An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s and an S-wave travels at 5000 m/s. If P-waves are received at a seismic station 2.00 minute before an S wave arrives, how far away is the earthquake center?

Answer:

a) t = 3.01s

b) 15th floor

Explanation:

First we need to know the distance the elevator has descended before the bolt fell.

[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]

Now we can calculate the time that passed before both elevator and bolt had the same position:

[tex]Y_{b}=Y_{e}[/tex]

[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]

[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex]   Solving for t:

t1 = -1.87s    t2 = 3.01s

In order to know how the amount of floors, we need the distance the bolt has fallen:

[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex]  Since every floor is 3m:

Floors = Yb / 3 = 15 floors.

Answer:920.31 J

Explanation:

Given

Volume of water (V)[tex]=14.1 cm^3 [/tex]

mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]

Temperature [tex]=8.4^{\circ} C[/tex]

Final Temperature [tex]=-7.2 ^{\circ}C[/tex]

specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]

Therefore heat required to removed is

[tex]Q=mc(\Delta T)[/tex]

[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]

[tex]Q=920.31 J[/tex]

Answer:

Explanation:

Resistance in series is given by the sum of all the resistor in series

value of Total Resistance is given by

[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]

Where [tex]R_{th}[/tex] is the total resistance

[tex]R_1,R_2[/tex] are the resistance in series

Current in series remains same while potential drop is different for different resistor

The value of net resistor is always greater than the value of individual resistor.

If a there is a defect in a single resistor then it affects the whole circuit in series.

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

Answer:

Thickness = 123.19 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of thin layer of magnesium fluoride = 1.38

The wavelength of the light = 680 nm

The thickness can be calculated by using the formula shown below as:

[tex]Thickness=\frac {\lambda}{4\times n}[/tex]

Where, n is the refractive index of thin layer of magnesium fluoride = 1.38

[tex]{\lambda}[/tex] is the wavelength

So, thickness is:

[tex]Thickness=\frac {680\ nm}{4\times 1.38}[/tex]

Thickness = 123.19 nm

Answer:

The second system must be set in motion [tex]t=0.70s[/tex] seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later

Answer:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

Due to increased width ,  position of a fringe  moves away from the centre.

Answer:

plane speed: 525mph, jetstream speed=75mph, in explanation it is solved with a linear equations system

Explanation:

First lets name each speed

vs:=speed of the jetstream

vp:=speed of the plane

Now when in the jetstream direction the speeds are added and on the opposite direction are subtracted, then we get these equations, that are linear.

1800 mi=(vp+vs)*3h

1800 mi=(vp-vs)*4h

which is a linear equation system equivalent to:

600 mph=vp+vs (1)

450 mph=vp-vs  (2)

Now from (2) vp= 450mph+vs (3), replacing this in (1) we get:

600mph=(450mph+vs)+vs=450mph+2*vs, then 2*vs=150mph or vs=*75mph, this is the jetstream speed, replacing this in (3) we get the plane speed too vp=450 mph +75mph = 525 mph

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]

[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]

The vertical distance fall down by the ball, h'  H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]

here, uy = 0

So,

[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]

v = 16.27 m/s

Answer:

[tex]t=2.8h[/tex]

[tex]t=10080s[/tex]

[tex]t=168 min[/tex]

Explanation:

From this exercise we have velocity and distance. Using the following formula, we can calculate time:

[tex]v=\frac{d}{t}[/tex]

Solving for t

[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]

[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]

[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]

Answer: 20 pm=20*10^-12 m

Explanation: To solve this problem we have to use the relationship given by:

λmin=h*c/e*ΔV= 1240/60000 eV=20 pm

this expression is related with the bremsstrahlung radiation when a flux of energetic electrons are strongly stopped hitting to a catode. The electrons give their kinetic energy to the atoms of the catode.

Answer:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]

[tex]0=H-g*\frac{t^{2}}{2}[/tex]    solving for t:

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

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex]   Replacing values:

[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]

Simplifying:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Answer:

The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.

Explanation:

Given that, the initial velocity of the snail is,

[tex]u=2m/s[/tex]

And the acceleration of the snail is,

[tex]a=1m/s^{2}[/tex]

And the time taken by the snail is,

[tex]t=5 sec[/tex]

Now according to first equation of motion,

[tex]v=u+at[/tex]

Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.

Now substitute all the variables

[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]

Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.

Now according to third equation of motion.

[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.

Substitute all the variables in above equation.

[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]

Therefore the distance of the snail is 22.5 m.

Answer:

a)11.6m

b)45.55s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

a)

for this problem

Vo=0

Vf=319m/min=5.3m/s

a=1.2m/s^2

we can use the ecuation number 1 to calculate the time

t=(Vf-Vo)/a

t=(5.3-0)/1.2=4.4s

then we use the ecuation number 3 to calculate the distance

X=0.5at^2

X=0.5x1.2x4.4^2=11.6m

b)second part

We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)

we will have the distance traveled in with constant speed.

With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate

X=218-11.6x2=194.8m

X=VT

T=X/v

t=194.8/5.3=36.75s

Total time=36.75+2x4.4=45.55s

Answer:

Distance between Karl and Joe is 38.467 m

Solution:

Let us assume that you are at origin

Now, as per the question:

Joe's tent is 19 m away from yours in the direction [tex]19.0^{\circ}[/tex] north of east.

Now,

Using vector notation for Joe's location, we get:

[tex]\vec{r_{J}} = 19cos(19.0^{\circ})\hat{i} + 19sin(19.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{J}} = 17.96\hat{i} + 6.185\hat{j} m[/tex]

Now,

Karl's tent is 45 m away from yours and is in the direction [tex]39.0^{\circ}[/tex]south of east, i.e.,  [tex]- 39.0^{\circ}[/tex] from the positive x-axis:

Again,  using vector notation for Karl's location, we get:

[tex]\vec{r_{K}} = 45cos(-319.0^{\circ})\hat{i} + 45sin(- 39.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{K}} = 34.97\hat{i} - 28.32\hat{j} m[/tex]

Now,  obtain the vector difference between [tex]\vec{r_{J}}[/tex] and [tex]\vec{r_{K}}[/tex]:

[tex]\vec{r_{K}} - \vec{r_{J}} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m[/tex]

[tex]\vec{d} = \vec{r_{K}} - \vec{r_{J}} = 17.01\hat{i} - 34.51\hat{j} m[/tex]

Now, the distance between Karl and Joe, d:

|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|

[tex]d = \sqrt{(17.01)^{2} + (34.51)^{2}} m[/tex]

d = 38.469 m

The distance between Karl's and Joe's tent is:

The distance between Joe's tent and Karl's tent is approximately 36.84 m.

To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:

Next, we can use the components to find the displacement from Joe's tent to Karl's tent:

Using the components, we can calculate the displacement from Joe's tent to Karl's tent:

Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:

Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m

Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.

Answer:67.45 m

Explanation:

Given

at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum  height

let u is the initial velocity of stone

v=u+at

0=u-gt

[tex]u=9.81\times 4=39.24 m/s[/tex]

Let us take h be the height of building

[tex]h=ut+\frac{-1}{2}gt^2[/tex]

[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]

h=67.45 m

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

Answer: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

Answer:[tex]3.874 m/s^2[/tex]

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

[tex]60mi/h \approx 26.8224m/s[/tex]

[tex]34mi/h \approx 15.1994 m/s[/tex]

we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]

[tex]a=\frac{15.1994-26.8224}{3}[/tex]

[tex]a=-3.874 m/s^2[/tex]

negative indicates that it is stopping the car

Distance traveled

[tex]v^2-u^2=2as[/tex]

[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]

[tex]s=\frac{488.419}{2\times 3.874}[/tex]

s=63.038 m

Answer:

[tex]a=4m/s^{2}[/tex]

Explanation:

From the concept of average acceleration we know that

[tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1}  }[/tex]

From the exercise we know that

[tex]v_{2}=30m/s\\v_{1}=10m/s\\t_{2}=5s\\t_{1}=0s[/tex]

So, the average acceleration of the car is:

[tex]a=\frac{30m/s-10m/s}{5s}=4m/s^{2}[/tex]

Answer:

1.  1.568 x 10^6 N m^2 / C

2. -  1.568 x 10^6 N m^2 / C

Explanation:

E = 4.9 x 10^4 N/C

Side of square, a = 8 m

Area, A = side x side = 8 x 8 = 64 m^2

Angle between lane of sheet and electric field = 30°

Angle between the normal of plane of sheet and electric field,

θ = 90°- 30° = 60°

The formula for the electric flux is given by

[tex]\phi = E A Cos\theta[/tex]

(1) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C

(2) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C   and q₂ = 11.0 n C

assume the distance be r from q₁  where the electric field is zero.

distance of point from q₂  be equal to 1.8 -r

now,

        E₁ = E₂

[tex]\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}[/tex]

[tex](\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}[/tex]

[tex]\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]

1.8 = 6 r

r = 0.3 m

let us assume  q₁  be negative so, distance from  q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

   E₁ = E₂

[tex]\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}[/tex]

[tex](\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}[/tex]

[tex]\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]

1.8 =4 r

r = 0.45 m

The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.

To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.

For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).

To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.

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Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

[tex]\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm[/tex]

Combined focal length is 19.642 cm

The focal length of the combination of lenses is approximately 19.64 cm.

To find the focal length of the combination of lenses in this scenario, we need to use the lensmaker's formula for thin lenses in combination.

Given:

Focal length of lens 1, [tex]\( f_1 = +11 \)[/tex] cm

Focal length of lens 2, [tex]\( f_2 = -25 \)[/tex] cm

The formula for the focal length of two thin lenses in contact is given by:

[tex]\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{+11 \text{ cm}} + \frac{1}{-25 \text{ cm}} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{11} - \frac{1}{25} \][/tex]

To subtract the fractions, find a common denominator, which is 275:

[tex]\[ \frac{1}{f} = \frac{25}{275} - \frac{11}{275} \][/tex]

[tex]\[ \frac{1}{f} = \frac{14}{275} \][/tex]

Now, invert both sides to solve for (f):

[tex]\[ f = \frac{275}{14} \][/tex]

[tex]\[ f \approx 19.64 \text{ cm} \][/tex]

The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice, due to the time it takes light and gravitational force to travel from the Sun to Earth. The Sun is approximately 1.50×10¹¹ meters away, and light travels at about 300,000 kilometers per second. Therefore, we see the Sun as it was 8 minutes ago.

If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice any change. This is because the Sun is approximately 1.50×10¹¹ meters away, and light, which travels at the speed of about 300,000 kilometers per second, takes 8 minutes to travel from the Sun to Earth.

Therefore, for 8 minutes, we would continue to see the Sun as it was before the change or disappearance occurred.

Similarly, if there were an immediate change in the gravitational force due to the Sun’s disappearance, it would also take 8 minutes for Earth to experience the effect because gravitational information, like light, propagates at the speed of light.

Hence, The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice

Answer:

a) 231 yards

b) 55.2 yards

c) 0 yards to 260 yards

d) Infinite values

Explanation:

This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:

x=Final Position - Initial Position

or

[tex]x=x_{f} - x_{i}[/tex]

a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:

[tex]x=260 y - 29 y=231 y[/tex]

b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:

[tex]x=260 y - 204.8  y=55.2  y[/tex]

c) If we take Meredith's initial position at her house  [tex]x_{i}=0 y[/tex] and her final position at the bus stop  [tex]x_{f}=260 y[/tex], the value of  [tex]x[/tex] varies from 0 yards to 260 yards.

d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]

The answers to the possible distance covered by Meredith at the various distances from her house are;

A) distance = 231 yards

A) distance = 231 yardsB) distance = 55.2 yards

A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

A) We are told that meredith walks from her house to a bus stop that is 260 yards away.

After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.

Distance left to reach bus stop = 260 - 29 = 231 yards

B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;

Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.

C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.

Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

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Answer:

a)  1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Explanation:

Given Data:

Distance between the plates, d = 4.67 mm

[tex]= (4.67) *10^{-3} m[/tex]

[/tex]= 4.67 *10^{-3} m[/tex]

Potential difference, V = 600 V

Solution:

(a) The  magnitude of the electric field between the plates is,

    [tex]E = \frac{V}{d}[/tex]  

[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]

  [tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]  

(b) Force on electron btwn the plates is,

   F = q E

 [tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]

 [tex]= 2.05 *10^{-14} N[/tex]  

(c) Work done on the electron is

   W = F * s

 [tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]

 [tex]= 4.83 *10^{-17} J[/tex]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

[tex]E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}[/tex]

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

[tex]E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}[/tex]

E(0.89) = 306500 N/C

Explanation:

Given that,

Initial sped of the runner, u = 0

Final speed of the runner, v = 30 ft/s

Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]

Let t is the time taken by the runner. It can be calculated using first equation of motion as :

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{30-0}{8}[/tex]

t = 3.75 seconds

Let s is the distance covered by the runner. Using the second equation of motion as :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]

s = 56.25 feet

Hence, this is the required solution.

Final answer:

To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.

Explanation:

The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.

Calculating Time to Reach Max Speed

To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.

Calculating Distance Traveled

To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.

Answer:

the critical angle of the flint glass is 37.04⁰

Explanation:

to calculate the critical angle for total internal reflection.

given,

wavelength of the flint glass =  λ  = 580.0 nm

                                                       = 580 × 10⁻⁹ m        

critical angle  = sin^{-1}(\dfrac{\mu_a}{\mu_g})

at the wavelength of 580.0 nm the refractive index of the glass is 1.66

refractive index of air = 1                        

critical angle  = sin^{-1}(\dfrac{1}{1.66})

                      = 37.04⁰              

hence, the critical angle of the flint glass is 37.04⁰