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CaHeK987 [17]
3 years ago
14

ALGEBRA Find x and the measures of the unknown sides of each triangle.

Physics
1 answer:
valentina_108 [34]3 years ago
3 0

9514 1404 393

Answer:

  12.  x = 11; sides are 29

  13.  x = 5; sides are 25

Explanation:

12. The marked sides are equal length, so ...

  3x -4 = 2x +7

  x = 11 . . . . . . . . add 4-2x

  2x+7 = 2(11)+7 = 29 . . . side lengths

__

13. The marked sides are equal length, so ...

  6x -5 = 5x

  x = 5 . . . . . . . add 5-5x

  5x = 5(5) = 25 . . . side lengths

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
3 years ago
Before the big bang, the universe was much ____ and _____ than it is now.
Dahasolnce [82]
The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.
5 0
3 years ago
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A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a
Doss [256]

Answer:

Part(a): The frequency is \bf{0.2~Hz}.

Part(b): The speed of the wave is \bf{0.5~m/s}.

Explanation:

Given:

The distance between the crests of the wave, d = 2.5~m.

The time required for the wave to laps against the pier, t = 5.0~s

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is \lambda = 2.5~m.

Also, the time required for the wave for each laps is the time period of oscillation and it is given by T = 5.0~s.

Part(a):

The relation between the frequency and time period is given by

\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Substituting the value of T in equation (1), we have

\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz

Part(b):

The relation between the velocity of a wave to its frequency is given by

v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting the value of \nu and \lambda in equation (2), we have

v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s

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3 years ago
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A population of deer is defined by:
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1. A
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6.B
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3 years ago
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