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3 years ago

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Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy

of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.

1 answer:

sashaice [31]3 years ago

4 0

Answer:

The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons

Explanation:

Binding energy can be given by the formula:

Binding energy = Binding energy per nucleon * Number of nucleons

This means that if the binding energy per nucleon = x MeV

Where x is a positive real number

The nucleus with 60 nucleons will have Binding energy = 60x MeV

The nucleus with 200 nucleons will have binding energy = 200x MeV

For a +ve x, 200x > 60x

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As you sit in a fishing boat, you noticed that 22 waves pass the boat every 60 seconds . If the distance from one crest to the n

Zinaida [17]

Answer:

0.1835m/s

Explanation:

The formula for calculating the speed of wave is expressed as;

v = fλ

f is the frequency - The number of oscillations completed in one seconds

If 22 waves pass the boat every 60 seconds,

number of wave that passes in 1 seconds = 22/60 = 0.367 waves

Therefore the frequency f of the wave is 0.367Hertz

λ (wavelength) is the distance between successive crest and trough of a wave

λ = 0.5m

Substitute the given values into the formula

v = fλ

v = 0.367 * 0.5

v = 0.1835

Hence the speed of the waves is 0.1835m/s

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2 years ago

What time period is jellyfish first appear

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According to scientists, the first fossil of a jellyfish first appeared approximately 500 million years ago. Or to be more exact, the Cambrian Period.

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3 years ago

24. Use Newton’s first law of motion to explain how wearing a seat belt in a moving car would help prevent injury (3 points)

barxatty [35]

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3 years ago

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

So

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

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2 years ago

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high

Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$y$ - Final height, measured in meters.

$y_{o}$ - Initial height, measured in meters.

$t$ - Time, measured in seconds.

$g$ - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

$y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}$

Given that $y = 0\,m$ , $v_{o} = 15\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 16\,s$ , the final height of the package is:

$y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}$

$y_{o} = 1015.296\,m$

The package was released at a height of 1015.296 meters.

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3 years ago