The bulk modulus of water is 2.2 x 109 Pa, and its density is 1.0 x 103 kg/m3. What is the speed of sound in water?

A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street

Gemiola [76]

Answer:

doppler shift's formula for source and receiver moving away from each other:

λ'=λ°√(1+β/1-β)

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

λ'=λ°√(1+β/1-β)

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

β=0.17

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

v=v₀+at

where v₀=0

so v=at

as we want to calculate t so:-

t=v/a v=0.17c ,c=3x10⁸ ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

t=1.73x10⁶

7 0

3 years ago

In an elastic collision, _______ energy is conserved.

Nonamiya [84]

Answer:

Kinetic energy and momentum are conserved.

Explanation:

4 0

2 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

What were some benefits to the american television from analog broadcast to digital broadcast?

Elanso [62]

Answer:

nothing no no no

Explanation:

nothing cause no

8 0

3 years ago

Read 2 more answers

What if the height is 85m, then how long long does it stay in the air?

wolverine [178]

Explanation:

hamburger cheesecake

6 0

3 years ago