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2 years ago

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C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-

14 has two more neutrons. C) C-14 has a larger atomic radius. D) C-14 has two additional valence electrons. ASAP

2 answers:

GuDViN [60]2 years ago

7 0

Answer: Correct answer is B) C-14 has two more neutrons

Explanation: usa test prep

mylen [45]2 years ago

4 0

Answer:

C-14 has two more neutrons.

Explanation:

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One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

2 years ago

a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help

Bezzdna [24]

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

60m by 240m

The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

The area of the parking lot is 14,400 meters squared.

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3 years ago

Read 2 more answers

Which statement supports Newton’s first law of motion?

Ipatiy [6.2K]

Answer: The statement that supports Newton's first law of motion is the one that says that planets can move at a varying speed due to forces exerted in space

6 0

2 years ago

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Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi

DerKrebs [107]

Answer:

Weight of the woman in Newton

$x = 698.6 \ N$

Mass of the woman in slug

$Mass = 4.86 \ slug$

Mass of the woman in kg

$Mass = 16 \ kg$

My weight in Newton

$W = 784 \ N$

Explanation:

From the question we are told that

The weight of the woman in pounds is $W = 157 \ lb$

Converting to Newton

1 N = 0.22472 lb

x N = 157

=> $x = \frac{157 * 1}{0.22472}$

=> $x = 698.6 \ N$

Obtaining the mass in slug

$Mass = \frac{W}{g}$

Here $g = 32.2 ft/s^2$

So

$Mass = \frac{157 }{32.2}$

$Mass = 4.86 \ slug$

Obtaining the mass in kilogram

$Mass = \frac{W}{g}$

Here $g = 9.8 \ m/s$

So

$Mass = \frac{157 }{9.8}$

$Mass = 16 \ kg$

Generally weight is mathematically represented as

$W = m * g$

Given that my mass is 80 kg then my weight is

$W = 80 *9.8$

$W = 784 \ N$

6 0

2 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0

2 years ago