If positive work is being done to an object...

A runner starts at point A, runs around a 1-mile track and finishes the run back at point A. Which of the following statements i

Lyrx [107]

This question is incomplete because the options are missing; here is the complete question:

A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?

A. The runner's displacement is 1 mile.

B. The runner's displacement is zero.

C. The distance the runner covered is zero.

D. The runner's speed was zero.

The answer to this question is B. The runner's displacement is zero

Explanation:

Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.

4 0

3 years ago

Nancys airplane trip took 4 hour. For one-fourth of that time, the airplane flew at a speed of 880 km/h, it flew at a speed of 6

9966 [12]

So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.

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4 0

3 years ago

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:

7 0

3 years ago

A student rolls these four items on the same surface and at the same speed. Which item will have the greatest kinetic energy?

dimulka [17.4K]

Basketball is the answer

7 0

3 years ago

Efficincy of simple machine is 60% what does it mean?<br>

umka2103 [35]

Explanation:

By the statement, Efficiency of a machine is 60% it is meant that 'The work output is 60% of the work input'. If you apply 100J of energy for a machine, it will be able to life load of 60 J only

7 0

3 years ago