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Assoli18 [71]
2 years ago
13

An object moves 4 meters north, then 3 meters east. What is the displacement of the object?

Physics
2 answers:
Lady bird [3.3K]2 years ago
4 0

The displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

<h3>What is displacement?</h3>

Displacement is change in the position of an object. It can be described as the shortest distance between the initial and final position of an object.

The displacement of the object is calculated as follows;

d² = 4² + 3²

d² = 25

d = 5 m

Thus, the displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

Learn more about displacement here: brainly.com/question/2109763

#SPJ2

ziro4ka [17]2 years ago
3 0

Answer:

5 meters

Explanation:

4  and  3   are legs of a right triangle with the hypotenuse = displacement

Use Pythagorean theorem

4 ^2 + 3^2 = (displacement)^2

displacement = 5 m

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<em>I must travel with a speed of 2.97 x 10^8 m/s</em>

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According to relativity, the time contraction on the spacecraft is gotten from

t = t_{0} /\sqrt{1 - \beta ^{2} }

where

t is the time that elapses on the spacecraft = 120 years

t_{0} = time here on Earth = 1 year

\beta is the ratio v/c

where

v is the speed of the spacecraft = ?

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

120 = 1/\sqrt{1 - \beta ^{2} }

squaring both sides of the equation, we have

14400 = 1/(1 - \beta ^{2} )

14400 - 14400\beta ^{2} = 1

14400 - 1 = 14400\beta ^{2}

14399 = 14400\beta ^{2}

\beta ^{2} =  14399/14400 = 0.99

\beta = \sqrt{0.99} = 0.99

substitute β = v/c

v/c = 0.99

but c = 3 x 10^8 m/s

v = 0.99c = 0.99 x 3 x 10^8 = <em>2.97 x 10^8 m/s</em>

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3 years ago
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Hope this helps you :)

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4 years ago
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A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

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Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
3 years ago
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