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Mazyrski [523]
3 years ago
13

Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s

ignificant figures.
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
You might be interested in
1 Point
irakobra [83]

Answer:

SECOND LAW OF NEWTON

Explanation:

When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

                I = F t = Δp

As we can see this expression is a form of Newton's second law

           F = m a

           a = dv / dt

           F = m dv / dt

           F dt = m dv

           p = mv

           F dt = dp

Therefore the station moves through the SECOND LAW OF NEWTON

7 0
2 years ago
A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a
Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

v_x = \frac{x}{t}

v_x = \frac{67}{4.5}

v_x = 14.89m/s

The vertical component of velocity is

-h = v_y t -\frac{1}{2} gt^2

Here,

h = Height

g = Gravitational acceleration

t = Time

v_y = Vertical component of velocity

-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2

-1.23= 4.5v_y - 99.225

v_y = 21.77m/s

The direction of the velocity will be given by the tangent of the components, then

tan\theta = \frac{v_y}{v_x}

\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{14.89^2 +21.77^2}

|V| = 26.37m/s

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0
2 years ago
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
2 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
Describe 2 ways that mass and weight are the same
navik [9.2K]
Anything that has mass has weight and anything that has weight has mass simple.
7 0
3 years ago
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