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Shkiper50 [21]
3 years ago
8

Which three characteristics of an object are represented by a motion map

Physics
1 answer:
nadya68 [22]3 years ago
8 0

There are three things that can be represented on a motion map

These three things are:

1)Motion

2)Acceleration

3)Velocity

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200 pages about string thirory
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Explanation:

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A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse
Sonja [21]
The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
\tan \alpha  =  \frac{v_y}{v_x}
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
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5 0
3 years ago
Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

L = L_0 \sqrt{1-\frac{v^2}{c^2} }

L = 1 light year

L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

v = 0.999999995 c

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0
1 year ago
A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1m/s^2 for 20s.
SSSSS [86.1K]

Answer:

A.) 1430 metres

B.) 80 seconds

Explanation:

Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:

U = acceleration × time

U = 1.1 × 20 = 22 m/s

It then proceeds at constant speed for 1100 m

Then, time t will be

Time = distance/ velocity

Time = 1100/22

Time = 50 s

before slowing down at 2.2m/s^2 until it stops at the station.

Deceleration = velocity/time

2.2 = 22/t

t = 22/2.2

t = 10s

Using area under the graph, the distance between the two stations will be :

(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)

220 + 1100 + 110

1430 m

The time taken between the two stations will be

20 + 50 + 10 = 80 seconds

6 0
3 years ago
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