The frequency of a wave is 10 Hz. If the speed of the wave is 5 m/s, what is the wavelength? A. 0.5 m B. 2 m C. 10 m D. 50 m

Which wave must have a medium to travel?

o-na [289]

Letter B

without a medium, there is nothing to compress, hence, no wave. A fast- medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.

6 0

2 years ago

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When a sideways force acts on a moving object, what effect does it have?

My name is Ann [436]

Answer:

it makes the object speed increase, decrease and change the direction of the object.

Hope it helps!

6 0

3 years ago

In which part of the ear is the sound wave converted into electrical impulse

Kryger [21]

The part of the ear where the sound wave converted into electrical impulse would be the cochlea. This part is the auditory portion of the inner ear which produces nerve impulses in response to sound vibrations. Hope this answers the question. Have a nice day.

3 0

3 years ago

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What is a free electron? Can someone help?

Aleksandr [31]

A free electron is one which has become detached from a covalent bond between two atoms and is able to move around from atom to atom and possibly take part in electric current flow.

3 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$