Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false

Answer:

Explanation:

Given data

frequency = 5 Hz

we know that the period T is expressed as

[tex]T= \frac{1}{f} \\[/tex]

Substituting we have

[tex]T= \frac{1}{5} \\T= 0.2s[/tex]

also the expression for angular velocity is

ω= [tex]\frac{2\pi}{T}[/tex]

Substituting we have

ω= [tex]\frac{2*3.142}{0.2}[/tex]

ω= [tex]\frac{6.284}{0.2} \\[/tex]

ω= [tex]31.42 rad/s[/tex]

Answer:

Explanation:

The equilibrium position will be in between q₁ and q₂ .

Let this position of third charge be x distance from q₁

q₁ will pull the third charge towards it with force F₁

F₁ =9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²

= 675 x 10⁻⁹ / x²

q₂ will pull the third charge towards it with force F₂

F₂ =9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ /( .40-x )²

= 270 x 10⁻⁹ / ( .40-x )²

For equilibrium

675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / ( .40-x )²

5  / x² = 2 / ( .40-x )²

( .40-x )² / x² = 2/5 = .4

.4 - x / x = .632

.4 - x = .632x

.4 = 1.632 x

x = .245 .

24.5 cm

so third charge must be placed at 24.5 cm away from q₁ charge.

Answer:

The correct answer to the following question will be "0.0562 rad/s".

Explanation:

[tex]r =\frac{3.9}{2}\times 1609.34[/tex]

  [tex]=3138.213\ m[/tex]

As we know,

⇒  [tex]\omega^2 \ r=g[/tex]

On putting the values, we get

⇒  [tex]\omega^2\times 3138.213=9.8[/tex]

⇒  [tex]\omega = \sqrt{\frac{9.8}{3138.213}}[/tex]

⇒  [tex]\omega = 0.0562 \ rad /s[/tex]

Answer:

a) v = 3.71m/s

b) U = 616.71 J

Explanation:

a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:

[tex]-W_f=\Delta K=\frac{1}{2}m(v^2-v_o^2)\\\\-F_fd=\frac{1}{2}m(v^2-v_o^2)[/tex]  

(the minus sign is due to the work is against the motion of the skier)

m: mass of the skier = 58.0 kg

v: final speed = ?

vo: initial speed = 6.00 m/s

d: distance traveled by the skier in the rough patch = 3.65 m

Ff: friction force = Mgμ

g: gravitational acceleration = 9.8 m/s^2

μ: friction coefficient = 0.310

You solve the equation (1) for v:

[tex]v=\sqrt{\frac{2F_fd}{m}+v_o^2}=\sqrt{\frac{2mg\mu d}{m}+v_o^2}\\\\v=\sqrt{-2g\mu d+v_o^2}[/tex]

Next, you replace the values of all parameters:

[tex]v=\sqrt{-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2}=3.71\frac{m}{s}[/tex]

The speed after the skier has crossed the roug path is 3.71m/s

b) The work done by the rough patch is the internal energy generated:

[tex]U=W_fd=F_fd=mg\mu d\\\\U=(58.0kg)(9.8m/s^2)(0.310)(3.50m)=616.71\ J[/tex]

The internal energy generated is 616.71J

Answer:

a) 0.5985 V

b) 0.05985 V

c) 0.00684 V

Explanation:

Given that

Number of turn in the coil, N = 60 turns

Magnetic field of the coil, B = 0.6 T

Diameter of the coil, d = 0.13 m

If area is given as, πd²/4, then

A = π * 0.13² * 1/4

A = 0.0133 m²

The induced emf, ε = -N(dΦ*m) /dt

Note, Φm = BA

Substituting for Φ, we have

ε = -NBA/t.

Now, we substitute for numbers in the equation

ε = -(60 * 0.6 * 0.0133)/0.8

ε = 0.4788/0.8

ε = 0.5985 V

at 8s,

ε = -(60 * 0.6 * 0.0132)/8

ε = 0.4788/8

ε = 0.05985 V

at 70s

ε = -(60 * 0.6 * 0.0132)/70

ε = 0.4788/70

ε = 0.00684 V

Answer:

Explanation:

mass of the girl m₁ = 47.3 kg

mass of the plank m₂ = 162 kg

velocity of the girl with respect to surface = v₁

velocity of plank with respect to surface = v₂

v₁+ v₂ = 1.36

v₂ = 1.36 - v₁

applying conservation of momentum law to girl and plank.

m₁v₁ = m₂v₂

47.3 x v₁ = 162 x ( 1.36 - v₁ )

47.3 v₁ = 220.32 - 162v₁

209.3 v₁ = 220.32

v₁ = 1.05 m /s

Explanation:

The initial velocity is 0 m/s.

The initial acceleration is -9.8 m/s².

The increase in the energy of an electromagnetic wave can be achieved only by decreasing the wavelength. Hence, option (d) is correct.

The given problem is based on the fundamentals of electromagnetic wave and the energy stored in an electromagnetic wave.

The mathematical expression for the energy carried out by the  electromagnetic waves is given as,

[tex]E = h \times \nu\\\\E = \dfrac{h \times c}{ \lambda}[/tex]

Here,

h is the Planck's constant.

[tex]\nu[/tex] is the frequency of the electromagnetic wave.

c is the speed of light.

[tex]\lambda[/tex] is the wavelength of wave.

Clearly, the energy of electromagnetic waves is directly proportional to the frequency of wave and inversely proportional to wavelength. So, decreasing the wavelength, we can easily increase the energy of electromagnetic wave.

Thus, we can conclude that the increase in the energy of an electromagnetic wave can be achieved only by decreasing the wavelength. Hence, option (d) is correct.

Learn more about the electromagnetic wave here:

https://brainly.com/question/3101711

Answer:

a) F = 0.0561 N

b) F = 2.86*W

Explanation:

a) The magnitude of the electric force between the plastic spheres is given by the following formula:

[tex]F=k\frac{q_1q_2}{r^2}[/tex]    (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2: charge of the plastic spheres = -50.0nC = -50.0*10^-9 C

r: distance between the plastic spheres = 2.0 cm = 0.02 m

You replace the values of the parameters in the equation (1):

[tex]F=(8.98*10^9Nm^2/C^2)\frac{(-50.0*10^{-9}C)^2}{(0.02m)^2}\\\\F=0.0561N[/tex]

The electric force between the spheres is 0.0561 N

b) To calculate the relation between weight and electric force, you first calculate the weight of one of the spheres:

[tex]W=mg[/tex]

m: mass = 2.0g = 2.0*10^-3 kg

g: gravitational acceleration = 9.8 m/s^2

[tex]W=(2.0*10^{-3}kg)(9.8m/s^2)=0.0196N[/tex]

The ratio between W and F is:

[tex]\frac{F}{W}=\frac{0.0561N}{0.0196N}=2.86\\\\F=2.86W[/tex]

The electric force is 2.86 times the weight

(a) The magnitude of the electric force on each sphere is [tex]5.625 \times 10^{-2} \ N[/tex].

(b)  The electric force on a sphere is larger than its weight by 2.87.

The given parameters:

The magnitude of the electric force on each sphere is calculated as follows;

[tex]F = \frac{kq^2}{r^2} \\\\F = \frac{9\times 10^9 \times (5 0 \times 10^{-9})^2}{(0.02)^2} \\\\F = 5.625 \times 10^{-2} \ N[/tex]

The weight of a sphere is calculated as follows;

[tex]W = mg\\\\W = 0.002 \times 9.8\\\\W = 0.0196 \ N[/tex]

Compare the electric force and the weight of a sphere;

[tex]= \frac{F}{W} = \frac{5.625 \times 10^{-2}}{0.0196} = 2.87[/tex]

Thus, the electric force on a sphere is larger than its weight by 2.87.

Learn more here:https://brainly.com/question/12533085

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

Answer:

microwave

Explanation:

Answer:

An electric power source is used to ionize fuel into plasma. Electric fields heat and accelerate the plasma while the magnetic fields direct the plasma in the proper direction as it is ejected from the engine, creating thrust for the spacecraft.

Explanation:

Answer:

1.) Vc = 1V

2.) Vc = 2.7V

3.) Time constant = 0.03

4.) V = 2.53V

Explanation:

1.) The value of Vc (in volts) just prior to the switch closing

The starting current = 1mA

With resistance R1 = 1000 ohms

By using ohms law

V = IR

Vc = 1 × 10^-3 × 1000

Vc = 1 volt.

2.) The value of Vc after the switch has been closed for a long time.

R2 and R3 are in parallel to each other. Both will be in series with R1

The equivalent resistance R will be

R = (R2 × R3)/R2R3 + R1

Where

R1 = 1000Ω,

R2 = 3000Ω,

R3 = 4000Ω

R = (4000×3000)/(4000+3000) + 1000

R = 12000000/7000 + 1000

R = 1714.3 + 1000

R = 2714.3 ohms

By using ohms law again

V = IR

Vc = 1 × 10^-3 × 2714.3

Vc = 2.7 volts

3.) The time constant = CR

Time constant = 10 × 10^-6 × 2714.3

Time constant = 0.027

Time constant = 0.03 approximately

4.) The value of Vc at t = 2msec (in volts). Can be calculated by using the formula

V = Vce^-t/CR

Where

Vc = 2.7v

t = 2msec

CR = 0.03

Substitute all the parameters into the formula

V = 2.7 × e^-( 2×10^-3/0.03)

V = 2.7 × e^-(0.0667)

V = 2.7 × 0.935

V = 2.53 volts

Answer:

reflected angle - secod mirror = 60°

Explanation:

I attached an image with the solution to this problem below.

In the solution the reflection law, incident angle = reflected angle, is used. Furthermore some trigonometric relation is used.

You can notice in the image that the angle of reflection in the second mirror is 60°

Answer:

a. 451.72 mi/ga

b. 1078.33 mi/ga

Explanation:

The computation is shown below:

a. The fuel economy for a person walking is

Given that

Walking at 3.20 mi/h requires about 220 kcal/h so it is equal to

[tex]= 220 \times 4186[/tex]

= 920920 j/hr

Now

[tex]= \frac{mi}{j}[/tex]

[tex]= \frac{3.2}{920920}[/tex]

So,

[tex]= \frac{mi}{ga}[/tex]

[tex]= \frac{3.2}{920920}\times 1.3 \times 100000000[/tex]

= 451.72 mi/ga

b. Now

Bicycling 12.5 mi/h requires about 360 kcal/h  energy so it is equal to

[tex]= 360 \times 4186[/tex]

= 1506960 j/hr

So,

[tex]= \frac{mi}{j}[/tex]

[tex]= \frac{12.5}{1506960}[/tex]

Now

[tex]= \frac{mi}{ga}[/tex]

[tex]= \frac{12.5}{1506960} \times 1.3 \times 100000000[/tex]

= 1078.33 mi/ga

We simply applied the above formula

Answer: The coefficient of static friction between the coin and the turntable is 0.13.

Explanation:

As we know that,

   Centripetal force = static frictional force

   [tex]\frac{mv^{2}}{r} = F_{s}[/tex]

or,  [tex]\frac{mv^{2}}{r} = \mu_{s} \times m \times g[/tex]

     v = [tex]\sqrt{\mu_{s} \times r \times g}[/tex]

or,  [tex]\mu_{s} = \frac{v^{2}}{rg}[/tex] ......... (1)

Here, it is given that

       r = 17 cm,      [tex]\omega[/tex] = 26 rpm,    

and  v = [tex]r \omega[/tex] ..........(2)

Putting equation (2) in equation (1) we get the following.

[tex]\mu_{s} = \frac{r^{2}\omega^{2}}{rg}[/tex]

            = [tex]\frac{17 \times 10^{-2} \times (26 \times [\frac{2 \times \pi}{60}]^{2})}{9.8}[/tex]

            = 0.128

            = 0.13 (approx)

Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.

Answer:

Direction of current = clockwise

Magnitude of current, I = 0.36 A

Explanation:

The magnetic field strength, [tex]B_{E} = 50 \mu T[/tex]

The angle of dip, ∅ = 56°

The net magnetic field in the center of the tank is:

[tex]B_{net} = (B_{E} cos \phi ) (\hat{x} ) + ( B + B_{E} sin \phi)(-\hat{y})\\B_{net} = (50 cos 56 ) (\hat{x} ) + ( B +50 sin 56)(-\hat{y})\\B_{net} = (28 \mu T ) (\hat{x} ) + ( B +41.4 \mu T)(-\hat{y})\\[/tex]

The direction of the net magnetic field is:

[tex]\phi = tan^{-1} \frac{B + 41.4 }{28} \\tan \phi = \frac{B + 41.4 }{28}\\\phi = 62^0\\tan 62 = \frac{B + 41.4 }{28}\\28 tan 62 = B + 41.4\\52.66 = B + 41.4\\B = 11.26 \mu T[/tex]

The magnetic field due to the coil:

[tex]B = \frac{\mu_{0}NI }{2r} \\11.26 * 10^{-6} = \frac{4\pi * 10^{-7} * 50 *I }{2 *1}\\I = \frac{2 * 11.26 * 10^{-6}}{4\pi * 10^{-7} * 50} \\I = 0.36 A[/tex]

The current must be in clockwise direction to produce the field in downward direction

Answer:

Explanation:

a)

A = 3m , B = 4m .

(A-B)² = A² + B² - 2ABcosθ where θ is angle between A and B.

4 = 9 + 16 - 2. 3.4 cosθ

cosθ = .875

θ = 29° .

b ) Unit vector in the direction of A X B will be k vector because A and B are in X-Y plane and A X B lies perpendicular to both A and B .

Answer:

Explanation:

Sry

Answer:

C = Q/V

where C is capacitance, Q is charge and V is voltage

C = (3×10)/50

C = 30/50

= 0.6F where F is in Farads

b) 16 cm

Magnification, m = v/u

3 = v/u

⇒ v = 3u

Lens formula : 1/v – 1/u = 1/f

1/3u = 1/u = 1/12

-2/3u = 1/12

⇒ u = -8 cm

V = 3 × (-8) = -24

Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm

Answer:

The speed of m2 is 0.6 m/s and its direction is to the right.

Explanation:

This numerical can be solved easily by applying law of conservation of momentum to it. According to law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where,

m₁ = Mass of 1st air glider = 0.25 kg

m₂ = Mass of 2nd air glider = 0.5 kg

u₁ =  Speed of 1st air glider before collision = 0.9 m/s

u₂ = Speed of 2nd air glider before collision = 0 m/s (at rest)

v₁ =  Speed of 1st air glider after collision = - 0.3 m/s (negative sign due to change in direction of velocity)

v₂ = Speed of 2nd air glider after collision = ?

Therefore,

(0.25 kg)(0.9 m/s) + (0.5 kg)(0 m/s) = (0.25 kg)(-0.3 m/s) + (0.5 kg)v₂

0.225 kg.m/s + 0.075 kg.m/s = (0.5 kg)v₂

v₂ = (0.3 kg.m/s)/(0.5 kg)

v₂ = 0.6 m/s

Positive sign indicates that v₂ is directed towards right

Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.

Explanation: hope this helped

Answer:

This is approximately 16 g's.

Explanation:

For the person’s head to stop falling, the rigid platform must exert a force that is equal to the sum of weight and force that caused the velocity to decrease from 6 m/s to 0 m/s.

Weight = m * -9.8

Let’s use the following equation to determine the acceleration.

vf^2 = vi^2 + 2 * a * d

0 = 36 + 2 * a * 0.12

a = -36 ÷ 0.24 = -150 m/s^2

The acceleration is negative, because it caused the velocity to decrease.

Total acceleration = -159.8 m/s^2

To determine the number of g, divide this by -9.8.

N g’s = -159.8 ÷- 9.8

This is approximately 16 g's.

Answer:

By exact formula

5076.59N/C

And by approximation formula

5218.93N/C

Explanation:

We are given that

Length of rod,L=2.6 m

Charge,q=98nC=[tex]98\times 10^{-9} C[/tex]

[tex]1nC=10^{-9} C[/tex]

a=13 cm=0.13 m

1 m=100 cm

By exact formula

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{a^2+\frac{L^2}{4}}}[/tex]

Where k=[tex]9\times 10^9[/tex]

Using the formula

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{(0.13)^2+\frac{(2.6)^2}{4}}}=5076.59N/C[/tex]

In approximation formula

a<<L

[tex]a^2+(\frac{L}{2})^2=\frac{L^2}{4}[/tex]

Therefore,the magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{\frac{L^2}{4}}}[/tex]

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{\frac{(2.6)^2}{4}}}=5218.93N/C[/tex]

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

[tex]T \approx e^{-2CL}[/tex]

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

[tex]0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)[/tex]

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

[tex]ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)[/tex]

Next, you divide the equation (3) into (4), and finally, you solve for L':

[tex]\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L[/tex]

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

Answer:

f₂ = 468.67 Hz

Explanation:

A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:

The higher frequency tone = f₁ = 470 Hz

No. of beats = n = 4 beats

Time period = t = 3 s

The lower frequency note = Frequency of Friend's Trombone = f₂ = ?

Beat Frequency = fb

So, the formula for beats per second or beat frequency is given as:

fb = n/t

fb = 4 beats/ 3 s

fb = 1.33 Hz

Another formula for beat frequency is:

fb = f₁ - f₂

f₂ =  f₁ - fb

f₂ = 470 Hz - 1.33 Hz

f₂ = 468.67 Hz

Answer:

a) [tex]t \approx 9.879\,s[/tex], b) [tex]v = 20.518\,\frac{m}{s}[/tex], c) [tex]t = 16.368\,s[/tex], d) [tex]v = 19.545\,\frac{m}{s}[/tex]

Explanation:

a) Since train is only translating in a straight line and experimenting a constant deceleration throughout the station, whose length is 210 meters. The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:

[tex]210\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

The following second-order polynomial needs to be solved:

[tex]-0.075\cdot t^{2} + 22\cdot t - 210 = 0[/tex]

Whose roots are presented herein:

[tex]t_{1}\approx 283.455\,s[/tex] and [tex]t_{2} \approx 9.879\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train.

b) The speed of the nose leaving the station is given by this expression:

[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (9.879\,s)[/tex]

[tex]v = 20.518\,\frac{m}{s}[/tex]

c) First, it is required to calculate the time when nose of the train reaches a distance of 130 meters.

[tex]130\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]-0.075\cdot t^{2} + 22\cdot t - 130 = 0[/tex]

Roots of the second-order polynomial are:

[tex]t_{1} \approx 287.300\,s[/tex] and [tex]t_{2} \approx 6.033\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train. Now, the speed experimented by the train at this instant is:

[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (6.033\,s)[/tex]

[tex]v = 21.095\,\frac{m}{s}[/tex]

The distance traveled by the end of the train throughout station is modelled after the following equation:

[tex]210\,m = \left(21.095\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]-0.075\cdot t^{2} + 21.095\cdot t - 210 = 0[/tex]

Roots of the second-order polynomial are:

[tex]t_{1} \approx 270.932\,s[/tex] and [tex]t_{2} \approx 10.335\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train. The instant when the end of the train leaves the station is:

[tex]t = 6.033\,s + 10.335\,s[/tex]

[tex]t = 16.368\,s[/tex]

d) The velocity experimented by the end of the train is:

[tex]v = 21.095\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}} \right)\cdot (10.335\,s)[/tex]

[tex]v = 19.545\,\frac{m}{s}[/tex]

The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all. The negative sign indicates that the train is moving in the opposite direction of its initial velocity. The end of the train leaves the station after approximately 14.98 seconds. The velocity of the end of the train as it leaves the station is approximately 19.753 m/s.

(a) Using the equation :

v² = u² + 2as

0 = (22.0)² + 2 × (-0.150 ) × s

s = -(22.0 )^2 / (2 × -0.150)

s = 325.3 m

The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all.

(b) The velocity of the train when the nose leaves the station is given by:

v = u + at

v = 22.0 + (-0.150 ) × 210

v ≈ 22.0 - 31.5

v ≈ -9.5 m/s

The negative sign indicates that the train is moving in the opposite direction of its initial velocity.

(c)

s = ut + (1/2)at²

210= 22.0 × t + (1/2) × (-0.150) × t²

0.075 t² - 22.0 t + 210 = 0

we find two solutions for t: t ≈ 14.98 s and t ≈ 3.01 s.

The end of the train leaves the station after approximately 14.98 seconds.

(d) The velocity of the end of the train as it leaves the station is given by:

v = u + at

v = 22.0 + (-0.150) × 14.98

v = 22.0 - 2.247

v = 19.753 m/s

So, the velocity of the end of the train as it leaves the station is approximately 19.753 m/s.

To know more about the velocity:

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Answer:

2.83×10⁻¹¹ N.

Explanation:

From the question,

Using

F = qvB....................... Equation 1

Where F = magnetic force acting on the duck, q = charge of the duck, v = velocity of the duck, B = magnetic field of the duck.

Given: q = 6.47×10⁻⁸ C, B = 4.09×10⁻⁵ T, v = 10.7 m/s.

Substitute these values into equation 1

F = 6.47×10⁻⁸×4.09×10⁻⁵×10.7

F = 2.83×10⁻¹¹ N.

Hence the magnetic force acting on the duck is  2.83×10⁻¹¹ N.

Answer:

Nicolaus Copernicus

Explanation:

Nevertheless, Copernicus began to work on astronomy on his own. Sometime between 1510 and 1514 he wrote an essay that has come to be known as the Commentariolus (MW 75–126) that introduced his new cosmological idea, the heliocentric universe, and he sent copies to various astronomers