIMURxx [149]

Answer:

C) The magnitude and direction

Explanation:

Velocity is a vector quantity, meaning that it has both magnitude and direction.

For the momentum, we look at both the direction of the ball (negative, positive) and the magnitude of the velocity (5 m/s, 10 m/s) when figuring out what to use for "v" in p = mv.

5 0
3 years ago
A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s
expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

\vec v_f=\vec v_o+\vec a.t

\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is 2.7 m/s^2 to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

v_f=v_o+a.t

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Solving the first one for t

\displaystyle t=\frac{v_f-v_o}{a}

We have

v_o=-6.5,\ v_f=-1.5,\ a=2.7

Using these values

\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

We now compute x

\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}

x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

5 0
3 years ago
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