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AleksAgata [21]
4 years ago
7

Calculate (in MeV) the total binding energy for 40Ar. Express your answer in mega-electron volts to four significant figures.

Physics
1 answer:
AleksAgata [21]4 years ago
4 0

Answer:

299.4 MeV

Explanation:

For 40 Ar

Number of neutrons = 22

Number of protons= 18

Mass of each proton = 1.007277 amu

Mass of each neutron= 1.008665 amu

Total mass of protons= 18 × 1.007277 amu = 18.130986 amu

Total mass of neutrons = 22 × 1.008665 amu = 22.19063 amu

Total mass of protons and neutrons= 18.130986 + 22.19063 = 40.321616 amu

Mass defect = 40.321616 amu - 40 amu

Mass defect = 0.321616 amu

Since 931 is the conversion factor from amu to MeV

Binding energy = 0.321616 amu × 931 = 299.4 MeV

Binding energy = 299.4 MeV

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Answer:

Explanation:

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6 0
4 years ago
A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a
IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}

Put the value into the formula

\dfrac{dM}{dt}=53.3+3.63

\dfrac{dM}{dt}=56.93\ kg/s

We need to calculate the thrust of the jet engine

Using formula of thrust

T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v

Put the value into the formula

T=56.93\times317-53.3\times260

T=4188.81\ N

Hence, The thrust of the jet engine is 4188.81 N.

7 0
4 years ago
student built a simple heat engine that could lift masses. If the heat engine takes in 300 J of heat and loses 50 J to the envir
k0ka [10]

Answer:

So it will lift the mass by h = 17 m

Explanation:

As per energy conservation we know that

Q_1 = W + Q_2

here we know that

Q_1 = 300 J

Q_2 = 50 J

now we have

W = 300 - 50

W = 250 J

so work done by the engine is 250 J

now we have

W = mgh

250 = 1.5 \times 9.81 \times h

h = 17 m

4 0
3 years ago
What is the volume of a marble with a diameter of 3.0
RideAnS [48]

Explanation:

You can solve for volume using radius or diameter.

Sphere Volume   =   4/3 • π • r³     =     ( π •d³)/6

We're given the diameter so let's use that.

Volume = PI * d^3 / 6

Volume = 3.14159 * 3.0^3 / 6

Volume = 3.14159 * 9 / 2

Volume = 14.137 cubic centimeters

7 0
3 years ago
A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is
Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration a_{c} is given by the following equation:  

a_{c}=\frac{V^{2}}{r} (1)

Where:  

a_{c}=17 m/s^{2} is the <u>centripetal acceleration</u>

V is the<u> tangential speed</u>

r=12 m is the <u>radius</u> of the circle

Isolating V from (1):

V=\sqrt{a_{c}r} (2)

V=\sqrt{(17 m/s^{2})(12 m)}

<u />

Finally:

V=14.28 m/s This is the girl's tangential speed

3 0
3 years ago
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