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4 years ago

How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.

Physics

1 answer:

pav-90 [236]4 years ago

8 0

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

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Scientists have organized all forms of energy into two types—kinetic and potential. Kinetic energy is the energy of motion, and

mihalych1998 [28]

Answer:

Potential energy only

Explanation:

at the top of its swing the pendulum stops moving , (therefore it has no KINETIC energy) thus all of the energy is stored as potential energy.

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3 years ago

Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R

jolli1 [7]

Answer:

$f=\frac{1}{2\pi \sqrt{LC}}$

Explanation:

We know that impedance of a RLC circuit is given by $Z=R+J(X_L-X_C)$

So $Z=\sqrt{R^2+(X_L-X_C)^2}$ here R is resistance $X_L$ is inductive reactance and $X_C$ is capacitive reactance

To minimize the impedance $X_L-X_C$ should be zero we know that $X_L=\omega L\ and \ X_C=\frac{1}{\omega C}$

So $\omega L-\frac{1}{\omega C}=0$

$\omega ^2=\frac{1}{LC}$

$\omega =\sqrt{\frac{1}{LC}}$

We know that $\omega =2\pi f$

So $\omega =2\pi f=\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

Where f is resonance frequency

8 0

3 years ago

A wave with a large amplitude has a lot of a.vibration b.speed c.energy

adoni [48]

<span>A wave with a large amplitude has a lot of a.vibration b.speed <u> c.energy</u></span>

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3 years ago

Read 2 more answers

Define liquid pressure

Stella [2.4K]

The pressure of a liquid on the surface of its container or on the surface of any body in the liquid is equal to the weight of a column of the liquid whose height equals the depth of the liquid at that certain point.

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3 years ago

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

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3 years ago