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3 years ago

Which of the following is not a popular surface for a tennis court?

Physics

1 answer:

vovikov84 [41]3 years ago

8 0

The answer is A. wood

because there are three different courts: clay court, grass court, and hard court. wood isn't in there so that would be the answer.

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Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees,

Iteru [2.4K]

Answer:

$x=65.55m$

Explanation:

Let x be the distance to the shore

From trigonometry properties:

$tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m$

3 0

3 years ago

A gas always spreads out to fill all available space.<br> True or False <br> Science Hw

GREYUIT [131]

Answer:

true

Explanation:

i know this im in 6th grade

6 0

3 years ago

Read 2 more answers

An Ethernet cable is 4.00m long. The cable has a mass of 0.200kg . A transverse pulse is produced by plucking one end of the tau

Mashcka [7]

Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.

<h3>How much tension is in the cable?</h3>

The tension in the cable can be found as:

= 4 x mass x length x frequency

Solving for the frequency is:

= 1 / (0.800 / 4)

= 1 / 0.20

= 5.0 Hz

The tension is therefore:

= 4 x 0.20 x 4.00 x 5

= 80N

Find out more on tension at brainly.com/question/14336853

#SPJ4

4 0

1 year 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

Does the cereal or the milk go first? <br><br>

vazorg [7]

Answer:

Cereal

Explanation:

Nobody puts the milk first that's just communism

7 0

3 years ago

Read 2 more answers