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ira [324]
2 years ago
14

What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2

=1.49×10−10j/u?
Physics
1 answer:
kkurt [141]2 years ago
6 0

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

\Delta E = c^2 \Delta m

where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

m(^4_2 He)=4.002603u is the mass of one nucleus of helium

So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

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tatyana61 [14]

Answer:your gay

Explanation:

gay

7 0
2 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

4 0
2 years ago
A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to
Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

0=y_{0}+\frac{1}{2}at^{2}

and solve for t:

t=\sqrt{\frac{-2y_0}{a}}

now we can substitute the values:

t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the  distance from the ship to point A to be:

x_{A}=V_{x}t

x_{A}=(100m/s)(3.19)

x_{A}=319m

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}

So we can substitute the given values:

x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}

Which yields:

x_{f}=329.18m

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0
2 years ago
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
What landforms form at convergent boundaries where two oceanic plates collide?
Bad White [126]
It forms mountains and sometimes islands.
4 0
2 years ago
Read 2 more answers
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