A certain shade of blue has a frequency of 7.06×1014 Hz. What is the energy E of exactly one photon of this light? Planck's constant h=6.626×10−34 J⋅s.

Answers

Answer 1

Answer:

Energy, [tex]E=4.67\times 10^{-19}\ J[/tex]

Explanation:

It is given that,

Frequency of blue shade is, [tex]f=7.06\times 10^{14}\ Hz[/tex]

We need to find the energy of exactly one photon of this light. The formula that is used to find the energy of photon is given by :

[tex]E=nhf[/tex]

Here, n is number of photon, n = 1

h is Planck's constant

So,

[tex]E=1\times 6.626\times 10^{-34}\times 7.06\times 10^{14}\\\\E=4.67\times 10^{-19}\ J[/tex]

So, the energy of this light is [tex]4.67\times 10^{-19}\ J[/tex].


Related Questions

An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg

Answers

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices

Answers

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  [tex]a_t[/tex] = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

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

where;

[tex]a_c[/tex] is the radial acceleration

[tex]a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2[/tex]

The radial acceleration relates to speed of particle in the following equations;

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

where;

v is the speed of the particle

[tex]v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s[/tex]

Therefore, the speed of the particle is 2.86 m/s

Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 meters) and back? Recall that depth in meters = ½ (1500 m/sec × Echo travel time in seconds). Round your answer to two decimal places.

Answers

Answer:

14.66secs

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?

Answers

Answer:

Each of the capacitor carries the same charge, Q

Explanation:

When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;

[tex]V_T = V_1 + V_2 + V_3[/tex]

We can also calculate this voltage in terms of capacitance and charge;

[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]

Therefore, each of the capacitor carries the same charge, Q

what help in reversing direction of current of current​

Answers

Answer:

To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.  

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

If theta is 30 degrees and there is no friction, what would be the block's acceleration down the incline, in meters per second squared?

Answers

Answer:

a= 4.9m/s²

Explanation:

Using Fnet= mgsintheta = ma

But a= gsintheta

a= 9.8xsin 30

= 4.9m/s²

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s

Answers

Answer:

the correct answer is c     v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

            v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

           a₁ = v² / 2x

           a₁ = 25² / (2 120)

           a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

           v₁² = 0 + 2 a1 x₂

           v₁ = Ra (2 2,6 60)

           v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c     v₁> 12.5 m / s

A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s

Answers

Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string

Answers

Answer:

The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]

Explanation:

From the question we are told that

   The length of the string is  [tex]l = 3.00 \ m[/tex]

    The  mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]

     The  tension on the string is  [tex]T = 500 \ N[/tex]

   

The  velocity of the pulse is mathematically represented as

      [tex]v = \sqrt{ \frac{T}{\mu } }[/tex]

Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as

       [tex]\mu = \frac{m}{l}[/tex]

substituting values

     [tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]

     [tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]

Thus  

     [tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]

    [tex]v = 547.7 m/s[/tex]

The time taken is evaluated as

    [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{3}{547.7}[/tex]

      [tex]t = 5.48 *10^{-3} \ s[/tex]

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system

Answers

Answer:

The mutual inductance is  [tex]M = 0.000406 \ H[/tex]

Explanation:

From the question we  are told that

    The  number of turns per unit length  is  [tex]N = 1800[/tex]

    The radius is  [tex]r = 0.0165 \ m[/tex]

     The  number of turns of the solenoid is  [tex]N_s = 210 \ turns[/tex]

   

Generally the mutual inductance of the  system is mathematically represented as

       [tex]M = \mu_o * N * N_s * A[/tex]

Where A is the cross-sectional area of the system which is mathematically represented as

       [tex]A = \pi * r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.0165)^2[/tex]

       [tex]A = 0.0008554 \ m^2[/tex]

also   [tex]\mu_o[/tex] is the permeability of free space with the value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

So  

      [tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]

      [tex]M = 0.000406 \ H[/tex]

A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.

Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?

Answers

Answer:

Peak current= 84.86 A

Area of each turn = 0.029 m^2

Explanation:

The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.

Hence;

Irms= 60.0A

Io= Irms/0.707

Io = 60.0/0.707

Io= 84.86 A

Vrms= 0.707Vo

Vo= Vrms/0.707= 170/0.707 = 240.45 V

From;

V0 = NABω

Where;

Vo= peak voltage

N= number of turns

B= magnetic field

A= area of each coil

ω= angular velocity

But ω= 2πf = 2×π×95= 596.9 rads-1

Substituting values;

A= Vo/NBω

A= 240.45/550×0.025×596.9

A= 0.029 m^2

what is electric field strength​

Answers

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

Please mark me as the brainliest!!!

Thanks!!!

At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.

Answers

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(ωt+ϕ), where ϕ, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of ϕ based on the initial conditions described in the problem. Express your answer in terms of q0 , L, and C. Use the cosine function in your answer.

Answers

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = Δ[tex]V_{C}[/tex]

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

                       

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

             

                  q = q₀ sin (wt)

10. A manufacturer knows from experience that the resistance of resistors she produces is normal with mean µ = 150Ω and the standard deviation σ = 5Ω. What percentage of the resistors will have resistance between 148 Ω and 152 Ω? Between 140 Ω and 160 Ω?

Answers

Answer:

Explanation:

Using the formula for finding the Z score

Z = x-µ/σ

x is the sample size

µ is the sample mean

σ is the standard deviation

For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown

P(148≤x152) = Z(152-150/5) - Z(148-150/5)

P(148≤x152) = Z(0.4)-Z( - 0.4)

P(148≤x152) = 0.6554-0.3446

The Z values are from the normal distribution table.

P(148≤x152) = 0.3108

The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%

Similarly for resistances between 140 Ω and 160 Ω

P(140≤x160) = Z(160-150/5) - Z(140-150/5)

P(140≤x160) = Z(2.0)-Z( - 2.0)

P(140≤x160) = 0.9775-0.02275

The Z values are from the normal distribution table.

P(140≤x160) = 0.9547

The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%

The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%

The percentage of the resistors will have resistance between 140 Ω and 160 Ω is  95.47%

Percentage of resistors:

To solve for the probability we will use the standard score of the Z score, which is given by:

Z = (x - µ)/σ

where x is the sample size

µ is the sample mean = 150Ω

σ is the standard deviation =

The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:

P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)

P(148 ≤ 152) = Z(0.4)-Z( - 0.4)

P(148 ≤ 152) = 0.6554-0.3446

P(148 ≤ 152) = 0.3108

So, the percentage will be:

0.3108×100% = 31.08%

Similarly for resistances between 140 Ω and 160 Ω

P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)

P(140 ≤ 160) = Z(2.0)-Z( - 2.0)

P(140 ≤ 160) = 0.9775-0.02275

P(140 ≤ 160) = 0.9547

The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%

Learn more about standard score:

https://brainly.com/question/25875323?referrer=searchResults

A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.

Answers

Answer:

e is correct

Explanation:

When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.

What is buoyant force?

The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.

The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.

When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.

Therefore, options (d), (e) are correct.

Learn more about buoyant force, here:

https://brainly.com/question/21990136

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An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?

Answers

Answer:

a) 1764.71 ohms

b) 1.73 H

Explanation:

From the question, we can identify the following parameters;

Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA

(a) Indcued emf V = L di/dt =0

From ohms law Vo = ioR

R = 12/6.80*0.001

R=1764.71 ohms

(b) For LR circuit

i =io (1-e^-t/T)

Time constant T = L/R

4.86 = 6.80 (1-e^-0.7*10^-3/T)

divide both side by 6.8

0.715 = 0.0007/T

L/R = 0.0007/0.715

L/R = 0.000979020979

Substitute R from above

L = 0.000979020979 * 1764.71

L =1.73 H


a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative density 0.2, calculate the mass of the fluid in the jar​

Answers

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s

Answers

Answer:

5.81 X 10^3 Ns

Explanation:

Given that

F = At² and F at t = 1.25 s is 781.25 N ?

A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²

d(Impulse) = Fdt

Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s

Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s

Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns

5.81 X 10^3 N.s

A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

Answers

the answer is c yw :D

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.

Answers

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

[tex]distance=v\,*\, t[/tex]

[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

Answers

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  [tex]f= -75 \ cm = - 0.75 \ m[/tex]

b

  [tex]P = -1.33 \ diopters[/tex]

Explanation:

From the question we are told that

    The  image distance is  [tex]d_i = -75 cm[/tex]

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  [tex]d_o = \infty[/tex]

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              [tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]

substituting values

             [tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]

=>         [tex]f= -75 \ cm = - 0.75 \ m[/tex]

Generally the power of the corrective lens is  mathematically represented as

        [tex]P = \frac{1}{f}[/tex]

substituting values

       [tex]P = \frac{1}{-0.75}[/tex]

        [tex]P = -1.33 \ diopters[/tex]

Which of the following is true about a planet orbiting a star in the uniform circular motion? B. The speed of the plant is always changing. C. The velocity of the plant is constant. D. The acceleration vector of the plant points towards the center of the circle.

Answers

Explanation:

If an object moves in a circular path, this type of motion is called uniform circular motion. The speed of the object is constant in this type of motion.As velocity is a vector quantity, it keeps on changing. It can be given by drawing tangent to the circle.Acceleration acts towards the center of circle. Hence, out of the given four options, the correct statement about a planet orbiting a star in uniform circular motion is (d) i.e. The acceleration vector of the plant points towards the center of the circle.

Answer:

D. The acceleration vector of the plant points towards the center of the circle. (Works for Apex)

Explanation:

Velocity is always changing because to have velocity you require direction and the direction is always changing when orbiting therefore A and C are not true. And B is not true because the speed of the planet is not accelerating when orbiting around a star.

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?

Answers

Answer:

The  angle is  [tex]\theta = 15.48^o[/tex]

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  [tex]d = 3.66 \ m[/tex]

     The time taken is  [tex]t = 0.455 \ s[/tex]

   

The  horizontal component of the speed of the dart is mathematically represented as

      [tex]u_x = ucos \theta[/tex]

where u is the the velocity at dart is lunched

  so

      [tex]distance = velocity \ in \ the\ x-direction * time[/tex]

substituting values

      [tex]3.66 = ucos \theta * (0.455)[/tex]

 =>   [tex]ucos \theta = 8.04 \ m/s[/tex]

From projectile kinematics the time taken by the dart can be mathematically represented as

         [tex]t = \frac{2usin \theta }{g}[/tex]

=>    [tex]usin \theta = \frac{g * t}{2 }[/tex]

       [tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]

      [tex]usin \theta = 2.23[/tex]

=>   [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]

       [tex]\theta = tan^{-1} [0.277][/tex]

      [tex]\theta = 15.48^o[/tex]

     

The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?

Answers

Answer:

God is omnipresent.

Explanation:

This means God is everywhere and He works where ever we are in the world

what is transmission of heat?​

Answers

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Conduction ConvectionRadiation.

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

Hope this helps ;) ❤❤❤

Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

Hope it helps!!Please mark me as the brainliest!!!

Thanks!!!!

When the reflected path from one surface of a thin film is one full wavelength different in length from the reflected path from the other surface and no phase change occurs, will the result be destructive interference or constructive interference?

Answers

Answer:

destructive interference

Explanation:

As we know that , when the phase difference between the path of two wavelength is 180°, then its known as destructive interference . And when the phase difference between the path of two wavelength is 0°, then its known as constructive interference.

In the constructive interference , the resulting amplitude will be maximum while in the destructive interference , the resulting amplitude will be zero(minimum).

Therefore the answer will be destructive interference.

Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)

Answers

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r²           [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.90 kg of ice and 1.10 kg of liquid water.

Required:
What was the initial temperature of the piece of copper?

Answers

Answer:

122°C

Explanation:

From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.

That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...

Heat lost by copper = Heat gained by ice + Heat of fusion

-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)

-> T(2730) = 334001

-> T = 122°C

An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.

Answers

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

 

The formula of the pressure of the static fluid

P1 = P2+ρgh

Where,

P1 -  pressure at depth 'h'

P2 -  atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] =  1670900 Pa

h - Depth = 160m  

ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]

g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]

The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]  

 

[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]  

 

 For an ideal gas,  

PV =nRT  

[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]  

[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]

So,

 

[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]  

Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

To know more air bubble volume,

https://brainly.com/question/10509397

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