Which of the following is an example of an electromagnetic wave?

a radio wave

a water wave

the oscillation of a spring

sound waves

Answer:

they were hunter gatherers

Explanation:

Answer:

bc earth rotates

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Explanation:

Answer:

The interaction between the solar wind and Earth's magnetic field, and the influence of the underlying atmosphere and ionosphere, creates various regions of fields, plasmas, and currents inside the magnetosphere such as the plasmasphere, the ring current, and radiation belts.

Explanation:

Answer:

m = 1.31 kg

Explanation:

Given that,

The speed of duck, v = 9.1 m/s

The magnitude of momentum, p = 12 kg-m/s

We need to find the mass of the duck. We know that the momentum of an object is given by :

p = mv

Where

m is the mass of the duck

[tex]m=\dfrac{p}{v}\\\\m=\dfrac{12\ kg-m/s}{9.1\ m/s}\\\\m=1.31 kg[/tex]

So, the mass of the duck is equal to 1.31 kg.

Answer:

72.57° North of east

Explanation:

From the given information:

We can compute the velocity plane that is related to the ground in air in the North direction as;

[tex]v^{\to} _{PG} = v \\ \\ v^{\to} _{PG,x} = 0 \\ \\ v^{\to} _{PG,y} = v[/tex]

However, the velocity of the wind-related to the ground from the NorthEast  direction is;

[tex]v^{\to}_{wG}=100 \ km/h \\ \\ \text{from North East} \\ \\ v_{wG,x} = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_{wG,y} = (-100 \ km/h ) sin 45 = -70.7 km/h[/tex]

Now,

Since the plane is moving with a 236 km/h speed in the Northeast direction;

Then;

[tex]v^{\to} _{pw} = 236 \ km/h \\ \\ v^{\to} _{pw.x} = (236 m/s) cos \theta \\ \\ v^{\to} _{pw,y} = (236\ m/s) sin \theta \\ \\ v_{pG,x} = v_{pw,x} + v_{w G,x} \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = \dfrac{70.7 \ km/h}{236 \ km/h} \\ \\ \theta = cos^{-1} (0.2996) \\ \\ \mathbf{\theta = 72.57}[/tex]

Answer:

Is this your question? Also I think work done is a scalar quantity.

Explanation:

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2π / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

Answer:

A. 4d

Explanation:

Let's begin with the formula for the magnetic field produced by a long wire.

[tex]B = \frac{\mu_0I}{2\pi d}[/tex]

So [tex]d=\frac{\mu_0 I}{2\pi B }[/tex]

at point d_{1} is

[tex]d_{1}=\frac{\mu_{0} i}{2 \pi B_{1}} \\ \frac{d_{1}}{d}=\frac{\frac{\mu_{0} i}{2 \pi B_{1}}}{\frac{\mu_{0} i}{2 \pi B}} \\ d_{2}=d\left(\frac{B}{B}\right) \\ =d\left(\frac{20 \mathrm{mT}}{5 \mathrm{mT}}\right) \\ =4 d[/tex]

Hence, option  (A) is correct answer

Answer:

8

Explanation:

(8√2)² = x² + x²

8² × √2² = 2x²

64 × 2 = 2x²

128 = 2x²

64 = x²

x = 8

give me brainliest please

Answer:

dvhd

Explanation:

xhxjjdvcbxhjddvifidid

Answer:

The units of frequency are called hertz (Hz). Humans with normal hearing can hear sounds between 20 Hz and 20,000 Hz. Frequencies above 20,000 Hz are known as ultrasound

Answer:

D) Report the problem to your supervisor.

Explanation:

This is probably the most efficient way get them to stop or to get them to follow the rules <3

Answer:

A

Explanation:

Answer:

1.17m

Explanation:

The formula to find distance is d=vt/2

This problem is asking for how far the reflecting object has moved so you need to find the distance from the motion sensor at both times.

(343)(0.115) / (2) = 1.97

(343)(0.0183) /(2) =3.14

After that, all you have to do is find the difference so

3.14 - 1.97

= 1.17

Answer:

Force,friction,inertia and momentum

Explanation:

The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Answer:

Push or pull of an object is considered a force. Push and pull come from the objects interacting with one another. Terms like stretch and squeeze can also be used to denote force.

In Physics, force is defined as:

The push or pull on an object with mass that causes it to change its velocity.

Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have ―

We have been asked to calculate the acceleration of the car.

[tex]\qquad \implies\boxed{\red{\sf{ F = ma }}}\\[/tex]

[tex] \quad \twoheadrightarrow\sf { 450 = 1300a} \\ [/tex]

[tex] \quad \twoheadrightarrow\sf {\cancel{ \dfrac{450}{1300}} = 1300a} \\ [/tex]

[tex]\quad\twoheadrightarrow\boxed{\red{\sf{0.346 \; ms^{-2} = a }}}\\[/tex]

Therefore, acceleration of the car is 0.346 m/s².

Answer:

In this conversation the Neil astronaut is right

Answer:

A. There is 100% cloud cover in that area.

Explanation:

Cloud cover is recorded on weather charts by shading in parts of the circle.

If there are no clouds, the circle is left white and if the sky is completely covered in cloud, the circle is shaded completely in which means 100% cloud cover in that area.

Answer:

[tex]0.475\ \text{V}[/tex]

Explanation:

n = Number of turns = 10

r = Radius = 1.5 cm

I = Current = 10 A

t = Time = [tex]6.25\times 10^{-6}\ \text{s}[/tex]

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2r}\\\Rightarrow B=\dfrac{4\pi 10^{-7}\times 10}{2\times 1.5\times 10^{-2}}\\\Rightarrow B=0.00042\ \text{T}[/tex]

EMF is given by

[tex]\varepsilon=\dfrac{nBA}{t}\\\Rightarrow \varepsilon=\dfrac{10\times 0.00042\times \pi (1.5\times 10^{-2})^2}{6.25\times 10^{-6}}\\\Rightarrow \varepsilon=0.475\ \text{V}[/tex]

The average induced emf is [tex]0.475\ \text{V}[/tex].

Answer:

N = 1.42 × 10⁴ cycles

Explanation:

Given that:

frequency f = 4.9 × 10¹⁴ Hz

Time = 2.9 × 10⁻¹¹ s

Speed = 2.3 × 10⁸ m/s

Recall that:

wavelength [tex]\lambda = \dfrac{c}{f} \\ \\[/tex]

Horizontal distance [tex]\Delta x = ct[/tex]

Number of wavelengths [tex](N) = \dfrac{\Delta x}{\lambda}[/tex]

[tex]N = \dfrac{ct}{c/f} \\ \\ N= ft[/tex]

N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)

N = 14210

N = 1.42 × 10⁴ cycles

Answer:

[tex]23^{\circ}[/tex]

Explanation:

n = Refractive index of air = 1

[tex]n_1[/tex] = Refractive index of contact lens = 1.6

[tex]n_2[/tex] = Refractive index of cornea = 1.4

[tex]n_3[/tex] = Refractive index of fluid = 1.3

From Snell's law

[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]

[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]

[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]

The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].

The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,

The given data in the problem is;

n is the refractive index of air = 1

n₁ is the refractive index of contact lens = 1.6

n₂ is the refractive index of cornea = 1.4

n₃ is the refractive index of fluid = 1.3

According to Snell's law. The formula for Snell's law is

[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]

For contact lenses;

[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]

For fluid;

[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]

Hence the angle is the light traveling in the fluid will be 23⁰.

To learn more about snell's law refer to the link;

https://brainly.com/question/10112549

Answer:

a) the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b)   #_photons = 2.6 10¹⁸ photons,  

c) he excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Explanation:

a) the energy of a photo is given by the planck relation

           E = h f

the speed of light is

          c = λ f

          f = c /λ

we substitute

          E = hc /λ

let's calculate the energy for the two photons

λ = 640 nm = 640 10⁻⁰ m

            E₁ = 6.63 10⁻³⁴ 3 10⁸/640 10⁻⁹

            E₁ = 3.1 10⁻¹⁹ J

λ = 520 nm = 520 10⁻⁹ m

             E₂ = 6.63 10⁻³⁴ 3 10⁸/520 10⁻⁹

            E₂ = 3.825 10⁻¹⁹  J

therefore the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b) For this part let's use a direct proportion rule (rule of three). If a photon (lam = 520 nm) has an energy of 3.825 10⁻¹⁹ J, how many photons have an energy of E = 1 10-3 J. Remember that the power is the energy per unit of time

        #_photons = 1 10⁻³ J (1 photon / 3.825 10⁻¹⁹ J)

        #_photons = 2.6 10¹⁸ photons

c) the excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Answer:

I = 1.093 x 10⁻⁴ kg.m²

Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.

Explanation:

The can which is filled with the soup can be modelled as a solid cylinder. The moment of inertia of this solid cylinder about its axis of rotation can be given by the following formula:

[tex]I = \frac{1}{2}mr^2[/tex]

where,

I = moment of inertia of can = ?

m = mass of can with soup = 215 g = 0.215 kg

r = radius of can = diameter/2 = 6.38 cm/2 = 3.19 cm = 0.0319 m

Therefore,

[tex]I = \frac{1}{2}(0.215\ kg)(0.0319\ m)^2 \\[/tex]

I = 1.093 x 10⁻⁴ kg.m²

Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.

Answer:

Table A

Measuring Current as a Function of Voltage with a 20 Ω Resistor

Voltage

(V)

Current: Calculated

(A)

Current: Experimental

(A)

1 0.05 0.05

5 0.25 0.25

10 0.50 0.50

20 1.00 1.00

50 2.50 2.50

Table B

Measuring Current as a Function of Resistance at 25 V

Resistance

(Ω)

Current: Calculated

(A)

Current: Experimental

(A)

10 2.50 2.50

20 1.25 1.25

100 0.25 0.25

200 0.12 0.12

Table C

Measuring Current in a Parallel Circuit

Resistor Set

(Ω)

Total  

Resistance

(Ω)

Calculated

Current

(A)

Observed  

Current

(A)

Observed Current  

through Each Resistor

(A)

20, 20, 20 6.67 3.75 3.74 1.25, 1.25, 1.25

20, 20, 200 9.52 2.63 2.62 1.25, 1.25, 0.12

Voltage needed to raise current to 3.75 A (20, 20, 200 resistor set):

Calculated: 35.7

Observed: 36

Table D

Calculating Power of Circuit ComponentsTeacher Guide (continued)

Observed Total Current

(A)

Current through Each Bulb

(A)

Power Usage per Bulb

(W)

2.00 0.67 6.7

Explanation:

got this from the teachers guide

Answer:

can you explain in Hindi language

because i learn hindi

Answer:

its a. and jusing that youl

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850  + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²

1.29 = 0.36125a[tex]_y[/tex]

a[tex]_y[/tex] = 1.29 / 0.36125

a[tex]_y[/tex] = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a[tex]_y[/tex]

where m₁ is the mass of the text book ( 2.10 kg )

a[tex]_y[/tex] is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

Answer:

a

Explanation:

so the answer is 200N

and I hope it is correct