The wavelength corresponding to light with a frequency of 4 x 10^14 Hz is 1.33m 0.075 mm 7500 nm 750 nm

Answer:

2.18 s

Explanation:

H = - 25 m ( downwards)

U = - 0.8 m/s

g = - 9.8 m/s^2

Let time taken is t.

Use second equation of motion

H = u t + 1/2 g t^2

- 25 = - 0.8 t - 1/2 × 9.8 × t^2

4.9 t^2 + 0.8 t - 25 = 0

By solving we get

t = 2.18 s

(a) 1) The field is in the direction of the electron's initial velocity

The electric field is in a direction opposite to the initial velocity of the electron.

Let's remind that, when an electric charge is immersed in an electric field:

- if the charge is positive, the charge experiences a force in the same direction as the electric field direction

- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction

In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.

(b) [tex]4.13\cdot 10^{-4} m[/tex]

We have:

Electron's initial velocity: [tex]u=5.25\cdot 10^6 m/s[/tex]

Electric field magnitude: [tex]E=1.9 \cdot 10^5 N/C[/tex]

Electron charge: [tex]q=-1.6\cdot 10^{-19} C[/tex]

Mass of the electron: [tex]m=9.11\cdot 10^{-31}kg[/tex]

The electric force exerted on the electron is:

[tex]F=qE=(-1.6\cdot 10^{-19} C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^{-14}N[/tex] (the negative sign means the direction of the force is opposite to its initial velocity)

The electron's acceleration is given by:

[tex]a=\frac{F}{m}=\frac{3.04\cdot 10^{-14} N}{9.11\cdot 10^{-31} kg}=-3.34\cdot 10^{16} m/s^2[/tex]

Now we can use the SUVAT equation:

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

where

v = 0 is the final speed (the electron comes to rest)

d is the total distance travelled by the electron

Solving for d,

[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(5.25\cdot 10^6 m/s)^2}{2(-3.34\cdot 10^{16} m/s^2)}=4.13\cdot 10^{-4} m[/tex]

(c) [tex]1.57\cdot 10^{-10}s[/tex]

We can use the following equation:

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

where we have

[tex]a=-3.34\cdot 10^{16}m/s^2[/tex] is the electron's acceleration

v = 0 is its final speed

[tex]u=5.25\cdot 10^6 m/s[/tex] is the initial speed

t is the time it takes for the electron to come at rest

Solving for t,

[tex]t=\frac{v-u}{a}=\frac{0-(5.25\cdot 10^6 m/s)}{-3.34\cdot 10^{16} m/s^2}=1.57\cdot 10^{-10}s[/tex]

(d) [tex]5.25\cdot 10^6 m/s[/tex]

This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.

So we have:

u = 0 is the initial speed of the electron

[tex]a=3.34\cdot 10^{16}m/s^2[/tex]

[tex]d=4.13\cdot 10^{-4} m[/tex] is the distance covered to go back

So we can use the following equation:

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

to find v, the new final speed:

[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2 + 2(3.34\cdot 10^{16} m/s^2)(4.13\cdot 10^{-4} m)}=5.25\cdot 10^6 m/s[/tex]

Answer:

H = 4.12 m

Explanation:

As we know that horizontal range is the distance moved in horizontal direction

Since horizontal direction has no acceleration

so here we have

[tex]Range = v_x T[/tex]

here we know that

[tex]v_x = vcos32[/tex]

so from above formula

[tex]92 = (vcos32)(6.4)[/tex]

[tex]v = 16.95 m/s[/tex]

now we have maximum height is given as

[tex]H = \frac{(vsin32)^2}{2g}[/tex]

[tex]H = \frac{(16.95 sin32)^2}{2(9.8)}[/tex]

[tex]H = 4.12 m[/tex]

Answer:

D]  speed

Explanation:

Answer:

2.4 x 10^21

Explanation:

V = 9 V, R = 1.4 ohm, t = 1 minute = 60 second

Use Ohm's law

V = I R

I = V / R

I = 9 / 1.4

I = 6.43 A

Now use Q = I t

Q = 6.43 x 60 = 385.7 C

Number of electrons passing in 1 minute , n

= total charge in one minute / charge of one electron

n = 385.7 / (1.6 x 10^-19) = 2.4 x 10^21

Answer:

Option (A)

Explanation:

A ball bearing is a device which is use to reduce the friction.

The outer rim of the bearing is fixed with the part of machine and inner rim is fitted into shafts. Now the shafts rotates and only the small spheres in the bearing will rotate. The friction can be further reduced by apply the oil or grease to the bearing.

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

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

[tex]k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2[/tex]

k = 1150  J

In two significant figure, [tex]k=1.2\times 10^3\ J[/tex]

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. [tex]P=\dfrac{W}{t}=\dfrac{\Delta K}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}[/tex]

P = 115 watts

In two significant figures, [tex]P=1.2\times 10^2\ Watts[/tex]

Hence, this is the required solution.  

The magnitude of the charge on the second object is 0.025 μC and its sign is negative because it is required work to bring the two charges together, suggesting these are opposite charges and repel each other.

The problem can be solved using the formula for the work done on a charge moving in an electric field, which is determined by the formula W = k * q1 * q2 / r, where k is the Coulomb's constant (8.99 × 10⁹ N · m²/C²), q1 and q2 are the charges, and r is the distance between them. From the problem, we know W = 32J, q1 = +35 μC, and r = 2.00 mm (the difference in the y-coordinates). Solving for q2 gives q2 = W * r / (k * q1) = 32J * 2.00 x 10⁻³m / (8.99 × 10⁹ N · m²/C² * 35 x 10⁻⁶C) = approximately -0.025 μC. Therefore, the magnitude of the charge on the second object is 0.025 μC, and its sign is negative because it takes work to bring the two charges together.

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Answer:

69.4 pA

Explanation:

n = number of sodium ions = 1473

e = magnitude of charge on each sodium ion = 1.6 x 10⁻¹⁹ C

t = time taken to flow across the membrane = 3.4 x 10⁻⁶ sec

Total Charge on sodium ions is given as

q = n e                                            eq-1

Current produced is given as

[tex]i = \frac{q}{t}[/tex]

Using eq-1

[tex]i = \frac{ne}{t}[/tex]

Inserting the values

[tex]i = \frac{(1473)(1.6\times 10^{-19})}{3.4\times 10^{-6}}[/tex]

i = 69.4 x 10⁻¹² A

i = 69.4 pA

Answer:

a)

16.33 Ω

b)

45009.18 A

Explanation:

a)

L = length of the line = 935 km = 935000 m

d = diameter of the line = 3.50 cm = 0.035 m

ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m

Area of cross-section of the line is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.035)²

A = 0.000961625 m²

Resistance of the line is given as

[tex]R=\frac{\rho L}{A}[/tex]

inserting the values

R = (1.68 x 10⁻⁸) (935000)/(0.000961625)

R = 16.33 Ω

b)

V = potential difference across the line = 735 kv = 735000 Volts

i = current carried by the wire

Using ohm's law, current carried by the wire is given as

[tex]i=\frac{V}{R}[/tex]

i = 735000/16.33

i = 45009.18 A

Answer:

4582 N

Explanation:

The initial speed of the test car is

u = 11.4 m/s

While the final speed is

v = 0

The displacement of the test car during the collision is

d = 0.78 m

So we can find the acceleration of the car by using the following SUVAT equation:

[tex]v^2 - u^2 = 2ad\\a=\frac{v^2-u^2}{2d}=\frac{0-(11.4)^2}{2(0.78)}=-83.3 m/s^2[/tex]

Now we can find the average force acting on the dummy by using Newton's second law:

F = ma

Where m = 55 kg is the mass. Substituting,

[tex]F=(55 kg)(-83.3 m/s^2)=-4582 N[/tex]

So the size of the average force is 4582 N.

Answer:

4 m/s2 in negative acceleration

Explanation:

If An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds, it is 4 m/s2 in negative acceleration.

Hope this helps!

Answer:

[tex]-4\frac{m}{s^2}[/tex]

Explanation:

The object changes its speed over some time, this means that there is an acceleration.

It has a uniformly accelerated movement.

The Formula for finding speed in a uniformly accelerated motion is

[tex]a=\frac{V_{f}-V_{o}}{t}[/tex]

[tex]V_{o}= 65\frac{m}{s}\\V_{f}= 25\frac{m}{s} \\t= 10s[/tex]

Replace

[tex]a=\frac{(25-65)\frac{m}{s} }{10 s}\\ a=\frac{-40\frac{m}{s} }{10s}\\a= -4\frac{m}{s^2}[/tex]

Acceleration gives us a negative value this means that it is slowing.

Answer:

Force required is 4387 N in the opposite direction of motion.

Explanation:

We have equation of motion v² = u² + 2as

      v = 0m/s , u = 15 m/s, s = 31.8 m

  Substituting

         0² = 15² + 2 x a x 31.8

          a = -3.54 m/s²

So, deceleration = 3.54 m/s²

Force = Mass x Acceleration

          = 1240 x -3.54 = -4387 N

So force required is 4387 N in the opposite direction of motion.

Answer:

2.0 s, 200 m

Explanation:

Time to hit the ground depends only on height.  Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.

Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance).  So it will travel a horizontal distance of 200 m.

Answer:

1) 2 seconds

2) 200 m

Explanation:

1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]

  Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]

  Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]

s = speed

t = time

h = height

Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,

Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]

i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s

Time taken to land = 2.0 s

2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m

   Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]

Therefore, since distance traveled is directly proportional to speed,

Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]

i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]

    [tex]2S_{1}[/tex] = 2 x 100 m = 200 m

Distance traveled = 200 m

Final answer:

The question inquires about the time needed for 80% of Pb-209 to decay, knowing its half-life is 3.3 hours. By applying the exponential decay formula, we calculate that approximately 7.39 hours are required for 80% of Pb-209 to decay.

Explanation:

The question involves the concept of radioactive decay and specifically asks how long it will take for 80% of Pb-209 to decay, given that its half-life is 3.3 hours. To find the time required for 80% of the lead to decay, we use the half-life formula and the property that radioactive decay is an exponential process. Since 80% decay means 20% remains, we set up the equation based on the exponential decay formula: N = N0(1/2)(t/T), where N is the remaining amount of substance, N0 is the initial amount, t is the time elapsed, and T is the half-life of the substance.

Substituting the given values and solving for t, we find:

N0 = 1 gram (100% initially)

N = 0.2 grams (20% remains)

T = 3.3 hours

Thus, the equation becomes 0.2 = 1(1/2)(t/3.3). Solving for t gives us the time required for 80% decay.

After calculations, the result is that it takes approximately 7.39 hours for 80% of the Pb-209 to decay. This showcases the practical application of exponential decay and half-life in determining the amount of a radioactive substance that remains after a given period.

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

Answer:1074.26

Explanation:just got it right on my accelus

Answer:

167.85 rev / s

Explanation:

r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N

F = m r w^2

where, w is the angular velocity.

4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2

w = 1054.1 rad / s

w = 2 π f

f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s

The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.

Recall :

We calculate the angular velocity, ω thus :

ω² = F/mr

ω² = [tex] \frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}[/tex]

ω = [tex] \sqrt{1.11 \times 10^{6}} = 1053.56 rad/s[/tex]

Frequency = 1053.56 ÷ (2π)

Frequency = 167.68 Hz

Therefore, the Number of revolutions per seconds would be about 167.8 Hz

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The planet's orbital period is about 388 days

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of the planet = m = 6.75 × 10²⁴ kg

mass of the star = M = 2.75 × 10²⁹ kg

radius of the orbit = R = 8.05 × 10⁷ km = 8.05 × 10¹⁰ m

Unknown:

Orbital Period of planet = T = ?

Solution:

Firstly , we will use this following formula to find the orbital period:

[tex]F = ma[/tex]

[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]

[tex]G M = \omega^2 R^3[/tex]

[tex]\frac{GM}{R^3} = \omega^2[/tex]

[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]

[tex]T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}[/tex]

[tex]T \approx 3.35 \times 10^7 \texttt{ seconds}[/tex]

[tex]T \approx 388 \texttt{ days}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

To calculate the orbital period of the planet, convert the radius to meters, substitute the mass of the star and the radius into Kepler's third law, solve for the orbital period squared, and convert to Earth days.

The question involves applying Kepler's third law of planetary motion to calculate the orbital period of a planet revolving around a star. The law relates the orbital period (T) to the radius (r) of the orbit and the mass (M) of the star around which the planet orbits. Kepler's third law can be expressed as:

T² = (4π²/GM) · r³,

where G is the gravitational constant (6.67430 × 10⁻¹¹ m³/kg · s²).

To find the orbital period of the planet:

The student will then have the orbital period of the planet in Earth days.

Answer:A

Explanation:

Initial velocity, u = 2m/s

Final velocity, v = 4m/s

Distance covered, s = 12m

Acceleration, a = ?

Using

v² = u² + 2as

2as = v² - u²

a = v²-u²/2s

a = 4²-2²/2 x 12

a = 16-4/24

a = 12/24

a = 0.5m/s²

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

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

[tex]T = \frac{2 \times 20 \times Sin35 }{9.8}[/tex]

T = 2.34 second

(b) The formula for the maximum height is given by

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

[tex]H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}[/tex]

H  = 6.71 m

(c) The formula for the range is given by

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

[tex]R = \frac{20^{2} \times Sin 2 \times 35}{9.8}[/tex]

R = 38.35 m

(d) It hits with the same speed at the initial speed.

Answer:

10.19 m

Explanation:

Energy is conserved, so elastic energy stored in spring = gravitational energy of block.

1/2 kx² = mgh

h = kx² / (2mg)

h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)

h = 10.19 m

Answer:

0.8 m

Explanation:

For "6 N" force :

F = magnitude of the force = 6 N

r = moment arm = 0.4 m

Torque due to "6 N" force is given as

τ = r F

τ = (0.4) (6)

τ = 2.4 Nm

For " 15 N" force :

F' = magnitude of the force = 15 N

r' = moment arm = ?

τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm

Torque due to "15 N" force is given as

τ' = r' F'

12 = r' (15)

r' = 0.8 m

So the moment arm for "15 N" force is 0.8 m

Answer with Explanation:

since the two blocks move (stick) together, the collision is inelastic, which does not conserve kinetic energy.  So do not use kinetic energy consideration.

Fortunately, in such a situation, momentum is still conserved.

Momentum of 1.0 kg block

= 1.0 * 3.0 = 3.0 kg-m/s

Momentum of second block

= 1.0 * 1.0 = 1.0 kg-m/s

Total mass after collision = 1.0+1.0 = 2.0 kg

Common velocity after collision

= total momentum / total mass

= (3.0+1.0)/2.0 = 2.0 m/s

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude

[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]

We have

[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]

where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so

[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so

[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]

Then the magnitude of the particle's acceleration is

[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]

The magnitude of the acceleration of the particle is approximately 70.55 m/s^2, calculated by using the formulas for combined radial and tangential acceleration in circular motion.

In this physics problem, the particle not only moves around in a circle but is also experiencing an increase in speed which is a case of combined radial and tangential acceleration. Radial acceleration, known as centripetal acceleration (ar), is the result of the change in direction of the velocity vector, while tangential acceleration (at) comes from changes in speed.

The total acceleration of an object in circular motion is given by:

a = sqrt((ar^2) + (at^2))

Centripetal acceleration can be calculated using the formula ar = v^2 / r, where: v = speed (37.2 m/s), r = radius of the circle (21 m). This gives us ar = (37.2^2) / 21, which approximately equals 66.62 m/s^2.

The tangential acceleration is given in the problem: at = 23.1 m/s^2.

We therefore calculate the total acceleration using the formula above which gives us:

a = sqrt((66.62^2) + (23.1^2)) which approximately equals 70.55 m/s^2.

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Answer:

[tex]8.6\cdot 10^{-18} m[/tex]

Explanation:

Initially, the electron is travelling undeflected at constant speed- this means that the electric force and the magnetic force acting on the electron are balanced. So we can write

q E = q v B

where

q is the electron's charge

[tex]E=8.3\cdot 10 V/m[/tex] is the electric field magnitude

v is the electron's speed

[tex]B=7.3\cdot 10^3 T[/tex] is the magnitude of the magnetic field

Solving for v,

[tex]v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s[/tex]

Then the electric field is turned off, so the electron (under the influence of the magnetic field only) will start moving in a circle of radius r. Therefore, the magnetic force will be equal to the centripetal force:

[tex]qvB= m \frac{v^2}{r}[/tex]

where

[tex]q=1.6\cdot 10^{-19} C[/tex] is the electron's charge

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron's mass

Solving for r, we find the radius of the electron's orbit:

[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m[/tex]

Answer:

Acceleration due to gravity value

[tex]=9.89m/ {s}^{2} [/tex]

Explanation:

Time period of simple pendulum is given by the expression

[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]

Here we have

T = 0.88 s

l = 19.4 cm = 0.194 m

Substituting

[tex]0.88=2\pi\sqrt{\frac{0.194}{g}}\\\\\frac{0.194}{g}=0.0196\\\\g=9.89m/s^2 \\ [/tex]

Acceleration due to gravity value

[tex]=9.89m/s^2 \\ [/tex]

Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:  

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

Answer:

82 pF

Explanation:

diameter = 8.2 cm, distance, d = 2.1 mm = 0.0021 m,

dielectric constant, k = 3.7

radius = 1/2 x diameter = 8.2 / 2 = 4.1 cm = 0.041 m

The formula for the capacitance of parallel plate capacitor is given by

[tex]C = \frac{k \varepsilon _{0}\times A}{d}[/tex]

[tex]C = \frac{3.7 \times 8.854\times 10^{-12} \times \pi \times 0.041 \times 0.041}{0.0021}[/tex]

C = 8.23 x 10 ^-11 F

C = 82 pF

Answer:

17 seconds

Explanation:

In the y direction:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 140 + (120 sin 38) t + ½ (-9.8) t²

4.9 t² - 73.9 t - 140 = 0

Solve with quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8

t = -1.7, 16.8

Since t can't be negative, t = 16.8.  Rounding to 2 sig-figs, the projectile lands after 17 seconds.

Answer:

[tex]2.1\cdot 10^{-3} C[/tex]

Explanation:

The initial charge stored on the capacitor is given by

[tex]Q_0 =C_0 V[/tex]

where

[tex]C_0 = 6.0 \mu F = 6.0 \cdot 10^{-6}F[/tex] is the initial capacitance

V = 100 V is the potential difference across the capacitor

Solving the equation,

[tex]Q_0 = (6.0 \cdot 10^{-6}F)(100 V)=6.0 \cdot 10^{-4}C[/tex]

The charge stored in the capacitor when inserting the dielectric is

[tex]Q = k Q_0[/tex]

where

k = 4.5 is the dielectric constant

Substituting,

[tex]Q=(4.5)(6.0 \cdot 10^{-4}C)=2.7\cdot 10^{-3}C[/tex]

So the additional charge is

[tex]\Delta Q=Q-Q_0 = 2.7 \cdot 10^{-3}C - 6.0 \cdot 10^{-4}C=2.1\cdot 10^{-3} C[/tex]

The additional charge that flows into the capacitor once it is immersed in transformer oil (of dielectric constant 4.5) and kept connected to the source is 2100 μC.

The question involves a capacitor that is connected across a 100-V voltage source and then submerged in transformer oil. The capacitor initially charges to capacity while in the air. When a capacitor is then immersed in a material with a dielectric constant, its ability to store charge improves. Here, the dielectric constant of the transformer oil is given as 4.5, suggesting that the capacitor's capacity to store charge will increase 4.5 times as compared to when it was in air.

The additional charge that flows into the capacitor can be calculated using the formula for the charge in a capacitor, Q = CV. The initial charge (Q1) on the capacitor when it was in air would be Q1 = CV1 = 6.0 μF * 100 V = 600 μC. After immersing in transformer oil, the capacitance would increase by a factor of 4.5, giving a new capacitance C2 = 4.5 * 6.0 μF = 27.0 μF. The new charge (Q2) would be Q2 = CV2 = 27.0 μF * 100 V = 2700 μC. Hence the additional charge that's flown from the source would be Q2 - Q1 = 2700 μC - 600 μC = 2100 μC.

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