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3 years ago

Does any part of our body vibrate when we speak ? Name the part

Physics

1 answer:

Nadusha1986 [10]3 years ago

7 0

When YOU speak, YOUR vocal cords vibrate, sending sound waves through the air and causing MY ear drums to vibrate.

When I speak, MY vocal cords vibrate, sending sound waves through the air an causing YOUR ear drums to vibrate.

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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

If element X has 99 protons, how many elctrons does it have?

slega [8]

It has 99 electrons because an element has the same number of protons and electrons

7 0

3 years ago

Read 2 more answers

A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.

Alex Ar [27]

Good. You can do some very interesting experiments with that equipment.

3 0

3 years ago

Which colors allowed the most electricity to be generated

Naddik [55]

Yellow and red hope that helped

3 0

3 years ago

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$

7 0

3 years ago