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3 years ago

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A positron and an electron annihilate each other upon colliding, thereby producing energy in the form of two gamma rays. Assumin

g that both gamma rays have the same energy (since both particles have the same mass), calculate the wavelength of the electromagnetic radiation used in pm. (1 pm = 10⁻¹² m; mass of electron (amu) = 0.000549)

1 answer:

kolbaska11 [484]3 years ago

6 0

Answer:

2.42631E-13m

Explanation:

First we find the mass defect

Which is m= 0-2(9.10939E-33kg)

= - 1.82188E-30kg

Now find the energy

S

E= mc²=( -18.82188E-30)(2.999792E8)²

= 1.63742E-13J

Thus energy per photon will be

1.63742E-13J/2= 8.18710E-14J

So wavelength is given as

Lambda= hc/E

= (6.62608E-34)(2.997E8)/8.18710J

= 2.42631E-13m

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Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is hi

Solnce55 [7]

Answer:

The weight at a distance 2 RE from surface of earth is <em>W/9</em>

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)² ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

<u>Wh = W/9</u>

6 0

3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

4 years ago

Read 2 more answers

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a force of 12.344 N

at distance r

and we know Electrostatic force is given

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{kQ\cdot Q}{r^2}$

$F=\frac{kQ^2}{r^2}$

Now the magnitude of charge is 2Q and is at a distance of $\frac{r}{2}$

$F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}$

F'=16F

F'=197.504 N

4 0

3 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

Propose an experiment that could be used to demonstrate that friction is a contact force?

vesna_86 [32]

Maybe push or pull an object with a large amount of mass? you are force a (pushing through object) aka making contact. i hope i helped not good with physics :)

3 0

3 years ago

Read 2 more answers