A mass m = 0.7 kg is released from rest at the origin 0. The mass falls under the influence of gravity. When the mass reaches point A, it is a distance x below the origin 0; when the mass reaches point B it is a distance of 3 x below the origin 0. What is vB/vA?

Answers

Answer 1

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]


Related Questions

If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)

Answers

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

A wire of 5.8m long, 2mm diameter carries 750ma current when 22mv potential difference is applied at its ends. if drift speed of electrons is found then:_________.
(a) The resistance R of the wire(b) The resistivity p, and(c) The number n of free electrons per unit volume.​

Answers

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

We are given;

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

A) Formula for resistance is;

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

B) formula for resistivity is given by;

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

C) Formula for current density is given by;

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

Read more at; brainly.com/question/17005119

Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion​

Answers

Last point, Newton’s first law of motion is the correct answer.

Hope this helps ya

The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is

Answers

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground

Answers

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b

Answers

Answer:

Will be equal to alpha x r; less than UsN

Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth

Answers

Answer:

Vrel= 0.75c

Explanation:

See attached file

When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass

Answers

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.

At what rate does this speaker produce energy?

What is the intensity of this sound 9.50 from the speaker?

What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?

Answers

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about

Answers

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

m=10kgVf=0m/sVi=10m/st=2s

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what is the energy of one of its photons in J

Answers

Answer:

The energy of one of its photons is 1.391 x 10⁻¹⁸ J

Explanation:

Given;

wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m

The energy of one photon of the UVC light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency of the light

f = c / λ

where;

c is speed of light = 3 x 10⁸ m/s

λ  is wavelength

substitute in the value of f into the main equation;

E = hf

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]

Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J

which category would a person who has an IQ of 84 belong ?

Answers

answer: below average

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.

Required:
What is the charge on the dust particle?

Answers

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

   

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

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

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T

Answers

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

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

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.

Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?

Answers

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.

Answers

Answer:

covalent

Explanation:

covalent bonds share electrons

3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.​

Answers

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.

Answers

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60

Answers

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Please help!
Much appreciated!​

Answers

Answer:

your question answer is 22°

I’m pretty sure the answer is 22

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?

Answers

Explanation:

Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,

[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]

[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]

[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]

[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]

after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,

[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]

[tex]\qquad \Phi_{B, f}=0[/tex]

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]

[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]

[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]

(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb

(b) The induced EMF in the coil is 0.583 mV

Flux and induced EMF:

Given that the coil has N = 250 turns

and an area of A = 14cm² = 1.4×10⁻³m².

It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T

(a) The flux passing through the coil is given by:

Ф = NBAcosθ

where θ is the angle between area vector and the magnetic field

The area vector is perpendicular to the plane of the coil.

So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil

So the initial flux is:

Φ = NABcos0° = NAB

Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb

Ф = 1.75 × 10⁻⁵ Wb

Finally, θ = 90°, and since cos90°, the final flux through the coil is 0

(b) The EMF induced is given by:

E = -ΔФ/Δt

E = -(0 - 1.75 × 10⁻⁵)/0.030

E = 0.583 × 10⁻³ V

E = 0.583 mV

Learn more about magnetic flux:

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A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.

Answers

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction

Answers

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.


Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?

Answers

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

A ball is thrown upward from a height of 432 feet above the​ ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. ​a) Find​ s(t), the function giving the height of the ball at time t. ​b) How long will the ball take to reach the​ ground? ​c) How high will the ball​ go?

Answers

Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet

By applying the integrations,

(a) [tex]S = 96t-16t^2+432[/tex]

(b) Time will be "t = 9".

(c) Height will be "576"

Given:

Height,

423 feet

Initial velocity,

96 feet/sec

According to the question,

(a)

Integrate v:

[tex]S = 96t-16t^2+C[/tex]

Initial Condition,

→ [tex]S = 96t-16t^2+432[/tex]

(b)

Hits the ground when,

S = 0

→ [tex]0=96t-16t^2+432[/tex]

→ [tex]t =9[/tex]

(c)

Maximum height when,

v = 0

→ [tex]0 = 96-32 t[/tex]

→ [tex]t = 3[/tex]

Now,

→ [tex]S = 96\times 3-16\times 3^2+432[/tex]

      [tex]= 576[/tex]

Thus the answer above is correct.

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Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.

Answers

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.

Answers

Answer:

a. to the west.

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.

If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answers

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answers

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q

Answers

Answer:

Upper plate Q/3

Lower plate 2Q/3

Explanation:

See attached file

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