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Maksim231197 [3]

2 years ago

14

Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

ow long will it take for him to hear his own voice echo off of the wall?

2 answers:

Sophie [7]2 years ago

8 0

Answer:

$\Delta t = 2.624\,s$

Explanation:

Sound is a form of mechanic wave, which needs a medium to propagate itself and has a constant speed. The time needed to hear his echo is:

$\Delta t = \frac{900\,m}{343\,\frac{m}{s} }$

$\Delta t = 2.624\,s$

nekit [7.7K]2 years ago

5 0

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

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The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4

BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

$\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})$

$\lambda=4.34\times 10^{-7}\ m$

or

$\lambda=434\ nm$

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0

2 years ago

Convert 15 joule into erg.

GuDViN [60]

Answer:

<h2><u>Joule</u><u>:</u></h2>

1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.

1 Joule= 1 Newton × 1 metre

1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²

So units of N is kgm/s²

So,

1 Joule

=1kgm/s² × m

=1kgm²/s²

<h2><u>Erg</u><u>:</u></h2>

1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.

1 Erg =1 Dyne × 1 cm

1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².

1 Erg=1 gmcm/s² × cm

1 Erg=1 gmcm/s² × cm=1gmcm²/s²

this is what you need to convert 1gmcm²/s² to 1kgm²/s²

<h3><u>what you need to know for conversion</u></h3>

[1gm=0.001kg

1cm²

=1cm ×1cm

=0.01 m × 0.01 m

=0.0001m²

second remains constant

]

So,

1gmcm²/s²

=0.001kg×0.0001m²/s²

=0.001kg×0.0001m²/s² =0.0000001kgm²/s²

Hence,

<h3><u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3><u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>

<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2> =1.5×10⁶ Erg</h2>

3 0

2 years ago

Read 2 more answers

Which of the following causes objects to orbit the Sun?

algol [13]

B. The suns gravity

hope this helps

7 0

3 years ago

Read 2 more answers

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a holl

Semenov [28]

Answer:

a) v=23.9 m/s

b) R=2646 N

Explanation:

According to Newton's second law the net force at the top is given by

∑F=-m*ac

according to

$a_{c}=\frac{v^2}{r}$

∑F= $-m*\frac{v^2}{r}$

a).

To lose contact this means that R=0 so the final equation is

$R-m*g=-m*\frac{v^2}{r}$

$-m*g=-m*\frac{v^2}{r}$

Solve to v

$v^2=g*r$

$v=\sqrt{g*r}=\sqrt{9.8*14.6}$

$v=11.96\frac{m}{s}$

b).

v is the twice of part a so

$v=2*11.96$

$v=23.9\frac{m}{s}$

$R_{m}-m_{m}*g-m_{p}=m*\frac{v^2}{r}$

Solve to Rm

$R_{m}=m*\frac{v^2}{r}+(m_{m}+m_{p})*g$

$R_{m}=m*\frac{23.9^2}{14.6}+(40+70)*9.8$

$R_{m}=2646 N$

5 0

2 years ago

1. If we place a box of 200 Kg on incline plane of 30 degree, does the weight of box and normal force are equal and opposite to

AfilCa [17]

Answer:

The weight on the box and the normal force on the box will not be exactly opposite to one another. The angle between these two forces will be $150^\circ$ .

If the box isn't moving:

$\begin{aligned}& \text{normal force} \\ &= \cos\left(30^\circ\right) \cdot \text{weight} \\ &= \frac{\sqrt{3}}{2} \cdot \text{weight}\end{aligned}$ .

Explanation:

The weight on an object on the surface of the earth should always point downwards towards the center of the planet.

On the other hand, as the name suggest, the normal force on an object is normal to the surface that exerted this normal force. In this question, that surface is a plane with a $30^\circ$ incline to the ground. The corresponding normal force will be at $\left(90^\circ - 30^\circ\right) = 60^\circ$ above the ground.

The angle between the normal force ( $60^\circ$ above the ground) and the weight (vertically downwards toward the ground.) would be $\left(60^\circ + 90^\circ\right) = 150^\circ$ .

How could the box possibly not move even though the normal force on it isn't exactly opposite to its weight? Refer to the second diagram (not to scale) attached; decompose the weight on the box into two components:

The first component is normal to the incline.
The second component is parallel to the incline.

Notice how the original weight and the two decomposed forces form a right triangle.

Component normal to the incline: $\displaystyle \left(\cos\left(30^\circ \right)\right) \cdot \text{weight} = \frac{\sqrt{3}}{2}\cdot \text{weight}$ .

Component parallel to the incline: $\displaystyle \left(\sin\left(30^\circ \right)\right) \cdot \text{weight} = \frac{1}{2}\cdot \text{weight}$ .

Notice that the vector sum of these two components is equal to the downward weight before the decomposition.

The normal force would be opposite to the component of the weight that is normal to the incline.The two forces should be equal in size. Therefore, the magnitude of the normal force should also be $\displaystyle \left(\left.\sqrt{3}\right/ 2\right)\cdot \text{weight}$ .

If the box is at equilibrium, then the friction on this box (parallel to the incline) should be equal in size to the component of weight that is parallel to the incline.

7 0

2 years ago