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3 years ago

when do you use cos and sin in situations like these? is horizontal always cos and vertical always sin?

Physics

1 answer:

Andreas93 [3]3 years ago

8 0

Answer:

yes

Explanation:

this is simple

the horizontal line is adjacent

the vertical line is opposite

recall that cos x=adj/hyp

adj=hyp(cos x)

while opp=hyp(sin x)

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A circuit contains a 305 ohm resistor, a 1.1 micro Farad capacitor, and a 42 mH inductor. At resonance, the impedance is determi

r-ruslan [8.4K]

Answer:

The resistance of the inductor at resonance is 258.76 ohms.

Explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

$\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L = \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }$

Where;

F₀ is the resonance frequency

$F_0 = \frac{1}{2\pi\sqrt{LC} } \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^{-6})} }\\\\F_0 =980.4 \ Hz$

The inductive reactance is given by;

$X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms$

Therefore, the resistance of the inductor at resonance is 258.76 ohms.

5 0

2 years ago

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.

(a) What is the magnitude of the satellite's velocity when the thruster turns off

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is: