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2 years ago

The binding energy of helium -4 is 28 MeV. a. What is the mass of a helium nucleus (round to 5 decimal places)?

Physics

1 answer:

lord [1]2 years ago

7 0

Answer:

4.001805 amu

Explanation:

931 Mev = 1 amu

28 Mev = 28 / 931 Mev

= .030075 amu

Helium nucleus consists of 2 proton and 2 neutron .

Binding energy in terms of amu

= total mass of 2 neutron and 2 proton - mass of nucleus of helium

mass of nucleus of helium = total mass of 2 neutron and 2 proton -Binding energy in terms of amu

mass of 2 neutron = 2 x 1.008664 = 2.017328

mass of 2 proton = 2 x 1.007276 = 2.014552

total = 4.03188

So mass of helium nucleus = 4.03188 - .030075

= 4.001805 amu.

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Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

5 0

2 years ago

Read 2 more answers

A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner

sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0

2 years ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at

Olegator [25]

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to $1 × 10^{-12}$ meter

1 micrometer is equivalent to $1 × 10^{-6}$ meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by $174 × 10^{-12}$ meter we get that the number of sheets will be as follows