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2 years ago

Find the energy released in the alpha decay⁹²²³⁸U → ⁹⁰²³⁴Th+²4He

Physics

1 answer:

KonstantinChe [14]2 years ago

7 0

The energy released in the alpha decay⁹²²³⁸U → ⁹⁰²³⁴Th+²4He is 4.25MeV.

<h3>Solution:</h3>

Q =(M U −M TH −M He ) $c^{2}$

=(238.05079−234.04363−4.00260)u× $c^{2}$

=(0.00456u)×c 2

=0.00456×( 931−MeV/ $c^{2}$ ). $c^{2}$ = 4.25MeV

Yes, the decay is spontaneous (since Q is positive).

To learn more about alpha decay, refer

brainly.com/question/1898040

#SPJ4

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A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. If the norma

Zina [86]

<u>Answer:</u>

The force F applied to the handle = 330.03 N

<u>Explanation:</u>

The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have

Horizontal component of force = F cos θ

Vertical component of force = F sin θ

In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.

So, F sin 29 = 160

F = 330.03 N

The force F applied to the handle = 330.03 N

6 0

4 years ago

Read 2 more answers

Can anyone answer this please?

Umnica [9.8K]

B is the correct answer according to my calculations.

5 0

3 years ago

Read 2 more answers

(b) Figure 3.32 shows the orbit of a planet around the Sun. Compare the linear speed of the planet at positions X, Y and Z.

Viefleur [7K]

At Z ... slowest speed

At Y ... fastest speed

At X ... medium speed

Wherever it is in its orbit, the line from the planet to the Sun smears over the same amount of area every second.

That's Kepler's second law of planetary motion.

The reason this happens is: That's how gravity works. (A better explanation is available, but first you have to be able to twirl calculus and solid geometry in the air on long sticks.)

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4 years ago

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

8 0

4 years ago

1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee

Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

7 0

4 years ago