Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 6.7 X 10 -27 Kg.

Answer:

The required wavelength is 1.19 μm

Explanation:

In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.

[tex]y_m = \dfrac{m \lambda D}{d}[/tex]

where;

m = fringe order

d = slit separation

λ = wavelength

D = distance between screen to the source

For the first bright fringe, m = 1, and we make (d) the subject, we have:

[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]

[tex]d = \dfrac{ \lambda D}{y_1}[/tex]

replacing the value from the given question, we get:

[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]

In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:

[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

The value of m in the dark fringe first order = 0

∴

[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

making λ the subject of the formula, we have:

[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]

[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]

Answer:

W = 354.9 J

Explanation:

Given that,

The mass of a bale of hay, m = 23 kg

The displacement, d = 3.9 m

The horizontal force exerted on the hay, F = 91 N

We need to find the work done. We know that,

We know that,

Work done, W = Fd

So,

W = 91 N × 3.9 m

W = 354.9 J

So, the required work done is 354.9 J.

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

Option (a).

Explanation:

Let the angular velocity is w.

The centripetal acceleration is given by

[tex]a = r w^2[/tex]

where, r is the distance between the axle and the spoke.

So, more is the distance more is the centripetal acceleration.

(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.

The statement is true.  

(b) The direction of centripetal acceleration is always towards the center, so the statement is false.

(c) It is false.

(d) It is false.

Option (a) is correct.

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

570k or 1050k

270k+290k= 570k

270k•290k = 1050k

Answer:

a. 59 ev. helpful answer

Answer:

I don't see anything on your question?

Answer:

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ds

Answer:

a) [tex]V=14.904m/s[/tex]

b) [tex]d = 191.49 m[/tex]

c) [tex]t= 25.696 s[/tex]

Explanation:

From the question we are told that:

Radius [tex]r =378m[/tex]

Acceleration [tex]a=0.580[/tex]

a)

Generally the  equation for speed of the car is mathematically given by

 [tex]a=\frac{v^2}{r}[/tex]

 [tex]V=\sqrt{a*r}[/tex]

 [tex]V=\sqrt{0.58*383}[/tex]

 [tex]V=14.904m/s[/tex]

b)

Generally the  equation for distance traveled of the car is mathematically given by

 [tex]V^2=u^2+2ad[/tex]

 [tex]d=\frac{V^2}{2a}[/tex]

 [tex]d=\frac{14.904^2}{2*0.58}[/tex]  

 [tex]d = 191.49 m[/tex]

c)

Generally the  equation for time of the car is mathematically given by

 [tex]V=u+at[/tex]

 [tex]t=\frac{V}{a}[/tex]

 [tex]t=\frac{14.904}{0.58}[/tex]

 [tex]t= 25.696 s[/tex]

Answer:

The product of mass and velocity is the correct answer

Explanation:

Momentum is defined as mass × velocity

p = mv

Answer:

The product of mass and velocity

Explanation:

I  just did it and got it right with a 100%

Posted 1/3/23

Answer:

the mass of the bullet is 10.5 g

Explanation:

Given;

initial velocity, u₁ = 280 m/s

final velocity of the bullet, v₁ = 70 m/s

final velocity of the block, v₂ = 0.2 m/s

mass of the block, m₂ = 11 kg

initial velocity of the block, u₂ = 0

let the mass of the bullet = m₁

Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

280m₁  +  11(0)  = 70m₁  +  11 x 0.2

280m₁ = 70m₁  + 2.2

280m₁ - 70m₁  = 2.2

210m₁ = 2.2

m₁ = 2.2/210

m₁ = 0.0105 kg

m₁ = 10.5 g

Therefore, the mass of the bullet is 10.5 g

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.

Hope it helps

Answer:

Newton's third law state that every action there is equal but opposite reactions

hope this will help you more

Answer:

a.49 n

b. 63 n

c. 112 n

Explanation:

a.10 times 9.8 from gravity/2 = 49 n

b. 49n times 4.5/8-4.5 = 63 n

c 49n + 63 n = 112 n

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Answer:

Solid is the state in which Matter maintains a fixed volume

Answer:

The state of matter that has a fixed volume is Solid.

Explanation:

Solid substances will maintain a fixed volume and shape.

Answer: θ would equal approximately 28.7°

This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.

Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°

Now if we multiply the range by 2, we get:

2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:

2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ

Thus, θ = 28.67780425

It's been awhile since I did this; though I hope it helped!

Answer:

The required ratio is 1.99.

Explanation:

We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.

We know that,

[tex]eV=\dfrac{1}{2}mv^2[/tex]

The LHS for both proton and an alpha particle is the same.

So,

[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]

So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.

Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

Given that

Blue chips =10

Red chips = 5

Yellow chips = 15

Number of the total samples =10+15+5=30

Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]

Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]

Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]

Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2

To know more about probability follow

https://brainly.com/question/24756209

Answer:

30.5°

Explanation:

Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from

n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).

Making D subject of the formula, we have

n = [sin(D + α)/2]/sin(α/2)

nsin(α/2) = [sin(D + α)/2]

(D + α)/2 = sin⁻¹[nsin(α/2)]

D + α = 2sin⁻¹[nsin(α/2)]

D = 2sin⁻¹[nsin(α/2)] - α

So, substituting the values of the variables into the equation, we have

D = 2sin⁻¹[nsin(α/2)] - α

D = 2sin⁻¹[1.42sin(60°/2)] - 60°

D = 2sin⁻¹[1.42sin(30°)] - 60°

D = 2sin⁻¹[1.42 × 0.5] - 60°

D = 2sin⁻¹[0.71] - 60°

D = 2(45.23°) - 60°

D = 90.46° - 60°

D = 30.46°

D ≅ 30.5°

False

It's not straight up

Answer:

Following are the solution to the given points:

Explanation:

For point a:

Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:

The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.

For point b:

[tex]160.2\ N[/tex]

First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that

where

[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]

Calculating the mass:

[tex]m = 145 g = 0.145 kg[/tex]

Calculating the force:

[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]

 For point c:

0.195 times the pitcher's weight

[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]

Solving for W:

[tex]W=84 \times 9.8= 823.2 \ N[/tex]

Now the force of Part B could be defined as the fraction of the mass of the pitcher:  

[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]

Answer:

The work done by the friction force is - 139927.5 J.

Explanation:

mass of diver, m = 45 kg

distance falls, h = 450 m

initial speed, u = 0 m/s

final speed = 51 m/s

According to the work energy theorem,

Work done by the gravity + work done by the friction force = change in kinetic energy

[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]