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2 years ago

14

How many protons, electrons and nurturing does krypton have NEED HELP ASAP THANK YOU

2 answers:

Assoli18 [71]2 years ago

8 0

Answer:

Krypton – Mass Number – Neutron Number – Kr 2020-11-21 by Nick Connor Krypton is a chemical element with atomic number 36 which means there are 36 protons and 36 electrons in the atomic structure. The chemical symbol for Krypton is Kr.

Atomic Number: 36

Element: Krypton

Element Category: Noble Gas

Symbol: Kr

Explanation:

LiRa [457]2 years ago

7 0

Answer:

your answer is in an image here

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3 years ago

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3 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0

3 years ago

A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

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

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

3 years ago

It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a

NARA [144]

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .

8 0

3 years ago