A car is moving along a road at 28.0 m/s with an engine that exerts a force of
2,300.0 N on the car to balance the drag and friction so that the car maintains a
constant speed. What is the power output of the engine?

Answer:

No change

Explanation:

First, we hare to understand what we mean by buoyant force.

Archimedes Principle states that

"the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object."

Hot air balloons rise into the air as a result of the density of the air inside the balloon is less dense i.e warmer air than the air outside the balloon i.e cooler air. This is basically how the balloons work. Now applying this to our question

Replacing Helium with less dense Hydrogen will make no difference to the buoyant force because the volume of the balloon did not change. The buoyant force depends on the weight of the displaced air, and not on the force causing the displacement.

Answer:

The correct answer is 88.84 mmHg.

Explanation:

The pressure differential between the brain and the heart while standing up will be 120 - rho × g (gravity) × h, here h is the distance from the brain to the heart. The h is 40 cm or 0.4 m.  

rho×g×h = 1060 kg/m³×9.8 m/s²×0.4m  

= 4155 Pa  

Now converting Pa to mmHg we get:  

4155 Pa × 760 mmHg / 1.01325 × 10⁵ Pa  

= 31.16 mmHg  

Thus, the pressure in the brain now is 120 - 31.16  

= 88.84 mmHg (hypotension)  

Answer:

Most metamorphic processes take place deep underground, inside the earth's crust.

Explanation:

During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)

Answer:

Answer in explanation

Explanation:

The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:

m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg

But, on moon:

m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg

Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.

Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

Refractive power is the measure of degree of convergence by a lens. The required refractive power for the given glasses is -4. 34 D.

Using lens formula  

[tex]\bold { \dfrac 1 v - \dfrac1 u = \dfrac {1}f}[/tex]

Where,

f-  focal point

v - distance of the image

u - distance of the object  

So,

[tex]\bold { \dfrac 1 {-17} - \dfrac1 {-65} = \dfrac {1}f}\\\\\bold { 0.434 = \dfrac {1}f}\\[/tex]

Since, [tex]\bold {power = \dfrac {- 100 }f}[/tex]

So,

[tex]\bold { power = - 100 \times 0.0434}}\\\\\bold { power = - 4.34\ D}[/tex]  

Therefore, the required refractive power for the given glasses is -4. 34 D.

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Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, [tex]q_1=54\ A[/tex]

Current in wire 2, [tex]q_2=45\ A[/tex]

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]

So, the magnetic force on a 4.3 m length of the wire  on both of currents is F=0.0189 N.

Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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Given that,

A particle to move with uniform circular motion in a uniform magnetic field.

Suppose, The conditions are,

(I). The charged particle has to be positive and it should be moving in a direction opposite to a uniform magnetic field.

(II). The charged particle  should be moving parallel to the magnetic force and perpendicular to the magnetic field.

(III). The magnetic field  should be uniform and charge particle should be moving perpendicular to the magnetic field.

We know that,

An particle to move with uniform circular motion.

Here, electric force is perpendicular to velocity of particle.

The electric field is defined as,

[tex]F_{c}=\dfrac{mv^2}{r}[/tex].....(I)

Suppose, there is magnetic field, if a charge moving with velocity and the magnetic field exerts a field.

The magnetic force is defined as,

[tex]F_{m}=qvB[/tex].....(II)

We need to find the magnetic field

Using equation (I) and (II)

[tex]F_{c}=F_{m}[/tex]

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

[tex]B=\dfrac{mv}{qr}[/tex]

Hence, The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.

(III) is correct option.

Answer:

A& D

Explanation:

See attached file

Answer:

There could be electric and magnetic fields, oriented in opposite directions

Explanation:

Lorentz force, is the force that may be exerted on a body of a specified magnitude of charge q, moving with a velocity v, in a magnetic field B and in an electric field of intensity E. This Lorentz force is given by; F= qE+qvBsin ϕ

However, if the motion of the particle is opposite to the magnetic field such a that ϕ = 0, then there is no net magnetic force on the charge and it moves freely, with a constant velocity and in a straight line. Hence, there is no magnetic field in the region.

The charge moves with constant speed due to same direction of magnetic and electric field.

There could be electric and magnetic fields that is oriented in the same direction or the other reason is that there is no magnetic field and electric field in that region where the charge moves. If the electric and magnetic field are present at the same direction then it means that it applies no force on the charge.

This is due to more distance from the charge as well as the charge travels away from the field occupies by the magnetic and electric field so we can conclude that the charge moves with constant speed due to same direction of magnetic and electric field.

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Answer:

A.) 909 cm/s

B.) 33075 N

Explanation:

A.) Given that the

Mass M = 43 g

Height h = 4.05 R

Radius r = R

At the top of the loop, the maximum potential energy P.E = mgh

Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2

P.E = 43 × 9610.517 × 4.05R

P.E = 1673671.536R J

According to conservative of energy

The maximum P.E = maximum K.E

But K.E = 1/2mv^2

1673671.536R = 1/2mv^2

Substitutes for mass m into the formula

1673671.536R = 1/2× 4.05R × v^2

The R will cancel out

Cross multiply

4.05 v^2 = 3347343.072

V^2 = 3347343.072 / 4.05

V^2 = 826504.4622

V = sqrt( 826504.4622)

V = 909 cm/s

B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,

MV^2/R = N + W

Make N the subject of formula

N = mv^2/ R - W

Where W = mg

Substitute all the parameters into the formula

N = (4.05R × 909^2) /R - 4.05R × 9610.517

N = 3346438.05 - 38922.59

N = 3307515 N

Answer:

Explanation:

v₁² = v₁ . v₁

= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )

= 12.25 + 8.41

= 20.66 m /s

a ) kinetic energy = 1/2 m v₁²

= 1/2 x .7 x 20.66

= 7.23 J

b )

changed velocity v₂ = v₂.v₂

= (6i - 5 j ) . (6i - 5 j )

= 36 + 25

= 61 m /s

kinetic energy = 1/2 m v₂²

= 1/2 x .7 x 61

= 21.35 J

Work done = change in energy

= 21.35 - 7.23

= 14.12 J .

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Answer:

The  time interval is  [tex]t = 3 \ s[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = 4.0 \ rad/s^2[/tex]

     The  time taken is  [tex]t = 4.0 \ s[/tex]

      The angular displacement is  [tex]\theta = 80 \ radians[/tex]

The angular displacement can be represented by the second equation of motion as shown below

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

where  [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval

So substituting values

        [tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]

=>    [tex]w_i = 12 \ rad/s[/tex]

Now considering this motion starting from the start point (that is rest ) we have

       [tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]

Where  [tex]w__{0}}[/tex] is the angular velocity at rest which is zero  and  [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s

        [tex]12 = 0 + 4 t[/tex]

=>       [tex]t = 3 \ s[/tex]

Following are the response to the given question:

Given:

[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]

[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]

To find:

[tex]\to \omega=?\\\\\to t=?\\\\[/tex]

Solution:

Using formula:

[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]  

It would be the angle for rotation at the start of the 4-second interval.

This duration can be estimated by leveraging the fact that the wheel begins from rest.  

[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]

Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".

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Answer:

The final velocity of the block after the bullet passes through is 0.66 meters per second.

Explanation:

The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:

[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]

Where:

[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.

[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.

[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.

The final speed of the block is cleared:

[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]

[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]

If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:

[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]

[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]

The final velocity of the block after the bullet passes through is 0.66 meters per second.

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

v₂ = 306.12 m/s

Answer:

The time taken is [tex]t = 0.0225 \ s[/tex]

Explanation:

From the question we are told that

    The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]

      The  refractive index is  [tex]n_f = 1.5[/tex]

The velocity of the signal is mathematically represented as

       [tex]v = \frac{c}{n_f }[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

 substituting values  

         [tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]

         [tex]v = 2.0*10^{8} \ m/s[/tex]

The time taken is mathematically evaluated as

      [tex]t = \frac{d}{v}[/tex]

substituting values  

      [tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]

      [tex]t = 0.0225 \ s[/tex]

Answer:

  x =  0.4654 cm

Explanation:

In this exercise we use the law of refraction

           n₁ sin θ₁ = n₂ sin θ₂

apply this formula to the first surface, where n₁ is the index of refraction of air (n₁ = 1) and n₂ is the index of refraction of glass (n₂ = n)

            θ₂ = sin⁻¹ (sin θ₁ / n)         (1)

having this angle we use trigonometry to find the value of the point where it comes out when we reach the other side

refracted ray

            tan θ₂ = x₂ / d

            x₂ = d tan θ₂

this value is the distance displaced by the refracted ray

now let's find the distance at which the incident beam should exit

           tan θ₁ = x₁ / d

           x₁ = d tan θ₁

the displacement of the ray is the difference between these two distances, we will call it x

           x = x₁ - x₂

            x = d tan θ₁ - d tan θ₂

           x = d (tan θ₁ - tan θ₂)        (2)

the easiest way to do the calculations is to find tea2 from the binding 1 and then perform the calculation with equation 2

calculate

            θ₂ = sin⁻¹ (sin 45 /1.5)

             θ₂ = 28.13º

             x = 1.0 (tan 45 - tan 28.13)

             x =  0.4654 cm

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

Answer:

Explanation:

Area of the loop = a b

current = I

magnetic moment of the loop M  = area x current

= ab I

Torque on the loop = MB sinθ

here θ = 90

Torque = MB

= abIB

In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .

Answer:

As a body moving upward

T=real weight + apparent weight

T=550+490

T=1040

hope u will get the answer:)

Explanation:

Answer:

the planet Earth is a good example

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

Answer:

Option A

Explanation:

From the graph, we came to know that Force and acceleration are in direct relationship.

Also,

Force = 0 when Acceleration = 0

Because Both are 0 at the origin.

Answer:

A. It will be 0 meters per second per second.

Explanation:

The force and acceleration is in a proportional relationship, that means the line goes through the origin.

On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

Answer:

7.2 kilowatts

Explanation:

Energy= power x time

where:

power = 1,200watts to kilowatts is 1.2kw

time = 6hours

therefore,

energy = 1.2kw x 6hrs

=7.2 kilowatts