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nadezda [96]
3 years ago
7

I'd: 9872093250, password: qqqqq, join the meeting​

Physics
1 answer:
crimeas [40]3 years ago
3 0
Why whats going on homie
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Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
morpeh [17]
Answer: In a longitudinal wave, the crest and trough of a transverse wave correspond respectively to the compression, and the rarefaction. A compression is when the particles in the medium through which the wave is traveling are closer together than in its natural state, that is, when their density is greatest.
4 0
3 years ago
Calculate the following forces for a box with a mass=20.0 kg sitting on an incline of 29.0 degrees and a coefficient of friction
Anna007 [38]
Vertical force on the box=mg 
<span>the component of gravity parallel=mg*SinTheta </span>
<span>the component of gravity normal=mg*CosTheta </span>
<span>frictional force up the plane: mg*cosTheta*mu max, but if it is sitting still, it is equal and opposite to mg*cosTheta (it cannot be greater than this or it would go up the plane).</span>
3 0
3 years ago
. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (T
s2008m [1.1K]

Answer:

Explanation:

The devices are connected in parallel, the same voltage will flow through them

1) I, current in the 1800 W toaster

P, power = I V

where p = 1800 W

1800 /120 = I

I drawn by the toaster = 15 A

2) I, current drawn by the 1400 W frying pan = 1400 / 120 = 11.667 A

3) I, current drawn by 75 W = 75 / 120 = 0.625 A

B) Total current drawn by the circuit = 15 A + 11.667 A + 0.625 A = 27.292 A which is greater than 15-A hence the combination will blow the fuse

6 0
3 years ago
Read 2 more answers
Se lanza una bala con una velocidad inicial de 200 m/s y con un ángulo de inclinación de 30º respecto a la horizontal. Si se con
puteri [66]

Answer:

El alcance de la bala es 3464,1 m.

Explanation:

El alcance de la bala se puede calcular como sigue:

y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}

En donde:

y: es la altura final = 0

y_{0}: es la altura inicial = 0

x: es el alcance

θ: es el angulo respecto a la horizontal = 30°

v_{0}: es la velocidad inicial = 200 m/s

g: es la gravedad = 10 m/s²

Entonces, tenemos:

y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}

x = \frac{2tan(\theta)*(v_{0}cos(\theta))^{2}}{g} = \frac{2tan(30)*(200 m/s*cos(30))^{2}}{10 m/s^{2}} = 3464,1 m

Por lo tanto, el alcance de la bala es 3464,1 m.

Espero que se te sea de utilidad!

8 0
3 years ago
When you do positive work on an object, you ______
gtnhenbr [62]

when you do positive work on an object, you ______  

A. decrease the object's energy

<u>B. keep the object's energy the same </u>

C. increase the object's energy

D. may increase or decrease the object's energy

3 0
3 years ago
Read 2 more answers
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