A boater is floating 1060 meters down a river. The person takes 40 minutes. At what speed is the boat moving?

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is

dimaraw [331]

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

$\:V\:=\:\frac{W}{q}$

substituting W = 3 and q = 10 in the formula

$V=\frac{3}{10}$

$V=0.3$ V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

4 0

3 years ago

Violet light (410 nm) and red light

Leokris [45]

Answer:10.03

Explanation:acellus

7 0

3 years ago

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic

shepuryov [24]

Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

$F=m\frac{v^2}{r}$ (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$

6 0

3 years ago

A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

3 years ago