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nalin [4]
2 years ago
7

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

uncertainty in the electron's momentum
Physics
1 answer:
Alina [70]2 years ago
4 0

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
  • Planck's constant (h) = 6.26 × 10⁻³⁴ Js

\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

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jeka94

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = \frac{m}{\frac{I_o}{r}  +mr} \ r v_o

let's calculate

       v = \frac{0.020}{\frac{1.4}{0.6  } + 0.020 \ 0.6  } \ 0.6 \ 4

       v = \frac{0.020}{2.345} \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

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Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

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And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
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