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marysya [2.9K]
3 years ago
5

This solid layer of the earth is made of mostly iron and nickel.

Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer

Explanation:

Yes, it's true that the solid layer of the earth is known as the most dense part as it is made up of the heavy metals like iron and nickel. Inner part is the hotter part due to the high pressure and temperature. It has the temperature of about 5,200°C and the pressure of 3.6 million atm but still the iron and nickel are present there in the solid form as they withstand such high temperature and pressure values.

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What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength
ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

\lambda  = \frac{h}{\sqrt{2 m K}}

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}

λ = 1.2826 x 10^-13 m

6 0
2 years ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
2 years ago
The hardest known natural material is _____. <br> jade<br> quartz<br> diamond<br> talc
kakasveta [241]
Its diamond i am 100% right!

8 0
3 years ago
Read 2 more answers
Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller element
Zolol [24]

Answer:

<em>Element C will be best for a nuclear fission reaction</em>

Explanation:

<em>Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process.</em> For the reaction to be continuous in a chain reaction,<em> the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission</em>. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

5 0
3 years ago
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

\theta=274rev

Explanation:

From the question we are told that:

Angular velocity \omega=510rpm

Mass m=40.kg

Diameter d 75=>0.75m

Off Time t=40.0s

Oscillation at Power off N=210

Generally the equation for Angular displacement is mathematically given by

 \theta_{\infty}=\frac{w+w_0}{t}t

 w=\frac{2*\theta_{\infty}}{t}-w_0

 w=\frac{28210}{40*(\frac{1}{60})}-510

 w=120rpm

Generally the equation for Time to come to rest is mathematically given by

 t=(\frac{\omega_0}{\omega_0-\omega})t

 t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})

 t=0.87min

Therefore Angular displacement is

 \theta =(\frac{120+510}{2})0.87

 \theta=274rev

6 0
2 years ago
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