What does an atomic nucleus give off a particle?

Sam is playing football. She kicks the ball with an average force of 75 N.

damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

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3 0

3 years ago

The shifting of the observed wavelength of light due to the motion of the source toward or away from the observer is called the

alexdok [17]

Answer:

doppler effect

Explanation:

When the relative motion of two bodies results in the wavelength becoming shorter this means that the bodies are getting closer. This is known as blue shift.

When the relative motion of two bodies results in the wavelength becoming longer this means that the bodies are getting farther. This is known as red shift.

Collectively this phenomenon is known as the Doppler effect.

7 0

3 years ago

1. Driving at slower speeds than traffic flow _____________ .

r-ruslan [8.4K]

The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge.

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance $d$ due to a static charge having charge $q$ is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$

Here, $E$ is the electric field intensity, $q$ is the amount of charge and $r$ is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge.

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords: Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0

3 years ago

Read 2 more answers

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

Photons have momentum, this was proved by he American physicist Arthur H. Compton after his experiments related to the scattering of photons from electrons (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

If a passenger train is traveling on a straight track with a negative velocity and a positive acceleration, is it speeding up or

lana66690 [7]

Answer:

Slowing down

Explanation:

Since a train with negative velocity and positive acceleration, this means its acceleration is in opposite direction with the velocity. As time progress, this opposite direction would decrease the velocity magnitude, making the train slowing down.

3 0

3 years ago