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3 years ago

Which is true about carbon-12 compared with carbon-13

1 answer:

7 0

E mass number of any given atom depends solely on the number of protons and neutrons in its nucleus. The mass number of any atom can be determined by adding the number of protons and neutrons. (Mathematically this is stated as Mass Number = Protons + Neutrons). For instance, a Carbon atom with 6 protons and 6 neutrons will have a mass number of 12AMU. However, a Carbon atom with 6 protons and 8 Neutrons will have a mass number of 14AMU. They are both Carbon atoms, however they each have a different mass number. Atoms of the same element that have different numbers of neutrons, and therefore, different mass numbers, are called isotopes. Isotopic symbols are used to indicate isotopes of the same elements. In the following isotopic symbols the lower number is the atomic number…it is the number of protons. The upper number is the mass number, it represents to sum of the protons and neutrons in the atoms nucleus.

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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

A 0.035 bullet strikes a 5kg wooden block that is stationary and gets stuck inside the block. The block and bullet move together

hjlf

Answer:

1240 m/s

Explanation:

[Do you mean 0.035 kg bullet?]

Since the momentum of the system is conserved,

Let u m/s be the initial velocity of the bullet

0.035u = 8.6(5+0.035)

u ~= 1237.1714 m/s

u = 1240 m/s (3 sig. fig.)

7 0

3 years ago

How many aluminium ions are needed to cross an ammeter if you want a current of 3A?

Reptile [31]

Assuming that the time needed to cross the ammeter is 1 hour or 3600s, then using coloumb's law,
I x t = F x m / (eq. wt)

Using the given values and formula:
3 x 3600 = 96500 x m / (27/3)

Solving for m,
m = 1.007 g of Al3+

6 0

4 years ago

The electric potential in a region of uniform electric field is -1600 V at x = -1.30 m and + 2000 V at x =+ 1.80 m . What is Ex?

Inga [223]

Answer:

-1161.29 V/m

Explanation:

The x component of electric field will be given by

$E_x= -\frac {\triangle V}{\triangle x}$

Here $\triangle V$ is the change in voltage while triangle x is the change in x

$E_x= -\frac {+2000 V--1600V}{\triangle 1.8--1.3}=-\frac {3600}{3.1}=-1161.29 V/m$

6 0

3 years ago

A pumpkin with a mass of 3.3 kg is launched from a catapult at an initial height of 3.8 m off the ground, with an initial speed

dimaraw [331]

Answer:

11.2m

Explanation:

Suppose the pumpkin is launched vertically and the speed of 13.9 m/s is absolutely vertical. As the pumpkin rises up, its kinetic energy is converted to potential energy, let the reference point be the round, we can create the following equation from the law of energy conservation:

$E_1 + P_1 = E_2 + P_2$