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AleksAgata [21]

3 years ago

8

Express the following in scientific notation:

1 answer:

DENIUS [597]3 years ago

5 0

Hello,

Here is your answer:

Here are the answers to your questions:

1. 6,000,000,000,000,000,000,000=.6×10^22

2. 547,000=5.47×10^5

3. 3,507,000,000=3.507×10^12

4. 0.0004650=4.650×10^-6

If you need anymore help feel free to ask me!

Hope this helps!

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A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 mm. The sphere initially had no net ma

grigory [225]

Answer:

$\omega=4.2704*10^-^5 rad/s$

Explanation:

Angular Momentum Formula For atoms= $L_{atom}=0.12Nm_{s}h$

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms= $\frac{N_{A}*m}{M}$

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be= $L_{atom}=0.12\frac{N_{A}m}{M} m_{s}h$

Angular Momentum of Sphere= $L_{sphere}=I\omega$

where:

$I=\frac{2mR^{2}}{5}$

So,Angular Momentum of Sphere= $L_{sphere}=\frac{2mR^{2}}{5}\omega$

Angular Momentum of sphere=Angular Momentum of atoms

$L_{sphere}$ = $L_{atom}$

$\frac{2mR^{2}}{5}\omega$ = $0.12\frac{N_{A}m}{M} m_{s}h$

For iron, $m_s =\frac{1}{2}$ . So above equation will become:

$\omega=\frac{0.12*5*N_{A}h}{4*M*R^{2} }$

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

$\omega=\frac{0.12*5*(6.022*10^{23})(6.63*10^{-34}/2*\pi)}{4*0.0558*(2*10^{-3})^{2}}$

$\omega=4.2704*10^-^5 rad/s$

5 0

3 years ago

A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?

ivolga24 [154]

Answer:

Explanation:

Given

mass of rock $m=40\ kg$

Elevation of Rock $h=10\ m$

Distance traveled by rock with time

$h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

$5=0\times t+\frac{1}{2}(9.8)(t^2)$

$t^2=\frac{10}{9.8}$

$t=1.004\approx 1\ s$

velocity at this time

$v=u+at$

$v=0+9.8\times 1.004$

$v=9.83\ m/s$

6 0

3 years ago

Which are considered noble gases

sladkih [1.3K]

Answer:Argon

Neon

Helium

Krypton

Xenon

4 0

3 years ago

Read 2 more answers

Electricity questions

egoroff_w [7]

Answer:

What Electricity questions?

Explanation:

5 0

3 years ago

A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?

seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0

3 years ago