A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 21.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.10 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 30.0 m before hitting the water.

After leaving the edge of the cliff how much time does the diver take to get to the water?



How far horizontally does the diver travel from the cliff face before hitting the water?



Remember that the angle is at a downward slope to the right.

Answer:

F_x = 750.7 N

Explanation:

Given:

- Length of the pipe between flanges L = 75 cm

- Weight Flow rate is flow(W) = 230 N/c

- P_1 = 165 KPa

- P_2 = 134 KPa

- P_atm = 101 KPa

- Diameter of pipe D = 0.05 m

Find:

The total force that the flanges must withstand F_x.

Solution:

- Use equation of conservation of momentum.

            (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = flow(m)*( V_2 - V_1)

- From conservation of mass:

                                       A*V_1 = A*V_2

                       V_1 = V_2 ( but opposite in directions)

- Hence,

             (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(m)*V_1

                                     flow(m) = flow(W) / g

                                     p*A*V_1 = flow(W) / g

                                    V_1 =  flow(W) / g*p*A    

Hence,      

             (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(W)^2 / g^2*p*A

Hence, compute:

   64*10^3 *pi*0.05^2 /4 + 33*10^3 *pi*0.05^2 /4 - F_x = - 2*(230/9.81)^2 / 997*pi*0.05^2 /4

                            125.6 + 64.7625 - F_x = -560.33

                                        F_x = 750.7 N

Answer:

It will go 96 times around the track.

Explanation:

Given that,

Distance covered by the race car, d = 765 km

Radius of the circular sprint track, r = 1.263 km

Let n times did it go around the track. It is given by :

[tex]n=\dfrac{d}{C}[/tex]

C is the circumference of the circular path, [tex]C=2\pi r[/tex]

[tex]n=\dfrac{d}{2\pi r}[/tex]

[tex]n=\dfrac{765}{2\pi \times 1.263}[/tex]

[tex]n=96.4[/tex]

Approximately, n = 96

So, it will go 96 times around the track. Hence, this is the required solution.

With a track radius of 1.263 km, the car completes approximately 96.50 laps.

The formula for the circumference (C) of a circle is: C = 2πr, where r is the radius of the circle, and π (pi) is approximately 3.14159. Using the given radius of 1.263 km, we can calculate the circumference of the track:

C = 2π(1.263 km) ≈ 2(3.14159)(1.263 km) ≈ 7.932 km (rounded to three decimal places).

Divide the total distance traveled by the circumference of the track:

Number of laps = Total distance traveled ÷ Circumference of the track
Number of laps = 765 km ÷ 7.932 km ≈ 96.50 laps.

Therefore, the race car would have completed approximately 96.50 laps around the track.

Answer:

Q'sphere=2.7*10^-9 C

Q'rod=-4.7*10^-9 C

Explanation:

given data:

charge on metallic sphere Qsphere=3.1*10^-9 C                ∴1n=10^-9

charge on rod Qrod =-4*10^-9 C  

no of electron n= 9.2×10^9 electrons

To find:

we are asked to find the charges Q'sphere on the sphere and Q'rod on the rod after the rod touches the sphere.

solution:

the total charge transferred when the rod touches the sphere equal to the no of electrons transferred multiplied by the charge of each electron:

Q(transferred)= nq_(e)

                       =(9.2×10^9)(1.6×10^-19)

                       =-1.312×10^-9 C

because electron are negative they move from the negatively charged rod to the positively charged rod so that new charged of the sphere is:

      Q'sphere =Qsphere+Q(transferred)

                       =(3.1*10^-9 )-(1.312×10^-9)

                       =2.7*10^-9 C

similarly the new charge of the rod is:

            Q'rod = Qrod-Q(transferred)

                      = (-6*10^-9 C)-(1.312*10^-9 C)

                      = -4.7*10^-9 C

∴note: there maybe error in calculation but the method is correct.

Final answer:

Upon contact, a metallic sphere and a negatively charged rod share their charges until equilibrium. The total charge of -0.9 nC is equalized, with 9.2×109 electrons changing the sphere's charge to +1.628 nC and the rod's charge to -2.528 nC.

Explanation:

When two charged objects come into contact, they share their charges until equilibrium is reached. This means each object will end up with the average charge. In the case of the metallic sphere with a charge of +3.1 nC and the negatively charged rod with a charge of -4.0 nC, the total charge before contact is (-4.0 nC) + (+3.1 nC) = -0.9 nC.

Since 9.2×109 electrons are transferred, we calculate the charge transferred using the charge of one electron, which is approximately -1.6×10-19 C. Multiplying the number of electrons by the charge of one electron gives us the total charge transferred: 9.2×109 × -1.6×10-19 C/electron ≈ -1.472 nC.

This charge is added to the metallic sphere and subtracted from the rod. So, the new charge on the sphere is +3.1 nC + (-1.472 nC) = +1.628 nC, and the charge on the rod is -4.0 nC - (-1.472 nC) = -2.528 nC. Both charges are now closer in magnitude, representing the sharing of charges due to contact.

Answer:

The correct option is 1.  720 m

Explanation:

Projectile Motion

When an object is launched in free air (no friction) with an initial speed vo at an angle [tex]\theta[/tex], it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.

[tex]\displaystyle V_o=250\ m/s[/tex]

[tex]\displaystyle \theta =-37^o[/tex]

The initial velocity has these components in the x and y coordinates respectively:

[tex]\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s[/tex]

[tex]\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s[/tex]

And we know the plane has an altitude of 600 m, so the package will reach ground level when:

[tex]\displaystyle y=-600\ m[/tex]

The vertical distance traveled is given by:

[tex]\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600[/tex]

We'll set up an equation to find the time when the package lands

[tex]\displaystyle -150.5t-4.9\ t^2=-600[/tex]

[tex]\displaystyle -4.9\ t^2-150.5\ t+600=0[/tex]

Solving for t, we find only one positive solution:

[tex]\displaystyle t=3.6\ sec[/tex]

The horizontal distance is:

[tex]\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m[/tex]

The correct option is 1.  720 m

The horizontal distance of the point of impact from the plane at the moment of the package's release is approximately 1760 m.

The time taken for the package to reach the ground can be found using the equation y = v0y * t + (1/2) * g * t2, where y is the initial altitude, v0y is the vertical component of the initial velocity, t is the time taken, and g is the acceleration due to gravity. Solving for t gives us a value of approximately 5 seconds. The horizontal distance traveled by the package can be found using the equation x = v0x * t, where x is the horizontal distance, v0x is the horizontal component of the initial velocity, and t is the time taken. Plugging in the values gives us x = 250 m/s * cos(37º) * 5 s, which simplifies to approximately 1760 m. So the horizontal distance of the point of impact is approximately 1760 m.

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Answer:

3.5 amperes

Explanation:

I = E/R

I = ?

E = 70volts

R = 20 Ohms

Therefore , I = 70/20

= 3.5 amperes

Answer: 7.2V

Explanation:

Peak values or peak to peak voltage are calculated from RMS values, which implies VP = VRMS × √2, (assuming the source is a pure sine wave).

Since it's a sinusoidal voltage and measuring from an oscilloscope, the peak to peak voltage is gotten thus:

No of division X Volts/divisions

So, 3.6 x 2V = 7.2V

Final answer:

The peak-to-peak voltage of a sinusoidal signal covering 3.6 divisions on an oscilloscope set to 2 volts per division is 7.2 volts.

Explanation:

The question involves calculating the peak-to-peak voltage of a sinusoidal signal observed on an oscilloscope where the vertical scale is set to 2 volts per division. Given that the signal covers 3.6 divisions from positive to negative peak, we calculate the peak-to-peak voltage by multiplying the number of divisions the signal spans by the voltage per division.

To find the peak-to-peak voltage, we use the formula: Peak-to-Peak Voltage = Number of Divisions × Voltage per Division. Thus, the peak-to-peak voltage of the signal is 3.6 divisions × 2 volts/division = 7.2 volts.

The work done by the electric force on the moving point charge is approximately -5.09 × 10^-5 J.

Work done by the electric force is given by the equation W = q1 * q2 * (1/r1 - 1/r2), where q1 and q2 are the charges, r1 is the initial distance, and r2 is the final distance.

In this case, q1 = 2.30 μC, q2 = -5.00 μC, r1 = 0.170 m, and r2 = 0.250 m. Plugging these values into the equation and solving for W, we get:

W = (2.30 μC) * (-5.00 μC) * [1/√(0.170^2) - 1/√(0.250^2 + 0.250^2)]

After simplifying, the work done is approximately -5.09 × 10^-5 J.

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Answer:

Explanation:

Given

mass of first child [tex]m_1=20\ kg[/tex]

mass of second child [tex]m_2=30\ kg[/tex]

Distance between two children is [tex]d=3\ m[/tex]

Suppose light weight child is placed at a distance of x m from Pivot point

therefore

Torque due to heavy child [tex]T_1=m_2g\times (3-x)[/tex]

Torque due to small child [tex]T_2=m_1g\times x[/tex]

Net Torque about Pivot must be zero

Therefore [tex]T_1=T_2[/tex]

[tex]30\times g\times (3-x)=20\times g\times x[/tex]

[tex]9-3x=2x[/tex]

[tex]9=5x[/tex]

[tex]x=\frac{9}{5}[/tex]

[tex]x=1.8\ m[/tex]

Answer:

0.0473m

Explanation:

345 ml = 0.000354 m3

6.5 cm = 0.065 m

20g = 0.02 kg

Since can is half filled with water, the water volume is 0.000354 / 2 = 0.000177 m cubed

Let water density be 1000kg/m3, the mass of this half-filled water is

1000*0.000177 = 0.177 kg

The total water-can system mass is 0.177 + 0.02 = 0.197 kg

For the system to stay balanced, this mass would be equal to the mass of the water displaced by the can submerged

The volume of water displaced, or submerged can is

0.197 / 1000 = 0.000197 m cubed

Then the volume of the can that is not submerged, aka above water level is

0.000354 - 0.000197 = 0.000157 m cubed

The base area of the can is

[tex]A = \pi r^2 = \pi (d/2)^2 = \pi (0.065)^2 = 0.003318 m squared[/tex]

The length of the can that is above water is

0.000157 / 0.003318 = 0.0473 m

The half-filled soda can displaces the equivalence of its own weight in water when placed in it. Half of the soda can's total volume will always be submerged since it is only half-filled i.e., half of the can's mass is displacing water.

The subject of this problem is the principles of buoyancy and volume. A half-filled soda can placed in water will displace its own weight of the water. The length of the can above the water level can be calculated using an understanding of volume and displacement.

First, calculate the volume of the can using the formula for the volume of a cylinder V = πr²h, where r is radius and h is height. Given the diameter of the can is 6.5 cm, the radius is 3.25 cm. The height can be calculated by rearranging the volume formula to find h. We know that the can's complete volume is 345 mL, so h (full can height) = V / (πr²).

From this, we can calculate the height of the can that is submerged in water. Since the can is half-filled, it displaces half its full weight in water. So half of the can's total volume will always be submerged. Therefore, the length of the can above the water will be half the total height of the can.

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Answer:

Average acceleration = - 2 m/s^2

Explanation:

Given data:

Initial velocity = 11 m/s

Final velocity = 5.0 m/s

duration of change in velocity = 3 sec

Average acceleration [tex]= \frac{v - u}{\Delta t}[/tex]

Average acceleration [tex]= \frac{5 - 11}{3} = -2 m/s^2[/tex]

Average acceleration = - 2 m/s^2

here negative sign indicate that acceleration is proceed in downward direction.

The sign of the point charge that produces a potential of -2.00 V at a distance of 1.00 mm is negative, and its magnitude is calculated to be approximately -2.22×10-13 C using Coulomb's law.

Explanation:

The question pertains to determining the sign and magnitude of a point charge based on the electric potential it produces at a specific distance. The electric potential (V) at a distance (r) from a point charge (q) is given by the equation V = k * q / r, where k is Coulomb's constant (k = 8.988×109 N·m2/C2). Since the potential is negative (-2.00 V), the point charge must have a negative sign. To find the magnitude, we rearrange the formula to solve for q: q = V * r / k.

Plugging in the values gives q = (-2.00 V * 1.00×10-3 m) / 8.988×109 N·m2/C2, which calculates to a charge magnitude of approximately -2.22×10-13 C.



The probable question is in the image attached.

Answer:

Cu = 1453.72J/Kg°C

The heat capacity of the liquid is 1453.72J/Kg°C

Explanation:

At equilibrium, assuming no heat loss to the surrounding we can say that;

Heat gained by ice + heat gained by cold water = heat loss by hot unknown liquid (24°C) + heat loss by aluminium calorimeter.

Given;

Mass of ice = mass of cold water = mc = 30g = 0.03kg

Mass of hot unknown liquid mh= 300g = 0.3kg

Mass of aluminium calorimeter ma= 100g = 0.1 kg

change in temperature cold ∆Tc = (4-0) = 4°C

Change in temperature hot ∆Th = 24-4 = 20°C

Specific heat capacity of water Cw= 4186J/Kg°C

Specific heat capacity of aluminium Ca = 900J/kg°C

Specific heat capacity of unknown liquid Cu =?

Heat of condensation of ice Li = 334000J/Kg

So, the statement above can be written as.

mcLi + mcCw∆Tc = maCa∆Th + mhCu∆Th

Making Cu the subject of formula, we have;

Cu = [mcLi + mcCw∆Tc - maCa∆Th]/mh∆Th

Substituting the values we have;

Cu = (0.03×334000 + 0.03×4186×4 - 0.1×900×20)/(0.3×20)

Cu = 1453.72J/Kg°C

the heat capacity of the liquid is 1453.72J/Kg°C

Answer:

22.8077659955 m deep

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 6.1+0^2}\\\Rightarrow v=10.9399268736\ m/s[/tex]

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{10.9399268736-0}{9.81}\\\Rightarrow t=1.11518112881\ s[/tex]

Time taken to fall through the foam

[tex]3.2-1.11518112881=2.08481887119\ s[/tex]

Distance is given by

[tex]s=vt\\\Rightarrow s=10.9399268736\times 2.08481887119\\\Rightarrow s=22.8077659955\ m[/tex]

The vat is 22.8077659955 m deep

The depth of the vat obtained is 44.076 m

H = ½gt²

H = ½ × 9.8 × 3.2²

H = 4.9 × 10.24

H = 50.176 m

Depth of vat = H – h

Depth of vat = 50.176 – 6.10

Depth of vat = 44.076 m

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Answer:

a) 2.67 m/s

b) 0.82 J

Explanation:

Amplitude A = 25 cm = 0.25 m

The period of the motion is the inverse of the frequency

[tex]T = \frac{1}{f} = \frac{1}{1.7} = 0.588 s[/tex]

So the angular frequency

[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{0.588} = 10.68 rad/s[/tex]

The speed at the equilibrium point is the maximum speed, at

[tex]v = \omega A = 10.68 * 0.25 = 2.67 m/s[/tex]

The spring constant can be calculated using the following

[tex]\omega^2 = \frac{k}{m} = \frac{k}{0.23}[/tex]

[tex]k = 0.23\omega^2 = 0.23*10.68^2 = 26.24 N/m[/tex]

The total energy of the oscillation is

[tex]E = kA^2 / 2 = 26.24*0.25^2 / 2 = 0.82 J[/tex]

Answer:

Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]                                                

Explanation:

Given that,

Electric field, E = 680 N/C

Speed of the proton, v = 1.3 Mm/s

We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :

[tex]qE=ma[/tex]

a and m are the acceleration and mass of the proton.

[tex]a=\dfrac{qE}{m}[/tex]

[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]

[tex]a=6.51\times 10^{10}\ m/s^2[/tex]

So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.

The mathematical description that fits to find the circumference of a circle (Approximation we will make to the earth considering it Uniform) is,

[tex]\phi = 2\pi r[/tex]

Here,

r = Radius

The radius of the earth is 6378.01 km or 6378010m

Replacing we have that the circumference of the Earth is

[tex]\phi = 2\pi (6378010m)[/tex]

[tex]\phi = 40074000 m[/tex]

[tex]\phi = 40074*10^3 m[/tex]

Therefore the circumference of the Earth in m with 5 significant figures is [tex]40074*10^3 m[/tex] and using only trailing zeros the answer will be [tex]40074000m[/tex]

Answer:

(a) Magnitude =3.18 km

(a) Angle =28.2°

Explanation:

(a) To find magnitude of this vector recognize

[tex]R_{x}=-2.8 km\\R_{y}=-1.5km[/tex]

Use Pythagorean theorem

[tex]R=\sqrt{(R_{x})^{2}+(R_{y})^{2} }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2} }\\ R=3.18km[/tex]

(b)To find the angle use the trigonometric property

[tex]tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}[/tex]

Answer:

[tex]1.15669\times 10^{-7}\ C[/tex]

Explanation:

[tex]\theta[/tex] = Angle with which the electric field is hung = 23°

m = Mass of ball = 5 g

E = Electric field = [tex]1.8\times 10^5\ N/C[/tex]

T = Tension

q = Charge

We have the equations

[tex]Tcos\theta=mg[/tex]

[tex]Tsin\theta=qE[/tex]

Dividing the equations

[tex]tan\theta=\dfrac{mg}{qE}\\\Rightarrow q=\dfrac{mgtan\theta}{E}\\\Rightarrow q=\dfrac{5\times 10^{-3}\times 9.81\times tan23}{1.8\times 10^5}\\\Rightarrow q=1.15669\times 10^{-7}\ C[/tex]

The magnitude of the charge is [tex]1.15669\times 10^{-7}\ C[/tex]

Answer:

(a) The energy  extracted from the hot reservoir (Qh) is 3316.67J

(b) The energy deposited in the cold reservoir (Qc) is 2321.67J

Explanation:

Part (a) The energy  extracted from the hot reservoir (Qh)

e = W/Qh

where;

e is the maximum efficiency of the system = 30% = 0.3

W is the the work done on the system = 995 J

Qh is the heat absorbed from the hot reservoir

Qh = W/e

Qh = 995/0.3

Qh = 3316.67J

Part (b) The energy deposited in the cold reservoir (Qc)

e = W/Qh

W = Qh - Qc

where;

Qc is the heat deposited in the cold reservoir

e = (Qh - Qc)/Qh

Qh - Qc = e*Qh

Qc = Qh - e*Qh

Qc = 3316.67J - 0.3*3316.67J

Qc = 3316.67J - 995J

Qc = 2321.67J

The angular displacement of the solid disk after 5 seconds is 0 rad.

To determine the angular displacement of the solid disk after 5 seconds, we can use the formula:

Δθ = ΔErot / (I * ω)

where Δθ is the angular displacement, ΔErot is the change in rotational kinetic energy, I is the moment of inertia of the disk, and ω is the angular velocity of the disk.

The moment of inertia of a solid disk rotating around an axis through its center perpendicular to its plane is given by:

I = (1/2) * m * r2

where m is the mass of the disk and r is the radius.

Given that ΔErot = 20 J/s, m = 4 kg, r = 2 m, and the disk starts from rest, we can calculate the angular displacement:

Δθ = ΔErot / (I * ω) = 20 / [(1/2) * 4 * 22 * ω]

Since the disk starts from rest, the initial angular velocity ω is 0. Therefore, the angular displacement after 5 seconds is:

Δθ = 20 / [(1/2) * 4 * 22 * 0] = 0 rad

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The Poisson's ratio definition is given as the change in lateral deformation over longitudinal deformation. Mathematically it could be expressed like this,

[tex]\upsilon = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}[/tex]

[tex]\upsilon = \frac{(\delta a/a)}{(\delta l/l)}[/tex]

Replacing with our values we would have to,

[tex]\upsilon = \frac{(2.89933-2.9/2.9)}{(9.70710-9.7/97)}[/tex]

[tex]\upsilon = 0.1977[/tex]

Therefore Poisson's ratio is 0.1977.

Answer:

(a) [tex]T_{1}=490N[/tex]

(b) [tex]T_{2}=240N[/tex]

Explanation:

For Part (a)

Given data

The box moves up at steady 5.0 m/s

The mas of box is 50 kg

As ∑Fy=T₁ - mg=0

[tex]T_{1}=mg\\T_{1}=(50kg)(9.8m/s^{2} ) \\T_{1}=490N[/tex]

For Part(b)

Given data

[tex]v_{iy}=5m/s\\ a_{y}=-5.0m/s^{2}[/tex]

As ∑Fy=T₂ - mg=ma

[tex]T_{2}=mg+ma_{y}\\T_{2}=m(g+ a_{y})\\T_{2}=50kg(9.8-5.0) \\T_{2}=240N[/tex]

The tension in the rope varies depending on whether the box is at rest, moving at a constant velocity, or accelerating. The tension equals the weight of the box when it's at rest or moving constantly, but it will be increased or decreased by the net force caused by acceleration when the box is speeding up or slowing down.

If the box is at rest, the tension in the rope is equal to the force of gravity. We can calculate this using the formula T = mg, where m is the mass of the box and g is the acceleration due to gravity. Therefore, T = (50 kg)(9.8 m/s²) = 490 N.

When the box moves upwards with a constant velocity, the tension in the rope also equals the weight of the box (T = mg), so the tension will stay the same at 490 N.

However, when the box is speeding up, the net force is the product of mass and acceleration. In this case, acceleration = 5.0 m/s². Using the equation Fnet = ma, we find that Fnet = (50 kg)(5 m/s²) = 250 N. The total tension now includes both the tension due to the box's weight and the additional force due to the acceleration. Therefore, T = T(g) + Fnet = 490 N + 250 N = 740 N.

Lastly, when the box is slowing down at 5.0 m/s², the net force acts in the opposite direction of the initial velocity. Using the same calculations, we find Fnet = 250 N. But this force now reduces the tension originally caused by the box's weight, so the total tension in the rope becomes T = T(g) - Fnet = 490 N - 250 N = 240 N.

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Explanation:

Given that,

The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :

Since,

1 meter = 3.28084 feet

632 m = 2073.49 feet

1 yard= 3 feet

710 yards = 2130 feet

V = lbh

[tex]V=2073.49 \ ft\times 2130\ ft\times 112\ ft[/tex]

[tex]V=494651774.4\ ft^3[/tex]

[tex]V=4.94\times 10^8\ ft^3[/tex]

Also,

[tex]V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3[/tex]

[tex]V=1.39\times 10^7\ m^3[/tex]

Hence, this is the required solution.

Answer:

27.1806500378 V

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

q = Charge = [tex]1.6\times 10^{-19}\ C[/tex]

r = Distance = [tex]5.292\times 10^{-11}\ m[/tex]

Voltage is given by

[tex]V=k\dfrac{q}{r}\\\Rightarrow V=8.99\times 10^9\dfrac{1.6\times 10^{-19}}{5.292\times 10^{-11}}\\\Rightarrow V=27.1806500378\ V[/tex]

The potential difference is 27.1806500378 V

Answer:

(a) 5.6 m/s, 7.2 m/s, and 8.8 m/s, respectively.

(b) 0.8 m/s^2

(c) 4.8 m/s

(d) 6 s

(e) 5.2 m

Explanation:

(a) The average velocity is equal to the total displacement divided by total time.

For the first 2s. interval:

[tex]V_{\rm avg} = \frac{\Delta x }{\Delta t} = \frac{25.6 - 14.4}{2} = 5.6~{\rm m/s}[/tex]

For the second 2s. interval:

[tex]V_{\rm avg} = \frac{40 - 25.6}{2} = 7.2~{\rm m/s}[/tex]

For the third 2s. interval:

[tex]V_{\rm avg} = \frac{57.6 - 40}{2} = 8.8~{\rm m/s}[/tex]

(b) Every 2 s. the velocity increases 1.6 m/s. Therefore, for each second the velocity increases 0.8 m/s. So, the acceleration is 0.8 m/s2.

(c) The sled starts from rest with an acceleration of 0.8 m/s2.

[tex]v^2 = v_0^2 + 2ax\\v^2 = 0 + 2(0.8)(14.4)\\v = 4.8~{\rm m/s}[/tex]

(d) The following kinematics equation will yield the time:

[tex]\Delta x = v_0 t + \frac{1}{2}at^2\\14.4 = 0 + \frac{1}{2}(0.8)t^2\\t = 6~{\rm s}[/tex]

(e) The same kinematics equation will yield the displacement:

[tex]\Delta x = v_0t + \frac{1}{2}at^2\\\Delta x = (4.8)(1) + \frac{1}{2}(0.8)1^2\\\Delta x = 5.2~{\rm m}[/tex]

The frequency of the alternating current is 394/2π Hz. The resistance of the bulb's filament can be determined using Ohm's Law. The average power delivered to the light bulb can be calculated using the formula P = IV.

(a) The frequency of the alternating current can be calculated using the angular frequency formula ω = 2πf. In this case, the angular frequency is 394 rad/s. So, we can rearrange the formula to find the frequency: f = ω/2π = 394/2π Hz.

(b) The resistance of the bulb's filament can be determined using Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is 120.0 V and the current is given by I = (0.644 A) sin [(394 rad/s)t].

(c) The average power delivered to the light bulb can be calculated using the formula P = IV, where I is the current and V is the voltage. In this case, the voltage is 120.0 V and the current is given by I = (0.644 A) sin [(394 rad/s)t].

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Answer:

A) The expression of the electric field halfway between the plates, if the plates are in the plane Y-Z is:

[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]

B) The expression for the magnitude of the electric field E₂ just in front of the plate two ends is:

[tex]|E_2|=\displaystyle \frac{q}{2LW\varepsilon_0}[/tex]

C) The charge density is:

[tex]\sigma_2=-4.2517\cdot10^{-3}C/m^2[/tex]

Completed question:

A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.

A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates

B) Input an expression for the magnitude of the electric field, E₂. Just in front of plate two END

C) If plate two has a total charge of q =-1 mC, what is its charge density, σ in C/m2?

Explanation:

A) The expression of the field can be calculated as the sum of the field produced by each plate. Each plate can be modeled as 2 parallel infinite metallic planes. Because this is a capacitor connected by both ends to a battery, the external planes have null charge (the field outside the device has to be null by definition of capacitor). This means than the charge of each plate has to be distributed in the internal faces. Because this es a metallic surface and there is no external field, we can consider a uniform charge distribution (σ=cte). Therefore in this case for each plane:

[tex]\sigma_i=\displaystyle \frac{q_i}{LW}[/tex]

The field of an infinite uniform charged plane is:

[tex]\vec{E_i}=\displaystyle \frac{\sigma_i}{2\varepsilon_0}sgn(x-x_{0i})\vec{x} =\frac{q_i}{2LW\varepsilon_0}sgn(x-x_{0i})\vec{x}[/tex]

In this case, inside the capacitor, if the plate 1 is in the left and the plate 2 is in the right, the field for 0<x<d is:

[tex]\vec{E_1}\displaystyle=\frac{q}{2LW\varepsilon_0}sgn(x})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]

[tex]\vec{E_2}\displaystyle=\frac{-q}{2LW\varepsilon_0}sgn(x-d})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]

[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]

[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]

B) we already obtain the expression of the field E₂ inside the space between the plates. Even if we are asked the expression just in front of the plate and not inside, the expression for |E₂| is still de same.

C) As seen above, we already obtain the charge density expression. Therefore we only have to replace the variables for the numerical values.

The electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.

Given:
Length, L = 0.49 m

Width, W = 0.48 m

Distance, d = 0.1 m

Here:

E = electric field

σ = surface charge density

ε₀ = permittivity of free space

We must determine the surface charge density on each plate since the plates are wired to a battery and charged so that the first plate has a charge of q.

The area of the plate is:

Area of each plate (A) = L x W = 0.49 m x 0.48 m = 0.2352 m²

The surface charge density is given by:

σ = q / A

The electric field is computed as:

E = (σ / ε₀)

Hence, the electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.

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#Complete question is:

A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.

A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates

Answer:

D) The potential difference between the plates increases.

Explanation:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

Where ϵ0 is the permittivity of free space.  

A capacitor with increased distance, will have a new capacitance C1=ϵ0kA/d1

Where d1 = nd  

since d1 > d

therefore n >1

n is a factor derived as a result of the increased distance

Therefore the new capacitance becomes:

              C1=ϵ0A/d1

        C1= ϵ0A/nd

        C1= C/n  -------1

Where C1 is the capacitance with increased distance.

This implies that the charge storing capacity of the capacitor with increased plate separation decreases by a factor of (1/n) compared to  that of the capacitor with original distance.

Given points

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric  Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Charge before plate separation increase same as after plate separation increase

Q   = Q1

CV = C1V1

  CV = C1V1  -------2

We derived C1=C/n in equation 1. Inserting this into equation 2

   CV = (CV1)/n

   V1 = n(CV)/C

        = n V

Since n > 1 as a result of the derived new distance, the new voltage will increase

The student determine the actual speed of the object when it reaches the ground as 12.52 m/s.

Given data:

The distance from the Earth's surface is, = 10 m.

Time taken to reach the ground is, t = 1.43 s.

The speed of object is, v = 14.3 m/s.

Experimental value of time interval is, t' = 3.2 s.

Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

[tex]h=ut+\dfrac{1}{2}a't'^{2} \\\\10=0 \times t+\dfrac{1}{2} \times a' \times 3.2^{2} \\\\a'=\dfrac{20}{3.2^{2}}\\\\a'= 1.95 \;\rm m/s^{2}[/tex]

Now, use the principle of conservation of total energy of system:

Potential energy - work done by air resistance = Kinetic energy

[tex]mgh-(ma) \times h=\dfrac{1}{2}mv^{2} \\\\gh-(a) \times h=\dfrac{1}{2}v^{2} \\\\v=\sqrt{2h(g-a)}[/tex]

Here, v is the actual speed of object while reaching the ground.

Solving as,

[tex]v=\sqrt{2 \times 10(9.8-1.95)}\\\\v=12.52 \;\rm m/s[/tex]

Thus, we can conclude that  the student determine the actual speed of the object when it reaches the ground as 12.52 m/s.

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The actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]

The actual speed of the object when it reaches the ground can be determined using the kinematic equation that relates initial velocity, acceleration, and time to the final velocity. Since the object is released from rest, the initial velocity [tex]\( u \)[/tex] is 0 m/s, and the acceleration [tex]\( a \)[/tex] is due to gravity, which is approximately [tex]\( 9.81 \, m/s^2 \)[/tex] near the Earth's surface.

The kinematic equation that relates these quantities to the final velocity [tex]\( v \)[/tex] is:

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

Given that [tex]\( u = 0 \, m/s \)[/tex] and [tex]\( a = 9.81 \, m/s^2 \)[/tex], and the time [tex]\( t \)[/tex] to reach the ground is [tex]\( 3.2 \, s \)[/tex], we can substitute these values into the equation to find the actual final velocity:

[tex]\[ v = 0 + (9.81 \, m/s^2)(3.2 \, s) \] \[ v = (9.81)(3.2) \, m/s \] \[ v = 31.392 \, m/s \][/tex]

Therefore, the actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]

Answer:

The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

[tex]E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}[/tex]

[tex]E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}[/tex]

Second equation is

[tex]E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}[/tex]

We need to calculate the ratio of [tex]E_{act}[/tex] and [tex]E_{app}[/tex]

Using formula

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}[/tex]

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}[/tex]

Put the value into the formula

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}[/tex]

[tex]\dfrac{E_{app}}{E_{act}}=0.9754[/tex]

Hence, The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754