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bagirrra123 [75]
2 years ago
8

A proton of mass m moving with a speed of 3.0 × 106 m/s undergoes a head on elastic collision with an alpha particle of mass 4m

at rest. What are the velocities of the two particles after the collision?
Physics
1 answer:
Sergio039 [100]2 years ago
4 0

Answer:6.0×10^5m/s

Explanation:

According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.

After their collision, the two bodies will move with a common velocity (v)

Momentum = mass × velocity

Let m1 be the mass of the proton = m

Let m2 be the mass of the alpha particle = m2

Let v1 be the velocity of the proton = 3.0×10^6m/s

Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).

Using the law,

m1v1 + m2v2 = (m1 + m2)v

m(3.0×10^6) + 4m(0) = (m + 4m)v

m(3.0×10^6) = 5mv

Canceling 'm' at both sides,

3.0×10^6 = 5v

v = 3.0×10^6/5

The common velocity v = 6.0×10^5m/s

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Alex787 [66]
Hey there!

The answer would be B. The sound moves from air to water.

Sound travels through different mediums. It goes fastest in solids, a little slower in liquids, and slowest in air. Sound is a very fast wave, but remember that mediums can differ that. In a vacuum space, there is no sound at all. (ex. outer space)

Hope this helps !
5 0
2 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0
3 years ago
The acceleration of an object depends on the ____________ exerted on it and it's_________ this is to do with newton's seconed la
MArishka [77]
Force is the first and mass is the second
6 0
3 years ago
Convert 0.700 atm of pressure to its equivalent in millimeters of mercury.
inna [77]

Answer:

532 millimeters of mercury

Explanation:

In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:

1 atm = 760 mm Hg

Therefore, we can solve the problem by setting up the following proportion:

1 atm : 760 mmHg = 0.700 atm : x

Solving for x, we find

x=\frac{(760 mmHg)(0.700 atm)}{1 atm}=532 mmHg

5 0
3 years ago
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