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3 years ago

Which statements are true about the speed of a wave?

Physics

2 answers:

Butoxors [25]3 years ago

5 0

Answer: C. it’s different in different materials !!!

Pie3 years ago

4 0

I guess b tell me if I’m wrong

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6. An excited dog runs full-speed toward his owner who has just returned home from

mixas84 [53]

Answer:

v = 8.65 m/s

Explanation:

Given that,

Distance covered by the doge, d = 45 m

Time taken, t = 5.2 s

We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,

$v=\dfrac{45\ m}{5.2\ s}\\\\=8.65\ m/s$

So, the average speed is 8.65 m/s.

4 0

3 years ago

A 68 kg object, starting from rest, travels from point A to point B at a rate of 30 m/s in 2 hours. What is the applied force on

Natasha2012 [34]

Answer:

$\huge\boxed{\sf F = 0.28\ N}$

Explanation:

<h3>Given Data:</h3>

Mass = m = 68 kg

Velocity = v = 30 m/s

Time = 2 hours = 2 × 60 × 60 = 7200 s

<h3>Required:</h3>

Force = F = ?

<h3>Formula to be used:</h3>

$\displaystyle F = \frac{mv}{t}$

<h3>Solution:</h3>

$\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}$

7 0

1 year ago

Find the work performed when the given force f f is applied to an object, whose resulting motion is represented by the displacem

Ahat [919]

work = force x distance

fd foot pounds

3 0

3 years ago

How can i calculate distances between objects by using the concepts of echo location

dangina [55]

Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return

5 0

3 years ago

Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$

The acceleration vector then has direction $\theta$ where

$\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ$

5 0

3 years ago