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3 years ago

6

How are different types of radiation arranged along the electromagnetic spectrum

1 answer:

Leona [35]3 years ago

7 0

In order of increasing frequency: Radio, Microwaves, Infrared, Visible light, Ultra-violet, X-rays, Gamma rays. To remember this try:
Rabbits
Mates
In
Very
Unusual
e(X)pensive
Gardens

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The maximum oxygen uptake is known as the

grandymaker [24]

<span>The maximum oxygen uptake is known as the VO max.</span>

3 0

4 years ago

A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁ = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Insert parameters and solve for v;

5 x 1 + 15 x 0 = v (5 + 15 )

5 = 20v

v = $\frac{5}{20}$ = 0.25m/s

5 0

3 years ago

What's the difference between the speed and velocity of an object?

Artemon [7]

Speed is just how fast the object is moving, while velocity is the speed and <em>direction </em>of the object.

edit* Direction would be something such as North, East, etc.

4 0

3 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

I have an object that I want to know when it was made. I calculate that the object contains 0.12g of Iodine-131 but started with

Sav [38]

Answer: 71.72 days

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=0.12 g$ is the final amount of Iodine-131

$A_{o}=60 g$ is the initial amount of Iodine-131

$t$ is the time elapsed

$h=8 days$ is the half life of Iodine-131

Knowing this, let's substitute the values and find $t$ from (1):

$0.12 g=(60 g)2^{\frac{-t}{8 days}}$ (2)

$\frac{0.12 g}{60g}=2^{\frac{-t}{8 days}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.12 g}{60g})=ln(2^{\frac{-t}{8 days}})$ (4)

$-6.21=-\frac{t}{8 days}ln(2)$ (5)

Finding $t$ :

$t=71.72 days$

7 0

4 years ago