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4 years ago

5

Hey can anyone help me out in dis pls

1 answer:

MissTica4 years ago

4 0

Answer: 1. Liquid, molecules aren't packed very closely, and take the shape of the bottom of the container 2. Gas molecules are far apart and fill the container completely

Explanation:

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A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently

mel-nik [20]

Answer:

$R=0.023m$

Explanation:

From the question we are told that:

Mass $m=1.16*10^{-26}$

Potential difference $V=523V$

Magnitude $m=0.370 T$

Generally the equation for Velocity is mathematically given by

$\frac{1}{2}mv^2=ev$

$v=\frac{2ev}{m}$

$v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}$

$v=12.22*10^4m/s$

Generally the equation for Force is mathematically given by

$F=qvBsin \theta$

Where

$qVB=m\frac{v^2}{R}$

$F=m\frac{v^2}{R}sin\theta$

Therefore

$R=\frac{mv}{qB sin \theta}$

$R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}$

$R=0.023m$

8 0

3 years ago

Calculate the frequencies corresponding to the wavelengths 500.00 nm and 500.10 nm. Use these to check the accuracy of equation

mel-nik [20]

Explanation:

It is given that,

Wavelength, $\lambda_1=500\ nm=5\times 10^{-7}\ m$

Wavelength, $\lambda_2=500.10\ nm=5.001\times 10^{-7}\ m$

We need to find the frequencies from corresponding wavelengths. The frequency of the light is given by :

$f=\dfrac{c}{\lambda}$

c is the speed of light

Frequency 1,

$f_1=\dfrac{c}{\lambda_1}$

$f_1=\dfrac{3\times 10^8\ m/s}{5\times 10^{-7}\ m}$

$f_1=6\times 10^{14}\ Hz$

Frequency 2,

$f_2=\dfrac{c}{\lambda_2}$

$f_1=\dfrac{3\times 10^8\ m/s}{5.001\times 10^{-7}\ m}$

$f_1=5.99\times 10^{14}\ Hz$

Hence, this is the required solution.

3 0

3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago

Which circuit would have the most electrical power?

myrzilka [38]

Answer:

c

Explanation:

3 0

3 years ago

Read 2 more answers

State the objects in the universe that reflect light from stars

stepan [7]

Everything in the universe that is not a star reflects light from Stars. Otherwise you can't see it.

7 0

4 years ago