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3 years ago

12

A wave has a wavelength of 5 meters and a frequency of 3 hertz. What is the speed of the wave?

1 answer:

otez555 [7]3 years ago

8 0

A sound wave<span> in a steel rail </span>has<span> a </span>frequency of<span> 620 </span>Hz<span> and a </span>wavelength<span> of 10.5 ... Find the </span>speed<span> of </span>a wave<span> with a </span>wavelength of 5<span> m and a </span>frequency of<span> 68 </span>Hz<span>.</span>

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What kind of energy does an unlit match have?

Rudiy27

It’s is none of the above answers.
It is d. Potential energy

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3 years ago

Describe how evaporation relates to heat regulation in your body

yanalaym [24]

When you sweat due to heat your sweat evaporates off your body taking the heat away

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3 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.

madam [21]

<h2>Answer:5 $ms^{-1}$ ,133.6 $m$ ,51.18 $ms^{-1}$ </h2>

Explanation:

Let $v_{x}$ , $v_{y}$ be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is $\frac{235}{47}=5ms^{-1}$

So, $v_{x}=5ms^{-1}$

Question b:

Time of flight= $\frac{2v_{y}}{g}$

So, $v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}$

Maximum height is given by $\frac{v_{y}^{2}}{2g}$

So,maximum height is $\frac{51.18^{2}}{2\times 9.8}=133.6m$

Question c:

The vertical velocity is already calculated in Question b.

$v_{y}=51.18ms^{-1}$

7 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

erma4kov [3.2K]

Answer:

8.162

Explanation:

W=mg

= 2.2 x 3.71

= 8.162 N

5 0

3 years ago