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3 years ago

An electron is emmited by an atomic nucleus in the process of...

Physics

2 answers:

Sliva [168]3 years ago

8 0

I’m pretty sure it’s Radioactive decay

harkovskaia [24]3 years ago

5 0

I think its some sort of decay

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A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee = 5kg

Final speed = 12m/s

Unknown:

Impulse of the frisbee = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

Impulse = mass (Final velocity - Initial velocity)

Impulse = 5(12 - 0) = 60kgm/s

3 0

3 years ago

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

1,230N

Explanation:

1. Name of the variables:

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

2. Formulae:

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

3. Solution (calculations)

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago

A pump is used to transport water to a higher reservoir. if the water temperature is 15ºc, determine the lowest pressure that ca

Reika [66]

Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation.

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at $15^{\circ}$ is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.

7 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

How is coulomb's law similar to newton's law of gravity?

Umnica [9.8K]

Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.

4 0

3 years ago