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Nina [5.8K]
2 years ago
10

How do you figure out cos and sin?

Physics
2 answers:
Nataly_w [17]2 years ago
6 0

Answer:

In a right triangle, you always want to use the right angle as a reference. Sin is opposite side over hypotenuse, and cos is adjacent side over hypotenuse. An easy way to remember is SOHCAHTOA (Sin Opposite Hypotenuse, Cos Adjacent Hypotenuse, Tan Opposite Adjacent), The acronym I use to remember this is "Some old hippy caught another hippy tripping on acid."

Explanation:

dexar [7]2 years ago
5 0

Answer:this is all i can give you

Explanation:

1.The sine of the angle = the length of the opposite side. the length of the hypotenuse.

2.The cosine of the angle = the length of the adjacent side. the length of the hypotenuse.

3.The tangent of the angle = the length of the opposite side. the length of the adjacent side.

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A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency
Vikki [24]

Answer:

Explanation:

f = \sqrt{T/(m/L)} / 2L

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = \sqrt{120/12} /(2(3)))

f = \sqrt{10\\}/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

5 0
2 years ago
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
Resistance of a light bulb with 0.33 A of current flowing from a 12V battery?
Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

<em>Resistance = 36.4 ohms</em>

8 0
3 years ago
What is the magnitude of the kinetic frictional force
Effectus [21]

The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

4 0
3 years ago
The electrical force on a 2-c charge is 60 n. the electric field where the charge is located is
Kazeer [188]
The electrical force acting on a charge q immersed in an electric field is equal to
F=qE
where
q is the charge
E is the strength of the electric field

In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
E= \frac{F}{q}= \frac{60 N}{2 C}=30 N/C
5 0
2 years ago
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