A motorcycle has a velocity of 15 m/s, due south as it passes a carwith a velocity of 24 m/s, due north. What is the magnitude anddirection of the velocity of the motorcycle as seen by the driverof the car?

a. 9 m/s, north
b. 9 m/s, south
c. 15 m/s, north
d. 39 m/s, north
e. 39 m/s, south

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

[tex]\omega=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]

ω =4.18 rad/s

The radius  r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

[tex]F=22\times 4.18^2\times 1.25\ N[/tex]

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

Answer:

a point of destructive interference.

Explanation:

the wavelength of the sound:

λ= v/f

v= velocity of sound =340 m/s

f= frequency of sound wave= 1800 Hz

L_1 = 4 m

then speaker is at the distance of

[tex]L_2 = sqrt(4^2+2^2)[/tex]

= 2√5 m  

ΔL = L_2-L_1

x = ΔL/λ

Now, if this result is an integer, the waves will add up  at the point. If it is nearly an integer + 0.5, the waves will have a destructive interference at the point. If it is neither of them , then  point is "something in between".

[tex]x= \frac{2\sqrt{5}-4 }{\frac{340}{1800} } =2.4995[/tex]

which is  close to 2.5, an integer + 0.5. So it's a point of destructive interference.

The result is within an integer value of +0.5, thus its a point of destructive interference.

The given parameters;

The resultant distance between the speakers is calculated as follows;

[tex]L = \sqrt{2^2 + 4^2} \\\\L = 4.47 \ m[/tex]

The wavelength of the sound wave is calculated as;

[tex]v = f\lambda\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{340}{1800} \\\\\lambda = 0.188 \ m[/tex]

Now, determine if the point is constructive interference, perfect destructive interference, or something in between?

[tex]x = \frac{\Delta L}{\lambda} \\\\x = \frac{4.47 - 4}{0.188} \\\\x = 2.5[/tex]

The result is within an integer value of +0.5, thus its a point of destructive interference.

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Answer:

6.71 × 10^8 mi/hr

Explanation:

Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.

To write it in the metric system, it has to be converted into miles/hour.

We know that,

1 minute = 60 seconds

60 minutes = 1 hour

1 kilometer = 1000 meter

1 miles = 1.6 kilometer

Now,

= [tex]\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}[/tex]

= 1.08 × 10^12 m/ hr (meter/hour)

= [tex]\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}[/tex]

= 6.71 × 10^8 mi/hr (miles/hour)

Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.

Final answer:

The speed of light in a vacuum is precisely known and has a fixed value of c = 2.99792458 × 10^8 m/s, roughly equal to 3.00 × 10^8 m/s. It's a fundamental constant in physics but is slower in materials, as defined by their index of refraction.

Explanation:

The speed of light in a vacuum is represented by the symbol c and has a fixed value of c = 2.99792458 × 108 m/s, which is essentially 3.00 × 108 m/s when rounded to three significant digits. This constant speed is a fundamental physical quantity and is crucial in many areas of physics including relativity and electromagnetism. It's important to note that the speed of light reduces when it passes through matter, which is characterized by the index of refraction n of the material.

Answer:

Total angular displacement will be 240 radian

Explanation:

In first case object starts from rest so initial angular speed [tex]\omega _i=0rad/sec[/tex]

Angular acceleration is given [tex]\alpha =2rad/sec^2[/tex]

Final angular speed[tex]\omega _f=24rad/sec[/tex]

From third equation of motion [tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]

So [tex]24^2=0^2+2\times 2\times \Theta[/tex]

[tex]\Theta =144[/tex] radian

Now in second case as the objects finally stops

So final velocity [tex]\omega _f=0rad/sec[/tex]

Angular acceleration [tex]\alpha =-3rad/sec^2[/tex]

So [tex]0^2=24^2-2\times 3\times \Theta[/tex]

[tex]\Theta =96[/tex] radian

So total angular displacement will be 96+144 = 240 radian

To find the angular displacement, we can analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. We can calculate the time and angular displacement in each phase using the formulas for angular speed and angular displacement. Finally, we can add the angular displacements from both phases to find the total angular displacement.

To find the angular displacement, we need to analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. In the first phase, the object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. We can use the formula:

Final Angular Speed = Initial Angular Speed + Angular Acceleration * Time

Using this formula, we can find the time taken in the first phase. Then, we can calculate the angular displacement during the first phase using the formula:

Angular Displacement = Initial Angular Speed * Time + 0.5 * Angular Acceleration * Time^2

In the second phase, the object decelerates at a rate of -3 rad/s^2 until it stops. We can use the same formulas to find the time and angular displacement in the second phase. Finally, we can add the angular displacements from both phases to get the total angular displacement.

The total angular displacement will be the sum of the angular displacements in the two phases.

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Answer:

Force = 3481.1 N.

Explanation:

Below is an attachment containing the solution.

Answer:

A. [tex]|\vec v_t|=52.42\ m/s[/tex]

B. [tex]\theta=256.74^o[/tex]

Explanation:

Velocity Vector

The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.

We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as

[tex]v_{xp}=|v_p|cos\alpha_p[/tex]

[tex]v_{yp}=|v_p|sin\alpha_p[/tex]

[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]

[tex]v_{xp}=39cos\ 270^o=0[/tex]

[tex]v_{yp}=39sin\ 270^o=-39[/tex]

Thus, the velocity of the plane is

[tex]\vec v_p=0\hat i-39\hat j[/tex]

The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):

[tex]v_{xw}=|v_w|cos\alpha_w[/tex]

[tex]v_{yw}=|v_w|sin\alpha_w[/tex]

[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]

[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]

Thus, the velocity of the wind is

[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]

Now we perform the vector addition to compute the plane's final speed

[tex]\vec v_t=\vec v_p+\vec v_w[/tex]

[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]

[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]

A) The magnitude of the total velocity is

[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]

[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]

B) The direction angle is given by

[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]

[tex]\theta=arctan\ 4.24[/tex]

[tex]\theta=76.74^o[/tex]

This angle is west of south, we must add 180° to express it in due reference, thus

[tex]\boxed{\theta=256.74^o}[/tex]

Answer

given,

width of trapezoidal channel, b = 6 ft

depth of water, d = 3 ft

discharge,Q = 250 ft³/s

now, we have to calculate Froude number

[tex]F_r = \dfrac{Q}{A\sqrt{gD}}[/tex]

Where D is the hydraulic radius

 [tex]D = \dfrac{A}{P}[/tex]

[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]

P is the width of the channel

P = b + 2 z d

P = 6 + 2 x 1 x 3

P = 13 ft

A = d(b + z d) = 3 (6 + 3) = 27 ft²

g = 32.2 ft/s²

now,

[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]

[tex]F_r = \dfrac{250}{27\sqrt{32.2\times \dfrac{27}{13}}}[/tex]

 F_r = 1.087

 F_r > 1

Froude number is greater than 1 so, the flow is Super critical flow.

Answer:

t= 22.9ºC

Explanation:

Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:

Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)

This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:

1) The heat needed to reach in solid state to 0º, as ice:

Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J

2) The heat needed to melt all the ice, at 0ºC:

Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J

3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:

Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)

So, the total heat gained by the ice  is as follows:

Qti = Qi + Qf + Qiw

⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)

As (1) and (2) must be equal each other, we have:

22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)

⇒ 22753 J + 240.9*t = 81684 J -2334*t

⇒ 2575*t = 81684 J- 22753 J = 58931 J

⇒ [tex]t= \frac{58931J}{2575 J/C} = 22.9C[/tex]

⇒ t = 22.9º C

Answer:

given,

reaction time.t_r = 0.50 s

deceleration of the car = 6 m/s²

initial speed,v = 20 m/s

distance at which the car stop = ?

distance travel by the car in reaction time

 d= v x t_r

 d = 20 x 0.5 = 10 m

using equation of motion

distance travel to stop the car

v² = u² + 2 a s

0² = 20² - 2 x 6 x s

 12 s = 400

 s = 33.33 m

Total distance travel by the car

D = 10 + 33.33

D = 43.33 m

Hence, the car stops before to avoid collision.

Answer:

38.6 mi/h

Explanation:

7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s

Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is

t = 1/v

Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means

t - 15 = 1/(v+0.00206)

We can substitute t = 1/v to solve for v

[tex]\frac{1}{v} - 15 = \frac{1}{v + 0.00206}[/tex]

We can multiply both sides of the equation with v(v+0.00206)

v+0.00206 - 15v(v+0.00206) = v

[tex]-15v^2 - 0.0308v + 0.00206 = 0[/tex]

[tex]v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}[/tex]

[tex]v= \frac{0.03083\pm0.35}{-30}[/tex]

v = -0.01278 or v = 0.01 0724 mi/s

Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h

Answer:

3.72 m/s²

Explanation:

Horizontal component of the force applied = Fcosθ = 86 cos 30 = 74.478 N

Force = mass × acceleration = 74.478 N

ma = 74.478 N

a = 74.478 / 20 since friction can be neglected = 3.72 m/s²

Answer:

[tex]3.7m/s^{2}[/tex]

Explanation:

The forces acting on the cart is displayed in the attached file below.

Since the motion is only along the horizontal, we can conclude that the force bringing about that motion is the horizontal force.And from second newton law of motion, we deduce that

F=ma,

F=force, m=mass and a=acceleration

from the given question,

F=86N, mass,m=20kg,

Hence we can write the horizontal component of the force as

[tex]F_{X}=Fcos\alpha \\F_{x}=86cos30\\F_{x}=74.48\\Hence \\a=\frac{F_{x}}{m} \\a=\frac{74.48}{20}\\ a=3.7m/s^{2}[/tex]

Answer:

The muzzle speed u =200 m/s

Explanation:

Let u be the muzzle speed.

maximum height s= 500 m

Assuming angle of elevation to be 30°

then initial vertical speed  = usin30° = u/2

moreover, when the bullet reaches the maximum height its vertical velocity will be zero.

then using

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

a= 10 m/s^2

u= u/2

v= 0

s=500

we get

[tex]0^2=u^2/4-2\times10\times500[/tex]

u= 200 m/s.

Therefore, muzzle speed = 200 m/s

Answer:

The muzzle speed is 197.9 m/s.

Explanation:

Given that,

Maximum height = 500 m

Suppose the angle is 30°.

We need to calculate the muzzle speed

Using formula of maximum height

[tex]h_{max}=\dfrac{u^2}{2g}[/tex]

Where, h = maximum height

g = acceleration due to gravity

u = initial vertical velocity

Put the value into the formula

[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]

[tex]u^2=8\times500\times9.8[/tex]

[tex]u=\sqrt{4\times500\times9.8}[/tex]

[tex]u=197.9\ m/s[/tex]

Hence, The muzzle speed is 197.9 m/s.

Answer:

V₂  =541.94 m L.

Explanation:

Given that

V₁ = 500 mL  

T₁ = 25°C  = 273 + 25 = 298 K

T₂ = 5°C = 273+50 =323  K

The final volume = V₂  

We know that ,the ideal gas equation  

If the pressure of the gas is constant ,then we can say that

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]

The final volume of the balloon will be 541.94 m L.

V₂  =541.94 m L.

The final volume of the balloon will be "541.94 mL".

Given:

Volume,

Temperature,

            [tex]= 273+25[/tex]

            [tex]= 298 \ K[/tex]

             [tex]=273+50[/tex]

             [tex]=323 \ K[/tex]

By using the Ideal gas equation, we get

→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]

or,

→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]

       [tex]= \frac{500\times 323}{298}[/tex]

       [tex]= 541.94 \ mL[/tex]

Thus the above approach is correct.  

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Answer:

a) 15.864s

b) 392.78m

c) 29.52 m/s

Explanation:

The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is

distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)

= 24 + 21 + 26 + 4.5 = 75.5 m

As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck

[tex]s = at^2/2[/tex]

[tex]75.5 = 0.6t^2/2[/tex]

[tex]t^2 = 251.67[/tex]

[tex]t = \sqrt{251.67} = 15.864s[/tex]

b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time

= 75.5 + 20*15.864 = 392.78 m

c) final speed of the car is the initial speed plus the change in speed

[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]

To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.

u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.

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Answer:

Explanation:

Given

[tex]v_x(t)=\alpha -\beta t^2[/tex]

[tex]\alpha =4\ m/s[/tex]

[tex]\beta =2\ m/s^3[/tex]

[tex]v_x(t)=4-2t^2[/tex]

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

[tex]\int dx=\int \left ( 4-2t^2\right )dt[/tex]

[tex]x=4t-\frac{2}{3}t^3[/tex]

acceleration of object

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

[tex]a=-4t[/tex]

(b)For maximum positive displacement velocity must be zero at that instant

i.e.[tex]v=0[/tex]

[tex]4-2t^2=0[/tex]

[tex]t=\pm \sqrt{2}[/tex]

substitute the value of t

[tex]x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}[/tex]

[tex]x=3.77\ m[/tex]

The definitions of acceleration and velocity allow to find the results for the questions about the motion of the particle are:

    A) the function of the acceleration is: a = -4t

         and the position function is:   x = 4 t - ⅔ t³

    B) The maximum displacement is: x = 3.77 m

Given parameters

To find

    a) position and acceleration as a function of time,

    b) maximum displacement,

The acceleration of defined as the change in velocity with time.

      a = [tex]\frac{dv}{dt}[/tex]  

Let's calculate.

      a = - 2βt  

      a = - 2 2 t

      a = -4 t

The speed is defined by the variation of the position with respect to time.

       v = [tex]\frac{dx}{dt}[/tex]  

       dx = v dt

We integrate.

      ∫ dx = ∫ v dt

      x - x₀ = ∫ (α - β t²)

      x-x₀ = αt - βt³/ 3

we substitute.

      x = 4 t - ⅔ t³

B) To find the maximum displacement we use the first derivative to be zero.

     [tex]\frac{dx}{dt}[/tex]  = 0

      4 - 2t² = 0        

      t² = 2

       t = √2 = 1.414 s

Let's find the position for this time.

     x = 4 √2 - ⅔ (√2)³

     x = 3.77 m

In conclusion using the definitions of acceleration and velocity we can find the result for the questions about the motion of the particle are:

    A) the function of the acceleration is: a = -4t

        and the position function is: x = 4 t - ⅔ t³

    B) The maximum displacement is: x = 3.77 m

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Answer:

The angle is 2.33°.

Explanation:

Given that,

Speed of ball = 12 m/s

Acceleration = 0.4 m/s²

We need to calculate the time

Using formula of time of flight

[tex]t=\dfrac{2u}{g}[/tex]

[tex]t=\dfrac{2v\cos\theta}{g}[/tex]

Put the value into the formula

[tex]t=\dfrac{2\times12\cos\theta}{9.8}[/tex]

[tex]t=2.44\cos\theta[/tex]

We need to calculate the angle

Using equation of motion along vertical direction

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

[tex]s=v\sin\theta\times t-\dfrac{1}{2}at^2[/tex]

Put the value in the equation

[tex]0=12\sin\theta\times2.44\cos\theta-\dfrac{1}{2}\times0.4\times(2.44\cos\theta)^2[/tex]

[tex]2\times12\sin\theta\times2.44=0.4\times(2.44)^2\cos\theta[/tex]

[tex]\tan\theta=\dfrac{0.4\times2.44}{2\times12}[/tex]

[tex]\theta=\tan^{-1}(0.04066)[/tex]

[tex]\theta=2.33^{\circ}[/tex]

Hence, The angle is 2.33°.

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

Answer:

Bulb A has a greater resistance.

Explanation:

Electric power (P) = V²/R

P = V²/R................ Equation 1

Where P = power, V = Voltage, R = Resistance.

Make R the subject of the equation

R = V²/P ................ Equation 2

For Bulb A,

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

For bulb B

Given: V =120 V, P = 100 W.

Substitute into equation 2

R = 120²/100

R = 14400/100

R = 144 Ω

Hence Bulb A has a greater resistance.

bulb A has the greater filament resistance.

This can be defined as the ability of a conducting material to oppose the flow of electric current.

To know the bulb with the greatest filament resistance, we use the formula below.

Formula:

making R the subject of the equation

Where:

For Bulb A,

Given:

Substitute these values into equation 1

For Bulb B,

Given:

Substitute these values into equation 1

Hence, From the above, it can be seen that bulb A has the greater filament resistance.

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Answer:

[tex]E=158.19\frac{kN}{m}[/tex]

Explanation:

Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:

[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]

The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):

[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]

Recall that the surface charge density is defined as:

[tex]\sigma=\frac{q}{S}[/tex]

Now, we get the electric field strength:

[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]

The electric field strength just outside the surface of a conducting sphere with surface charge density of 1.4 μC/m^2 can be calculated using Gauss' Law. This gives an electric field strength of approximately 1.58 x 10^5 N/C.

The electric field strength just outside the surface of a conducting sphere, with a surface charge density of 1.4 μC/m2, can be calculated using Gauss' Law, which states that the electric field is equal to the surface charge density divided by the permittivity of free space, ε0.

It is given by the formula:

E = σ/ε0

Where E is the electric field strength, σ is the charge density, and ε0 is the permittivity of free space. Using the given value for σ (1.4 x 10-6 C/m2) and the known value for ε0 (8.85 x 10-12 C2/N.m2), we find that:

E = (1.4 x 10-6) / (8.85 x 10-12)

This simplifies to approximately E = 1.58 x 105 N/C.

Therefore, the electric field strength just outside the surface of the conducting sphere is approximately 1.58 x 105 N/C.

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Final answer:

The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.

Explanation:

To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.

To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.

Answer:

324.066096 m.

Explanation:

Given that

height of the tower ,h= 324 m

The increase in temperature ,ΔT = 17°C

We know that coefficient of thermal expansion for steel ,α= 12 x 10⁻⁶ C⁻¹

The increase in height is given as

Δ h = α h ΔT

Now by putting the values in the above equation we get

Δ h= 12 x 10⁻⁶ x 324 x 17 m

Δ h=66096 x 10⁻⁶ m

Δ h=0.066096 m

Therefore the height of the tower become 324.066096 m.

Due to thermal expansion, the Eiffel Tower, made of steel, would increase in height by approximately 0.066 meters or 6.6 cm over a day when the temperature increases by 17°C.

The height of the Eiffel Tower increases due to the phenomenon of thermal expansion, which is an increase in volume, including height, in response to an increase in temperature. The amount of expansion can be calculated using this formula: ΔL = α * L_original * ΔT. Let's apply the given values:

So ΔL = 0.000012 * 324 * 17, which equates to approximately 0.066048 m. Therefore, the Eiffel Tower would increase in height by about 0.066 meters (or 6.6 cm) over the course of a day when the temperature increases by 17°C.

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Answer:

a) 66.4 relative to the west in the south-west direction

b) 5.455 hours

Explanation:

a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of

[tex]cos(\alpha) = 40 / 100 = 0.4[/tex]

[tex]\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0[/tex] relative to the West

b) The south-component of the geese velocity is

[tex]100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h[/tex]

The time it would take for the geese to cover 500km at this rate is

t = 500 / 91.65 = 5.455 hours

Answer:

[tex]\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]

Explanation:

The x- and y- components of the velocity vector can be written as following:

[tex]\vec{v}_x = ||\vec{v}||\cos(\theta)\^i[/tex]

[tex]\vec{v}_y = ||\vec{v}||\sin(\theta)\^j[/tex]

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

[tex]\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]

The rate constant at 75°C is calculated using the two-point form of the Arrhenius equation. The original conditions, the new temperature, and the activation energy are substituted into the equation and solved for the new rate constant, k2. The result is k2 = 0.048, or 4.8 x 10^-2 s^-1.

For calculating the rate constant at a different temperature, we can use the Arrhenius equation: k = Ae^(-Ea/RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the temperature in Kelvin.

To find the new temperature constant, we can transform the Arrhenius equation into the two-point form: ln(k2/k1) = (-Ea/R)(1/T2 - 1/T1).

Given:
k1 = 7.8 × 10−3 s−1,  T1=25°C = 25 + 273 = 298K
Ea = 33.6 kJ/mol = 33,600 J/mol,  R = 8.314 J/(mol·K)
T2 = 75°C = 75 + 273 = 348K

Substituting these values and solving for k2 (rate constant at 75°C), you get k2 = 0.048 or 4.8 x 10^-2 s^-1.

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Complete question:

Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer:

Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV

Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz

Explanation:

The kinetic energy (KE) of the emitted photon:

KE = 0.5mv²

m is mass of electron = 9.1 X 10⁻³¹ kg

KE = 0.5 * 9.1 X 10⁻³¹  * (460000)² = 9.628 X 10⁻²⁰ J

in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV

The photon energy of the incoming radiation:

E = hf = hc/λ

c is speed of light (photon) = 3 x 10⁸

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)

E = 31.805 X 10⁻²⁰ J

in eV = 31.805 X 10⁻²⁰ J x  6.242 X 10¹⁸ ev = 1.985 eV

Part (a) the work function of the surface

KE = hf - W

where;

W is work function

W = hf - KE

W =  31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J

in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV

Part(b) the cutoff frequency for this surface

W =hf

f = W/h

f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)

f = 3.347 x 10¹⁴ Hz

The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.

The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.

The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.

The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected

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Answer:

a)[tex]a=5 i+2t j - 6\ t^2k[/tex]

b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k[/tex]

[tex]a=5 i+2t j - 6\ t^2k[/tex]

Therefore the acceleration function a will be

[tex]a=5 i+2t j - 6\ t^2k[/tex]

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

[tex]a=\sqrt{5^2+4^2+24^2}\ m/s^2[/tex]

a= 24.83 m/s²

The direction of the acceleration a is given as

[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]

a)[tex]a=5 i+2t j - 6\ t^2k[/tex]

b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]

Answer:

[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]

Explanation:

Given that,

The ray makes an angle of 32 degrees with respect to the normal of the surface.

The refractive index of air, [tex]n_1=1[/tex]

The refractive index of water, [tex]n_2=1.33[/tex]

Snell's law is given by :

[tex]n_1\ sin\theta_1=n_2\ sin\theta_2[/tex]

[tex]sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}[/tex]

[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]

So, option (4) is correct. Hence, this is the required solution.    

The answer is option d.

The correct expression for the angle (relative to the normal to the surface) for the ray in the water is d) θ2 = asin (sin(θ1).n1/n2), based on Snell's law of refraction.

The question addresses the refraction of light, specifically the change in angle as light moves from air to water. According to Snell's law, which is used to calculate the angle of refraction, the correct expression in your options is d) θ2 = asin (sin(θ1).n1/n2). Here's a step by step process:

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Answer:

Explanation:

Option a is correct  

If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .

Option e is correct

If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.

Friction will cause the energy to dissipate in heat.

To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.

A nanosecond is one billionth of a second while and a millisecond is one millionth of a second

[tex]\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6[/tex]

Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds

Word magnitude means the extent of something. It is the property that determines the object is larger or smaller than the other. The object's magnitude can be arranged into class.

The correct answer is 100,00,00.

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