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RideAnS [48]
3 years ago
8

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

4×10-27 kg. What is the kinetic energy of an α particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects.
Physics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

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Ivanshal [37]

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

<u>AT t = 0 s</u>

α(0) = 2(1.5)(0)

<u>α(0) = 0 rad/s²</u>

<u></u>

<u>AT t = 5 s</u>

α(5) = 2(1.5)(5)

<u>α(5) = 15 rad/s²</u>

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3 years ago
Which statement is correct?
miv72 [106K]
'A' is correct. B, C, and D are false statements.
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A 1.0 kg rock is thrown straight upward with an initial speed of 8.0 m/s. What is its speed
Ronch [10]

Answer:5.7m/s

Explanation:

Mass=1kg

Initial velocity=u=8m/s

height=h=1.6m

Final velocity =v

Acceleration due to gravity=g=9.8m/s^2

v^2=u^2-2xgxh

v^2=8^2-2x9.8x1.6

v^2=8x8-2x9.8x1.6

v^2=64-31.36

v^2=32.64

Take the square root of both sides

√(v^2)=√(32.64)

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Speed at the height of 1.6m is 5.7m/s

8 0
3 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
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