Now imagine that a beam of moving electrons enters the quadrupole, with velocity parallel to the charged rods. When the beam enters the device, its cross-sectional profile is circular. What will the beam look like when it exits the quadrupole on the other side

Answer:

4. None

Explanation:

Applying,

a(t) = dV(t)/dt

Where a(t) = Acceleration of the ship at a given time.

From the question,

Given: V(t) = 5-2t

Therefore,

a(t) = dV(t)/dt = 5 m/s²

Hence it's acceleration is 5 m/s²

From the question,

The right option is 4. None

Answer:

Explanation:before the phase change the substance is a particle.

Answer:

I just noticd i dont speak this launguage

Explanation:

Answer:

C. 5730 years

Explanation:

N(t) = N(0)e^-kt

The half-life is T = 5730 years,

e^-kT = 1/2

→ k = - ln(1/2) / T

→ - ln(1/2) / 5730

→ 1.209681 x 10^-4 years^-1

The amount present dropped to 50%.

Then one half-life has elapsed, so the age is 5730 years.

Answer:

a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]  

b) attached below

c) 0.037 m/s

Explanation:

a) Determine the magnetic "drag" force acting  at the moment

speed = v

first step: determine current in the loop

I = [tex]\frac{\pi d^2}{16pl} B lv[/tex]   ----- ( 1 )

given that the current will induce force on the three sides of the loop found in the magnetic field

forces on vertical sides = + opposite

we will cancel out

hence equation 1 becomes

F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]   ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )

b) Determine the formula for Vt

attached below

c) Find Vt

given :

B = 0.80 T

density of copper = 8.9 * 10^3 kg/m^3

resistivity of copper = 1.68 * 10^-8 Ωm

∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2

       = 0.037 m/s

Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field

Explanation:

While interpreting the data in NMR, the positions of signals are studied.

The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.

The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.

So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]

Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons

So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]

Answer: Colonialism is a big topic, but it can only be understood by looking at human experiences. Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony”.

Explanation: hey, hope this hlps! oh, btw you picked the wrong subject for this question it should have been history insteat of phiscics!

Answer:

Electric force is the attractive force between the electrons and the nucleus. It works the same way for a negative charge, you also have an electric field around it. ... Now, like charges repel each other and opposite charges attract.

The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. ... This is called Newton's Universal Law of Gravitation.

electric force

This table shows clearly that the electric force dominates the motion of electrons in atoms. However, on a macroscopic scale, the gravitational force dominates. Since most macroscopic objects are neutral, they have an equal number of protons and electrons.

Explanation:

Answer:

The answer is "Option D".

Explanation:

Please find the complete question in the attached file.

As plate separation increased to 2d the capacitance get halred but the change remain same

[tex]\therefore V=\frac{Q}{C}[/tex]

The voltage doubles are now electric field remain same because both the distance and voltage get doubled.

[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]

So,

[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]

OPTION C is the correct answer.

The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.

The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.

The isotope decay of an atom is given by the equation:

ln [A] = -kt + ln [A]₀

The rate constant, k is:

k = ln 2 / Half life

k = ln 2 / 4.96 x 10³

k = 1.40 × 10⁻⁴ s⁻¹

t = 1.98 x 10⁴

[A]₀ = 3.21 x 10¹⁷

ln [A] = -1.40 × 10⁻⁴  ×  1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538

[A] = 2.02 x 10¹⁶ nuclei

Thus the correct option is C.

To know more about half-life, visit;

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Answer:

Explanation:

From the given information:

The initial PE [tex](PE)_i[/tex] = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = [tex]P.E_f -P.E_i[/tex]

ΔP.E = 0 - [tex](PE)_i[/tex]

ΔP.E = [tex]-P.E_i[/tex]

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]

[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]

this can be re-written as:

[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]

[tex]\Delta U =(0.70) (490.5)[/tex]

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]

[tex]v_1t= 0.35[/tex]  ...........(i)

[tex]v_0t= 1.09[/tex]  ...........(ii)

Dividing the equation (ii) by (i)

[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]

∴ [tex]v_0=3.11 \ v_1[/tex]

Now loss of energy  = change in the kinetic energy

[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]

[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]

[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]

If f is average friction force, then

(f)(L) = W

(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

(f)  = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

The Average force of friction is ( F )  = 7.307 * 10⁻³ v₀²

Given data:

Predicted range ( v₁t ) = 0.3503 m

Actual range ( v₀t ) = 1.09 m

mass = 16.3 g

First step : Determine the value of  V₀

[tex]t = \sqrt{\frac{2H}{g} }[/tex]    ,    v₁t  =  0.3503 ,    ( v₀t ) = 1.09 m

To obtain the value of  V₀  

Divide ( v₀t ) by ( v₁t )  =  1.09 / 0.3503 = 3.11 v₁

∴ V₀ = 3.11 v₁

Next step : Determine the average force of friction ( f )

given that loss of energy results in a change in kinetic energy

W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]

    = 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]

∴ W = 7.307 * 10⁻³ v₀²

Average force of friction = W / Actual length

                                         = 7.307 * 10⁻³ v₀² / 1  

∴ Average force of friction ( F )  = 7.307 * 10⁻³ v₀²

Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²

Learn more about average force of friction : https://brainly.com/question/16207943

Your question has some missing data below are the missing data related to your question

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Answer:

The average speed is 1 m/s

The average velocity is 0

Explanation:

Given;

length of the pool, L = 50 m

time taken for the motion, t = 100 s

The total distance = 50 m + 50 m

The total distance = 100 m

The average speed = total distance / total time

                                  = 100 / 100

                                  = 1 m/s

The average velocity = change in displacement / change in time

change in displacement = 50 m - 50 m = 0

The average velocity = 0 / 100

The average velocity = 0

Answer:

For (a): The chemical shift is [tex]2.08\delta[/tex]

For (b): The chemical shift is [tex]9.85\delta[/tex]

For (c): The chemical shift is [tex]7.5\delta[/tex]

Explanation:

To calculate the chemical shift, we use the equation:

[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]

Given value of spectrometer frequency = 200 MHz

Given peak position = 416 Hz

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]

Given peak position = [tex]1.97\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]

Given peak position = [tex]1.50\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]

Answer:

ans: 2250 meters

Explanation:

initial velocity (U)= 15 m/s

final velocity (V) = 0m/s , since need to come in rest

total time taken (T) = 5 min= 300 seconds

total distance covered (S)= UT + 1/2 aT^2 ,

a= acceleration

S= 15 × 300 + 0.5 ×(0 - 15) × 300

since a = (V - U)/ T

S = 4500 - 2250

S= 2250 m

Let v be the dog's initial velocity. Then

(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)

==>   v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)

==>   v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))

==>   v ≈ 5.86 m/s

Answer:

To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)

Explanation:

Answer:

There was 7 continents in africa

Answer:

1 cal/s =4.184w

p=50 cal/s =2093w

v=12v

P = V*I

I =P/V

I = 17.43 A

P =1²*R

R = P/I²

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

Answer:

Q = M * (Cf +  C * 100 + Cv)

Cf and Cf are heats of fusion and vaporization and C is the heat required to heat mass M of water 1 deg

Q = .001 kg ( 3.36 * E5 + 100 deg * 4200 + 2.26 * E6) J

Q = .001 kg ( 3.36 J / kg + 4.2 J / kg + 22.6 J /kg) * 10E5

Q = .001 kg * 30.2 * 10E5 J / kg = 3020 J

Answer:

the answer is option B because opposit sides of the magnets attract each other

Answer:

Because the surface area of her skis are greater than the surface are of her shoes

Explanation:

the reason for this is that the weight per in is too heavy crushing the snow blower but with the skies the weight is distributed to the point were the snow can support her weight

Answer:

31475404.8 kg/day

Explanation:

From the information given:

The power plant capacity W = 3570 MW

Energy content of the coal = 28000 kJ/kg

let assume that the thermal efficiency = 35%

Recall that:

1 kw = 3600 kJ/hr provided that the energy conversion is 100% efficient

But assuming the thermal efficiency = 35%.

Then:

Heat input = 3600/0.35 = 10286 kJ/kw.hr

Now, for producing 1 kw.hr, the quantity of the required coal = 10286/28000

= 0.36736 kg

For 3570 MW, the amount of coal that must be input is:

= 0.36736 kg × 3570000

= 1311475.2 kg/hr

= 1311475.2 × 24 kg/day

= 31475404.8 kg/day

Answer:

We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.

Explanation:

5 ways our governments can confront climate change

PROTECT AND RESTORE KEY ECOSYSTEMS

SUPPORT SMALL AGRICULTURAL PRODUCERS

PROMOTE GREEN ENERGY

COMBAT SHORT-LIVED CLIMATE POLLUTANTS

BET ON ADAPTATION, NOT JUST MITIGATION

False

It's not straight up

Answer:

uhm oxygen tank, a suit (like an astronaut suit) uh food, and a space ship

Explanation: