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3 years ago

Which statement about an atom is correct?

Physics

1 answer:

Alexus [3.1K]3 years ago

7 0

Answer:

$\boxed{ \sf{The \: electron \: has \: a \: negative \: charge \: and \: is \: found \: outside \: of \: the \: nucleus}}$

Option A is correct

Explanation:

Let's check the options:

A: The electron has a negative charge and is found outside of the nucleus.

$\mapsto{}$ Yeah! It's TRUE . An electron is a negatively charged particle and is located outside the nucleus of an atom.

B : The neutron has a negative charge and is found in the nucleus.

$\mapsto{}$ No! It's FALSE . A neutron carries no charge. i.e it is a neutral particle and found inside the nucleus.

C : The proton has no charge and is found in the nucleus.

$\mapsto{}$ No! It's FALSE.A proton is a positively charged particle present inside nucleus of an atom.

D : The neutron has no charge and is found outside of the nucleus.

$\mapsto{}$ I agree that the neutron has no charge. But it is found inside the nucleus not outside . So, this statement is FALSE .

Hence, we found our answer! :D

A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.

Hope I helped!

Best regards! :D

~ $\text{TheAnimeGirl}$

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A 0.3-m-radius automobile tire rotates how many revolutions after starting from rest and accelerating at a constant 2.13 rad/s2

Aloiza [94]

Answer:

The automobile tire rotates 91 revolutions

Explanation:

Given;

angular acceleration of the automobile, α = 2.13 rad/s²

time interval, t = 23.2-s

To calculate the number of revolutions, we apply the first kinematic equation;

$\theta = \omega_i \ + \frac{1}{2} \alpha t^2$

the initial angular velocity is zero,

$\theta =0\ + \frac{1}{2} (2.13) (23.2)^2\\\\\theta = 573.2256 \ Rad$

Find how many revolutions that are in 573.2256 Rad

$N = \frac{\theta}{2 \pi} = \frac{573.2256}{2\pi} \\\\N = 91 \ revolutions$

Therefore, the automobile tire rotates 91 revolutions

3 0

4 years ago

What is the frequency of a wave travelling at 45.45 m/s with a wavelength of 2 m?

HACTEHA [7]

Answer:

22.73 Hz

Explanation:

Frequency = 45.45/2 = 22.73 Hz

4 0

3 years ago

A proton initially at rest is accelerated by a uniform electric field. The proton moves 4.76 cm in 1.10 x 10^-6 s. Find the volt

iVinArrow [24]

Answer:

The voltage drop through which the proton moves is 39.1 V.

Explanation:

Given that,

Distance = 4.76 cm

Time $t=1.10\times10^{-6}\ s$

We need to calculate the acceleration

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

Where, s = distance

a = acceleration

t = time

Put the value in the equation

$4.76\times10^{-2}=\dfrac{1}{2}\times a \times(1.10\times10^{-6})^2$

$a=\dfrac{2\times 4.76\times10^{-2}}{(1.10\times10^{-6})^2}$

$a=7.87\times10^{10}\ m/s^2$

We need to calculate the voltage drop

Using formula of electric field

$F=qE$

$F = q\dfrac{V}{d}$ ....(I)

Using newton's second law

$F = ma$ ....(II)

Put the value of F in equation (I) from equation (II)

$ma=\dfrac{qV}{d}$

$V=\dfrac{mad}{q}$

Where, q = charge

a = acceleration

d = distance

m= mass of proton

Put the value into the formula

$V=\dfrac{1.67\times10^{-27}\times7.87\times10^{10}\times4.76\times10^{-2}}{1.6\times10^{-19}}$

$V=39.1\ V$

Hence, The voltage drop through which the proton moves is 39.1 V.

8 0

3 years ago

A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of

Leto [7]

Answer: $2.27\ m/s^2$

Explanation:

Given

Length of the race track $L=200\ m$

the radius of curvature of the track $r=29.5\ m$

time taken to run on track is $t=24.4\ s$

Speed of runner is

$\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s$

Centripetal acceleration is

$\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2$

5 0

3 years ago

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

sweet-ann [11.9K]

Answer:

See attachment for complete solving.

Explanation:

Given that:The brief pulses of energy absorbed by the retina welds the detached portion back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 μm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.A) If the laser pulses are each 1.5 ms, long, how much energy is delivered to the retina with each pulse?B) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot?C) What is the wavelength of the laser light inside the vitreous humor of the eye?D) What is the frequency of the laser light inside the vitreous humor of the eye?E) What is the maximum value of the electric field in the laser beam?F) What is the maximum value of the magnetic field in the laser beam.

See attachment for further solving.

4 0

3 years ago

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