A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse

Answers

Answer 1

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.


Related Questions

A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."

Required:
a. What is the field's magnitude?
b. What is the field's direction?

Answers

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒  Magnetic force = Copper wire's weight

So,

⇒   [tex]B\times I\times L=M\times g[/tex]

On putting the estimated values, we get

⇒  [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]

⇒  [tex]50B=69.03297\times 10^{-3}[/tex]

⇒  [tex]B=1.38\times 10^{-3} \ T[/tex]

(b)...

As we know,

⇒  [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]

⇒      [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]

⇒      [tex]=2240\times \pi \ 0.000001[/tex]

⇒      [tex]=7.037\times 10^{-3} \ kg[/tex]

The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor

Answers

Answer:

  B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.

Answers

Answer:

C) less than 129 lb.

Explanation:

Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards  with magnitude a .

If R be the reaction force

For the elevator is going downwards with acceleration a

mg - R = ma

R = mg - ma

R measures its apparent weight . Spring scale will measure his apparent weight.

So its apparent weight is less than 129 lb .

A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is

Answers

Answer:

v = 1.4 m/s

Explanation:

This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):

[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex]  (1)

m1: mass of the toy car = 2 kg

m2: mass of the toy truck = 3 kg

v1: speed of the toy car = 1 m/s

v2: speed of the truck car = 3 m/s

v: speed of both car and truck after the collision = ?

In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.

You solve the equation (1) for v, and you replace the values of all variables involved:

[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]

this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck

Hence, the speed of both car and truck toys is 1.4 m/s

28 points!! please help

Answers

7(a) transpiration is faster in warmer dry air
(b)(I)xylem
(ii) 1. They are stacked end-to-end
2. Consists of dead cells

Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m

Answers

Answer:

The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

Explanation:

From the question we are told that

    The potential energy is  [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]

The force on the mass can be mathematically evaluated as  

      [tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]

The negative sign shows that the force is moving in the opposite  direction of the potential energy

       [tex]F = - 6 x^2 + 30x - 36[/tex]

At critical point

      [tex]\frac{d U(x)}{dx} = 0[/tex]

So  

     [tex]- 6 x^2 + 30x - 36 = 0[/tex]

     [tex]- x^2 + 5x - 6 = 0[/tex]

Using quadratic equation formula to solve this we have that

       [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

               

A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.

Answers

Answer:

Work is done by friction = -165 J

Explanation:

Given:

Mass of block (m) = 15 kg

Ramp inclined = 28°

Friction force (f) = 30 N

Distance (d) = 5.5 m

Find:

Work is done by friction.

Computation:

Work is done by friction = -Fd

Work is done by friction = -(30)(5.5)

Work is done by friction = -165 J

The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?

Answers

Answer:

A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules

Explanation:

Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules

Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.

Answers

Answer:

a) v = -30 - 32 t ,  s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s

c) v (1) = -62 ft / s,  v (3) = -126 ft / s , d) t = 7.13 s , e)  v = -258.16 ft / s

Explanation:

a) For this exercise they give us the function of the position of the ball

          s (t) = s (o) + v_o t - 16 t²

notice that you forgot to write the super index

indicate the initial position of the ball

        s (o) = 600 ft

also indicates initial speed

        v_o = - 30 ft / s

let's substitute in the equation

        s (t) = 600 - 30 t -16 t²

to find the speed we use

       v = ds / dt

       v = v_o - 32 t

       v = -30 - 32 t

b) To find the average speed, look for the speed at the beginning and end of the time interval

t = 1 s

     v (1) = -30 -32 1

     v (1) = - 62 ft / s

t = 3 s

     v (3) = -30 -32 3

     v (3) = -126 ft / s

the average speed is

    v = (v (3) -v (1)) / (3-1)

    v = (-126 +62) / 2

    v = -32 ft / s

c) instantaneous speeds, we already calculated them

    v (1) = -62 ft / s

    v (3) = -126 ft / s

d) the time to reach the ground

in this case s = 0

    0 = 600 - 30 t -16 t²

     t² + 1,875 t - 37.5 = 0

we solve the quadratic equation

     t = [-1,875 ±√ (1,875² + 4 37.5)] / 2

     t = [1,875 ± 12.39] / 2

     t₁ = 7.13 s

     t₂ = negative

Since the time must be positive, the correct answer is t = 7.13 s

e) the speed of the ball on reaching the ground

     v = -30 - 32 t

     v = -30 - 32 7.13

      v = -258.16 ft / s

4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor

Answers

Answer:

Overall charge still remains zero on conductor until touched by charged rod.

Explanation:

Here, we want to know what has happened to the overall charge on the conductor.

Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.

Thus, overall charge still remains zero on conductor until touched by charged rod.

A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?

Answers

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]

The rms voltage (in V) measured across the secondary coil is calculated as;

[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

Which symbol in a chemical equation separates the reactants from the products?

Answers

Answer:

the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔

Explanation:

i hope this will help you

arrow symbol
Explanation: The arrow symbol in a chemical equation separates the reactants from products

g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?

Answers

Answer:

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).

Explanation:

The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.

From Rayleigh's formula,

Angular spread = 1.22 (wavelength ÷ diameter)

                          = 1.22 (λ ÷ D)

Given that:

diameter, D = 0.18 m and wavelength, λ = 490 nm, then;

Angular spread of the laser beam = 1.22 (λ ÷ D)

                         = 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]

                         = 1.22× 2.7222 × [tex]10^{-6}[/tex]

                        = 3.3211 × [tex]10^{-6}[/tex] rad

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.

Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?

Answers

Answer:

A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B.  Lebron James hits at an angle of 17.48° North -East.

The direction of Stephen is   = 20° due to East of North

Explanation:

Let [tex]r ^ {\to[/tex] represent the position vector of the hole;

Also; using  the origin as starting point. Let the east direction be along the positive x axis and the North direction  be + y axis

Thus:

[tex]r ^ {\to[/tex]  = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]

[tex]r ^ {\to[/tex]  = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]

Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot

So;

[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]

[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]

Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot

[tex]r_2 ^ \to[/tex]  [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]

[tex]r_2 ^ \to[/tex]  = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]

However;

[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]

Also;

[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]

Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B .

For Lebron James ;

The angle can be determine using the trigonometric function:

[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]

Thus  Lebron James hits at an angle of 17.48° North -East.

For Stephen Curry;

[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]

The direction of Stephen is  = 90° - 70° = 20° due to East of North

An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?

Answers

Answer:

Ff = 33.4N

Explanation:

To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.

The frictional force is given by:

[tex]F_f=\mu_kN[/tex]         (1)

Ff: frictional force = ?

µk: coefficient of kinetic friction = 0.167

N: normal force of the object = 200N

You replace the values of the parameters in the equation (1):

[tex]F_f=(0.167)(200N)=33.4N[/tex]

The frictional force, while the objects is moving, is 33.4N

n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.

Answers

Answer:-

-1 m/s^2

Explanation:

The average acceleration is given by dividing the change in velocity by change in time;

[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]

[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]

the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.

70 pointss yall !!! helpp

Answers

A: the type of plant

B: how tall the plant is

Answer:

A= The type of plant

B= How tall the plant is

Explanation:

Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?

Answers

Explanation:

The relation between resistance and resistivity is given by :

[tex]R=\rho \dfrac{l}{A}[/tex]

[tex]\rho[/tex] is resistivity of material

l is length of wire

A is area of cross section of wire

Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is

Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.

Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?

Answers

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

[tex]y_m=\frac{m\lambda D}{d}[/tex]    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]

For λ = 470 nm = 470*10^-9 m

[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

Answers

Answer:

[tex]\large \boxed{42\, \mu \text{C}}$[/tex]

Explanation:

The formula for the force exerted between two charges is

[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

[tex]F=k\dfrac{q^{2}}{r^2}[/tex]

[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]

As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.

Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?

Answers

Answer:

a.[tex]5.66ms^{-2}[/tex]

b.55 m

Explanation:

We are given that

Mass ,m=0.9 kg

Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]

1 m=100 cm

[tex]\theta=30^{\circ}[/tex]

R=55 m

a.Centripetal acceleration

[tex]a_c=gtan\theta[/tex]

[tex]a_c=9.8tan30^{\circ}[/tex]

[tex]a_c=5.66 m/s^2[/tex]

Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]

b. Radius of curve,R=55 m

An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________

Answers

Answer:

either +z direction or -z direction.

Explanation:

The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.

You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction

In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.

A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?

Answers

Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s

Explanation: Please see the attachment below

The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

The given parameters:

Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 rad

The angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]

where;

[tex]\omega_i[/tex] is the initial angular velocity

[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]

Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

Learn more about angular velocity here: https://brainly.com/question/540174

A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.

Answers

Answer:

the magnitude is 7 and sign of the point charge on the surface shell is -13

Explanation:

Of one of the planets becomes a black hole , what would the escape speed be?

Answers

Answer:

If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.

Explanation:

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Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg

Answers

Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.

250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *​

Answers

Answer:

1008.57kg/m3

Explanation:

Now the mass of fresh water is 250×1000 /1000000 = 0.25kg

Now the mass of salt water is

100×1030 /1000000 = 0.103kg

Note Density = mass / volume

Mass = volume × density

Note that converting from cm3 to m3 we divide by 1000000

Total mass = 0.25kg +0.103kg= 0.353kg.

Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3

Hence the density of the mixture= total mass / total volume

0.353kg/35 × 10^{-5}m3=1008.57kg/m3

Unit conversion

The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units

Answers

Answer:

  1.234567 kA

Explanation:

The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...

  1.234 kA + 0.000567 kA = 1.234567 kA

Which of the following best describes the current age of the Sun?

A.) It is near the end of its lifespan.

B.) It is about halfway through its lifespan.

C.) It is early in its lifespan.

D.) We do not have a good understanding of the Sun's age.

Answers

Answer:  Its b, The only problem with this is is there supposed to be a picture?

Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.


please help! i will be giving 50 points, this is for my psychology class.

Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted

Answers

Answer:

D

Explanation:

the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)

Answer:

D

Explanation:

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