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3 years ago

SOH-CAH-TOA is used to solve for the ________

Physics

1 answer:

Misha Larkins [42]3 years ago

3 0

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

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a hiker walks 200 m west and then walks 100 m north in what direction is her resulting displacement draw to show your answer

Sliva [168]

Answer:

The direction of the displacement is in North-West.

Explanation:

Resultant displacement D is

$=\sqrt{(200)^{2} + (100)^{2} } \\=223.60m$

Here the direction is

$\Theta =tan^{^{-1}}\left ( \frac{100}{200} \right )\\\Theta =26.6^{o}$

Then the direction is $26.6^{o}$ North-west.

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2 years ago

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3 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

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3 years ago

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$