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A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0

3 years ago

The reaction mixture represented above is at equilibrium at 298 K. The value of the equilibrium constant for the reaction is 16

andre [41]

Answer : The equilibrium concentration of T(g) is 0.5 M

Solution :

Let us assume that the equilibrium reaction be:

The given equilibrium reaction is,

$R(g)+2T(g)\rightleftharpoons 2X(g)+Z(g)$

The expression of $K_c$ will be,

$K_c=\frac{[Z][X]^2}{[R][T]^2}$

where,

$K_c$ = equilibrium constant = 16

[Z] = concentration of Z at equilibrium = 2.0 M

[R] = concentration of R at equilibrium = 2.0 M

[X] = concentration of X at equilibrium = 2.0 M

[T] = concentration of T at equilibrium = ?

Now put all the given values in the above expression, we get:

$16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}$

$[T]=0.5M$

Therefore, the equilibrium concentration of T(g) is 0.5 M

8 0

4 years ago

If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object

777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0

3 years ago

A sports car of mass 1000 kg can accelerate from rest to 27 m/s in 7.0 s. what is the average forward force on the car?

Marysya12 [62]

The average force on the car=3857 N

Explanation:

mass of car= 1000 kg

initial velocity= Vi=0

Final velocity= Vf=27 m/s

time= t= 7 s

using the kinematic equation,

Vf= Vi + at

27=0+a (7)

a=3.857 m/s²

Now the force is given by F= m a

F=1000 (3.857)

F=3857 N

8 0

4 years ago

Looking straight downward into a rain puddle whose surface is covered with a thin ﬁlm of gasoline. you notice a swirlingpattern

Helen [10]

Answer:

Explanation:

In case of colored film we apply the theory of interference.

Condition that colored fringe is formed ( constructive interference)

2μt = λ/ 2 ( minimum order)

t = λ/ 4μ

= 540 / 4 x 1.38

= 97.83 nm

4 0

3 years ago