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3 years ago

What is the value of x in the triangle? a 45-45-90 triangle with leg length x and hypotenuse length 4

Physics

2 answers:

mrs_skeptik [129]3 years ago

5 0

Answer:

$x = 2\sqrt 2$

Explanation:

Given

$Hypotenuse = 4$

Required

Find x

Since the triangle is a 45-45-90 triangle, the following relationship exists

$x^2 + x^2 = 4^2$ --- i.e. the other legs are equal

So, we have:

$2x^2 = 16$

Divide both sides by 2

$x^2=8$

Take square roots of both sides

$x = \sqrt 8$

Simplify

$x = 2\sqrt 2$

myrzilka [38]3 years ago

4 0

Answer:

7√2

Explanation:

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A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-

wariber [46]

Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

$F_n = \frac{mv^2}{R} - mg$ [/tex]

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$

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3 years ago

An ocean wave has a wavelength of 15 meters and is moving at 2.5 m/s. what is the frequency

Tasya [4]

Answer:

frequency is .167/s for this wave

8 0

3 years ago

Read 2 more answers

If the distance between two asteroids is halved, the gravitational force they exert on each other will

mamaluj [8]

Answer:

e) Be four times greater

Explanation:

Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses ( $m_{1}$ & $m_{2}$ ) and the distance between them ( $r$ ) in the next way:

$F=G\frac{m_{1}m_{2}}{r^{2}}$ (2)

Now if distance between asteroids is halved:

$F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}}$

$F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}}$

Note that $G\frac{m_{1}m_{2}}{r^{2}}$ because (1) is F so:

$F_{2}=4F$

It's four times greater!

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3 years ago

A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the

ioda

Answer:

Force, F = 44 N

Explanation:

Given that,

Initial speed of the football, u = 0

Final speed, v = 15 m/s

The time of contact of the ball, t = 0.15 s

The mass of football, m = 0.44 kg

We need to find the average force exerted on the ball. It is given by the formula as :

$F=ma\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{0.44\times 15}{0.15}\\\\F=44\ N$

So, the average force exerted on the ball is 44 N. Hence, this is the required solution.

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3 years ago

Three flat layers of transparent material are stacked upon one another. The top layer has index of refraction n1, the middle has

serg [7]

Answer:

a. the bottom medium

Explanation:

it has the least index of refraction and hence most rarer.

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3 years ago