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lidiya [134]
3 years ago
14

According to newton's second law of motion, how much force will be required to accelerate an object at the same rate if its mass

is reduced by half?
A.No change in the amount of force is needed.
B.Four times the amount of force is needed.
C.Twice the amount of force will be required.
D.Half the amount of force will be required.
Physics
1 answer:
guajiro [1.7K]3 years ago
5 0

<u>D.</u>

For every action ther is an equal and opposite reaction.

equal in force, opposite in direction

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Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

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3 years ago
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

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The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

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2 years ago
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

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de Broglie wavelength = 6.0 nm

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Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

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3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

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Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

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If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

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Importance of electric circuit​
Feliz [49]

Answer:

It is important because it carries useful energy through your house that you can use for a variety of tasks.

Explanation:

Hope this helped !

8 0
3 years ago
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