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3 years ago

No links please, links never work, this is science not physics

Physics

1 answer:

stiks02 [169]3 years ago

7 0

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

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Most of the compunds that make up organisms contain __

8090 [49]

All living creatures contain carbon

6 0

3 years ago

Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What

IgorLugansk [536]

Answer:

Frequency is $213.04\ s^{-1}.$

Explanation:

Distance between source 1 from the receiver , $S_1 =\sqrt{10^2+22^2}=24.17\ m.$

Distance between source 2 from the receiver , $S_2=\sqrt{5^2+22^2}=22.56\ m.$

Now ,

Path difference , $r = S_1-S_2=24.17-22.56=1.61\ m.$

We know, for constructive interference path difference should be integral multiple of wavelength .

Therefore, $r=n\times \lambda$

It is given that n = 1,

Therefore, $\lambda=1.61\ m.$

Frequency can be found by , $\nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}= 213.04\ s^{-1} .$

Hence, this is the required solution.

5 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

$f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f$

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

$f=f_{n+1}-f_n=480Hz-400Hz=80Hz$

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

$f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}$

So the tension is:

$T=\mu(2Lf)^2$

Which for our values is:

$T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N$

6 0

3 years ago

Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago