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3 years ago

A wave travels through the air. The particles of the wave move at right angles to the motion of the energy of the wave.

Physics

1 answer:

maria [59]3 years ago

6 0

Transverse Waves will be your answer I just saw the answer

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Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0

3 years ago

Https://phet.colorado.edu/sims/html/balloons-and-static-electricity/latest/balloons-and-static-electricity_en.html

Katarina [22]

Answer:

there is not pic

Explanation:

5 0

3 years ago

A ball is thrown upward from the ground with an initial speed of 22.0 m/s; at the same instant, another ball is dropped from a b

Vikki [24]

They will be together in 8 minutes

3 0

3 years ago

A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S

amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

v is image distance

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm$

If h' is the height of the image. Magnification is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm$

So, the height of the image of the candle is 20 cm.

3 0

3 years ago

Two positive point charges that are equal in magnitude are fixed in place, one at x = 0.00 m and the other at x = 1.00 m, on the

olga55 [171]

Answer:

0.5 m

Explanation:

Two charges each of magnitude q

Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.

The force on Q due to q at origin is balanced by the charge on Q due to the charge q placed at x = 1 m.

So,

$\frac{KQq}{x^{2}}=\frac{KQq}{\left ( 1-x \right )^{2}}$

1 - x = x

1 = 2x

x = 0.5 m

Thus, the third charge is placed at x = 0.5 m .

6 0

3 years ago