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BlackZzzverrR [31]

3 years ago

14

You travel 60 meters to the left in 20 seconds and then you travel 60 meters to the right in 30 seconds; what is your average ve

locity?
What is the answer and what are the steps to arrive at the answer? I need help figuring this out.

1 answer:

vovikov84 [41]3 years ago

4 0

Average velocity = displacement / time

In this case, since you move left 60, then right 60, you return to the starting point making the displacement zero.

average velocity = 0 / 60 sec = zero

the average speed is calculated using total distance traveled.

average speed = distance / time

average speed = 120 meters / 60 sec = 2 m/s

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What is the SI unit of acceleration and work?

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A=m/s^2(meter per second square)
Work=joule

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

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andrezito [222]

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Read 2 more answers

In loading a lorry, a man lifts boxes each of weight 100N through a height of 1.5m.

liq [111]

Answer:

The amount of work done in order to lift the box is $150\ Joule$

Explanation:

Given the weight of each box is $100N$

And the man lifts boxes at a height of $1.5m$

We need to find the amount of work done.

The amount of work done is the product of applied force $F$ that causes the displacement $d$ .

In our problem force is $100N$ and displacement is $1.5m$ .

Now, work done

$=force\times displacement\\=100\times1.5\\=150\ Joule$

So, the amount of work done in order to lift the box is $150\ Joule$

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3 years ago