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

Answer:

Initial velocity, u = 60 km/h = 16.7 m/s

Final velocity, v = 72 km/h = 20 m/s

time, t = 2 sec

From first equation of motion:

[tex]{ \bf{v = u + at}}[/tex]

Substitute the variables:

[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]

Answer:

Quantum theory gives the concept of

Answer:

[tex]F=202650N[/tex]

Explanation:

From the question we are told that:

Area [tex]a=50m^2[/tex]

Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]

Generally the equation for Force is mathematically given by

[tex]F=dP*A[/tex]

[tex]F=4053*50[/tex]

[tex]F=202650N[/tex]

The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

W = ∆K = K

where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

W = 1/2 mv ²

where m is the mass of the object and v is the speed you want to find. Solving for v, you get

v = √(2W/m) ≈ 0.46 m/s

Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

′

=7+5+12=24Ω

Now R

1

′

and R

5

′

are in parallel. So,

R

2

′

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

′

=6ohm.

Now R

2

′

,R

1

and R

3

are in series. So,

R=R

2

′

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Answer:

false?

Explanation:

The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.

Answer:

I think the answer is False.

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Answer:

imma try and fail again and again

Answer:

The correct answer is "121.36 meters and 2.40 MHz".

Explanation:

Given:

Red shift,

[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]

Wavelength,

[tex]\lambda = 125 \ meters[/tex]

The observed frequency will be:

⇒ [tex]f = \frac{c}{\lambda}[/tex]

      [tex]=\frac{3\times 10^8}{125}[/tex]

      [tex]=24\times 10^5[/tex]

      [tex]= 2.40\times 10^6[/tex]

      [tex]=2.40 \ MHz[/tex]

hence,

The Emitted wavelength will be:

⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]

   [tex]\frac{125}{\lambda_0}-1 =0.03[/tex]

         [tex]125=1.03 \ \lambda_0[/tex]

           [tex]\lambda_0=\frac{125}{1.03}[/tex]

                [tex]=121.36 \ m[/tex]

Explanation:

All the energy in oil, gas, and coal originally came from the sun captured through photosynthesis. In the same way that we burn wood to release energy that trees capture from the sun, we burn fossil fuels to release the energy that ancient plants captured from the sun.

Consulta los dibujos adjuntos

Answer:

[tex]Emf=24.4V[/tex]

Explanation:

From the question we are told that:

Angular speed [tex]\omega=40*\omega'[/tex]

Radius [tex]r=0.33m[/tex]

No. Turns [tex]N=163[/tex]

Area [tex]A= 3.23 * 10^{-3} m^2[/tex]

Magnetic field. [tex]B=0.0948T[/tex]

Acceleration [tex]a=0.60m/s^2[/tex]

Time [tex]t=6.72s[/tex]

Generally the equation for momentum is mathematically given by

[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]

[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]

[tex]W=12.22rad/s^2[/tex]

Therefore

[tex]\omega=Aw[/tex]

[tex]\omega=12.22*40[/tex]

[tex]\omega=488.7rads/s[/tex]

Generally the equation for Peak emf is mathematically given by

[tex]Emf=NBA \omega[/tex]

[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]

[tex]Emf=24.4V[/tex]

Answer:

a)  t = 2.0 s,  b)  x_f = - 24.56 m,  Δx = 16.56 m

Explanation:

This is an exercise in kinematics, the relationship of position and time is indicated

          x = 5 t³ - 9t² -24 t - 8

a) ask when the velocity is zero

speed is defined by

         v = [tex]\frac{dx}{dt}[/tex]

let's perform the derivative

        v = 15 t² - 18t - 24

        0 = 15 t² - 18t - 24

let's solve the quadratic equation

      [tex]t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}[/tex]

       t = [tex]\frac{18 \pm 42}{30}[/tex]

       t1 = -0.8 s

      t2 = 2.0 s

the time has to be positive therefore the correct answer is t = 2.0 s

b) the position and distance traveled for a = 0

acceleration is defined by

       a = dv / dt

       a = 30 t - 18

       a = 0

       30 t = 18

       t = 18/30

       t = 0.6 s

we substitute this time in the expression of the position

       x = 5 0.6³ - 9 0.6² - 24 0.6 - 8

       x = 1.08 - 3.24 - 14.4 - 8

       x = -24.56 m

we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s

the position for t = 0

       x₀ = -8 m

the position for t = 0.6 s

      x_f = - 24.56 m

the distance

     ΔX = x_f - x₀

     Δx = | -24.56 -(-8) |

     Δx = 16.56 m

Answer:

A

Explanation:

So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

Answer:

Muscular endurance is how many times you can move a weight without getting tired.

Muscular strength is the amount force you can put out.

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{\rm total}=\Delta K[/tex]

or

[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]

Compute the work performed by friction:

[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]

[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]

[tex]\implies \boxed{\mu\approx0.45}[/tex]

The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{total}=\Delta K[/tex]

or

[tex]W_{friction}+W_{spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]

Compute the work performed by friction:

[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{friction}=-f(0.20)[/tex]

[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]

[tex]\mu =0.45[/tex]

Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

To know more about Coefficient of friction follow

https://brainly.com/question/136431

Answer:

induction using ground connection

Explanation:

This process is referred to as induction and it is done using ground process.

The process can be carried out as follows;

-An object that has a positive charge would be brought close to another object that is neutrally charged. then the electrons of the neutral object would be attracted and +ve charges would be repelled. This would cause an alignment at the two opposite corners.

- After this is done, the object would be grounded from the side that has the positive charges. This would bring about electron flowing from the ground to the earth. bringing about a neutralization of the positive charges.

-then the grounding would be taken away resulting in electrons leaving in majority.

-At the end, after the positively charged object has been removed, the charges would reassemble themselves. this would cause the object to be negatively charged.

therefore negative charge can be induced on a different object that is neutral using positive charge, through the induction process and grounding.

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer: 129.5 m

Explanation:

310 + 115 + 25 + 68 = 518

518 / 4 = 129.5 m

i think. Sorry if this is wrong

:)

Answer:

   x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]

Explanation:

This is an exercise in fluid mechanics, let's find the speed of the water in the hole

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole

the pressure inside and outside the tank is the same

       P₁ = P₂

we must measure the distance from the same reference point, let's locate it on the surface of the water

       y₁ = 0

       y₂ = h

Suppose the gap is small compared to the diameter of the tank

        v₁ «v₂

       v₂² = 2 g (0-h)

This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.

Let's use the projectile launch ratios

         vₓ = [tex]\sqrt{2g |h|}[/tex]

         y = y₀ + v₀ t - ½ g t²

the height when reaching the floor is y = 0,

the initial height is measured from the floor therefore y₀ = H-h

        0 = (H-h) + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2(H-h)}{g} }[/tex]

we look for the distance x traveled

        x = vₓ t

        x = [tex]\sqrt{2g |h| } \ \sqrt{ 2(H-h)/g}[/tex]

        x = [tex]\sqrt{4 h (H-h)}[/tex]

        x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]

Answer:

If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.

Explanation:

Answer:

The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.

It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is

∑ F = (force from engine) - (resistive forces) = 0

Answer:

[tex]D'=2.933*10^{18}m[/tex]

Explanation:

From the question we are told that:

Distance [tex]D=310ly[/tex]

Where

[tex]1ly=9.46*10^{15}[/tex]

Generally the equation for  meters will you traverse to reach your vacation destination is mathematically given by

[tex]D'=D*1ly[/tex]

[tex]D'=310*9.46*10^{15}[/tex]

[tex]D'=2.933*10^{18}m[/tex]

Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

Answer:

[tex]l=12916.5m[/tex]

Explanation:

Distance [tex]d=3600km[/tex]

Since

Density of steel [tex]\rho=7900kg/m^3[/tex]

Stress of steel [tex]\mu= 1*10^9[/tex]

Generally the equation for Stress on Cable is mathematically given by

[tex]S=\frac{F}{A}[/tex]

[tex]S=\frac{\rho Alg}{A}[/tex]

Therefore

[tex]l=\frac{s}{\rhog}[/tex]

[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]

[tex]l=12916.5m[/tex]

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

Explanation:

It isn't possible to learn an entire language fast. For an examination, memorise some of the easy words. Practice mew minutes speaking French and keep speaking it till you know it.