An electromagnetic wave has a frequency of 1.0 x 10^5. What is the wavelength of the wave?

Answer:

Use some Glue.. . . Aluminium is not magnetic, so your magnet can't sticl to aluminium without using Glue. You cannot get a magnet to stick to aluminum, no matter how hard you try. Aluminum is diamagnetic, which means it repels a magnetic field.

Explanation:

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brainliest please

Answer:

Wind is a renewable energy source. Overall, using wind to produce energy has fewer effects on the environment than many other energy sources. Wind turbines do not release emissions that can pollute the air or water (with rare exceptions), and they do not require water for cooling.

Explanation:

Answer:

V2 = 37.5 L

Explanation:

Given the following data;

Initial volume, V1 = 25 L

Initial temperature, T1 = 20°C

Final temperature,T2 = 30°C

To find the final volume V2, we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

[tex] \frac {V}{T} = K[/tex]

[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]

Making V2 as the subject formula, we have;

[tex] V_{2}= \frac{V_{1}}{T_{1}} * T_{2}[/tex]

[tex] V_{2}= \frac{25}{20} * 30 [/tex]

[tex] V_{2}= 1.25 * 30 [/tex]

V2 = 37.5 L

Answer:

a) [tex] v = 2.36 \cdot 10^{7} m/s [/tex]

b) [tex] B = 3.80 \cdot 10^{-4} T [/tex]

c) [tex] f = 1.06 \cdot 10^{7} Hz [/tex]

d) [tex] T = 9.43 \cdot 10^{-8} s [/tex]

Explanation:

a) We can find the electron's speed by knowing the kinetic energy:

[tex] K = \frac{1}{2}mv^{2} [/tex]

Where:    

K: is the kinetic energy = 1.59 keV

m: is the electron's mass = 9.11x10⁻³¹ kg

v: is the speed =?

[tex] v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s [/tex]

b) The electron's speed can be found by using Lorentz's equation:

[tex] F = q(v\times B) = qvBsin(\theta) [/tex]   (1)

Where:

F: is the magnetic force

q: is the electron's charge = 1.6x10⁻¹⁹ C

θ: is the angle between the speed of the electron and the magnetic field = 90°

The magnetic force is also equal to:

[tex] F = ma_{c} = m\frac{v^{2}}{r} [/tex]   (2)

By equating equation (2) with (1) and by solving for B, we have:

[tex] B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T [/tex]

c) The circling frequency is:

[tex] f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r} [/tex]

Where:

T: is the period = 2π/ω

ω: is the angular speed = v/r

[tex] f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz [/tex]

d) The period of the motion is:

[tex] T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s [/tex]

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Answer:

Cumulus, Stratus, and Cirrus. There are three main cloud types.

Explanation:

hopes it help^_^

Answer:

medium

Explanation:

I am not really sure

Answer:

The angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Given;

mass of the ball, m = 1.8 kg

number of the ball's rotation per minute, n = 53 RPM

The angular speed of the ball in radians per second is calculated as follows;

[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]

Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Use the formula:

[tex] e = mc \delta \theta[/tex]

e is the energy released,

m is the mass of water,

c is the specific heat capacity,

δθ is the change in temperature ( 100 - 22)

Answer:

Energy = 7.83 x 10⁻¹⁹ J

Energy = 6.63 x 10⁻¹⁹ J

Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

[tex]Energy = \frac{hc}{\lambda}\\[/tex]

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ Js

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\[/tex]

Energy = 7.83 x 10⁻¹⁹ J

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\[/tex]

Energy = 6.63 x 10⁻¹⁹ J

Answer:

All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the speed c in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves

Answer:

C

Explanation:

The said wave takes the shortest time to move/get transmitted

An action which could increase the pair's angular velocity is the figure skater on the left pointing his right arm down instead of up.

Angular velocity simply refers to the rate of change of angular displacement of a physical object (body) with respect to time.

This ultimately implies that, angular velocity is a measure of how fast and quickly a physical object (body) rotates with respect to another point or how its angular displacement (position) changes with respect to time.

In this scenario, an action which could increase the pair's angular velocity is when the figure skater on the left points his right arm down instead of up.

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the skater on the left pulls the skater on the right closer to him

Explanation:

apex bliches

Answer:

the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

Explanation:

Given the data in the question;

slit separation d = 0.165 mm = 0.165 × 10⁻³ m

wavelength λ = 560 nm = 560 × 10⁻⁹ m

distance between the screen and slits D = 4.05 m

now,

for fifth-order bright fringe path difference = mλ

where m is 5

so, the difference in path lengths from each of the slits will be;

Δr = mλ

we substitute

Δr = 5( 560 × 10⁻⁹ m )

Δr = 28 × 10⁻⁷ m

Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

Answer:

Block, incline, spring and gravity.

Explanation:

For us to have an energy analysis involving gravitational and spring potential energy, we will need to have a block with specific mass that is held at rest on a frictionless incline plane. Now, the block will compress the spring by a specific length from its equilibrium position and then the block is released to travel a distance right up the slope.

So basically, we will need Block, incline, spring and then gravity for it to move.

Answer:

Option B

Explanation:

Forces acting on the skier-

F1 [tex]= -mg sin(30)[/tex] down the slope  

F2 [tex]= -mg cos(30)[/tex]

F3 = friction force  [tex]= 0.74 mg cos(30)[/tex]  

Net force, down the slope  

[tex]= -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m[/tex]

Acceleration [tex]= F/m= 1.38[/tex] m/s2  

Acceleration remains constant, initial speed is 20 m/s

Speed at time t is [tex]1.38t- 20[/tex] m/s

Distance down the slope at time t is [tex]0.69t^2- 20t[/tex]

When the skier stops, her speed is 0. Thus,  

, [tex]1.38t- 20= 0\\t= 20/1.38\\= 14.5[/tex] seconds

Distance travelled in 14.5 seconds [tex]= (0.69)(14.52- 20(14.5)= -145 m[/tex](negative because it is down the slope).

Option B is correct

In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.

As given in the problem the  Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,

As given in the table marks is 1.5 AU far from Mars,

1 AU = 100 meters

1.5 AU = 1.5 × 100

           = 150 meters

Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.

Learn more about the unit of measurement here, refer to the link ;

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Answer:

its true

Explanation:

The titan mission scheduled for 2026 is a reconnaissance to explore the origin of life. Because, some scientific studies based on titan has showed that there is the presence of some biomolecules. Hence, the statement is true.

The distinctive, abundantly organic planet of Titan is where we will travel next in the solar system, according to NASA. The Dragonfly expedition will fly numerous sorties to sample and inspect locations around Saturn's icy moon, advancing our hunt for the components of life.

In 2026, Dragonfly will take off, and it will land in 2034. The rotorcraft will travel to a great number of Titan's promising sites in search of prebiotic chemical processes that are similar to those on Titan and Earth.

Dragonfly, which has eight rotors and flies like a big drone, is the first multi-rotor vehicle NASA will operate for science on another world. To become the first spacecraft to ever fly its full science payload, it will take advantage of Titan's dense atmosphere, which is four times denser than Earth's. Hence, the statement is true.

Find more on Titan mission :

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Answer:

80

Explanation:

800/10

=80

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Answer:

Centripetal acceleration = 3.125 m/s²

Explanation:

Given the following data;

Radius, r = 50 m

Velocity, V = 12.5 m/s

To find the centripetal acceleration;

Centripetal acceleration = Velocity²/radius

Centripetal acceleration = 12.5²/50

Centripetal acceleration = 156.25/50

Centripetal acceleration = 3.125 m/s²

Therefore, the centripetal acceleration of the greyhound running on a circular track is 3.125 meters per seconds square.

Answer:

1

Explanation:

Answer:

[tex]Q_net=534.67\frac{w}{m}[/tex]

Explanation:

From the question we are told that:

Steam line diameter [tex]D=89mm \approx 0.089m[/tex]

Surface emissivity [tex]\mu=0.8[/tex]

Steam temp [tex]T_s=200\textdegree C \approx 200+273=473k[/tex]

Surrounding temp [tex]T_a=20 \textdegree C \aprrox 20+273= 293k[/tex]

Generally the equation for heat loss per unit length due to radiation [tex]Q_{net}[/tex] is mathematically given by

[tex]Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)[/tex]

[tex]Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)[/tex]

[tex]Q_net=534.67\frac{w}{m}[/tex]

Answer:

I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}

Explanation:

The moment of inertia is a scalar quantity, therefore additive, therefore we can find the moments of inertia of each body with respect to the point and add them.

Let's use the parallel axis theorem for the moment of inertia.

          I = [tex]I_{cm}[/tex] + m d²

the moments and inertia of the bodies are

disk                    Icm = ½ m R²

rod                     Icm = 1/12 m L²

rectangle           Icm = 1/12 m (a² + b²)

where a and b are the sides of the rectangle

Let's fix a reference frame on the point body, the length of the rectangle is x and its height y, the total moment of inertia is

          I_total = I_point + I_disco + I_rod + I_rectangular

the moment of inertia of the point is

           I = m r² = m 0

           I_point = 0

disk moment of inertia

suppose it is on the y-axis with x = 0

           I_disco = ½ m R² + m y²

moment inertia rod

located in the opposite corner

The distance from the point to the center of the mass of the rod is

           R = [tex]\sqrt{x^2 +y^2 }[/tex]

           I_varrilla = 1/12 m L² + m ([tex]\sqrt{ x^2 + y^2 }[/tex])

rectangle moment inertia

located on the x axis

            I_rectangle = ½ m (a² + b²) + m x²

we substitute

         I_total = 0 + ½ m R² + m y² + 1/12 m L² + m (Ra x ^ 2 + y²) + ½ m (a² + b²) + m x²

         I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) + m √(x^2 + y²)

         I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}

Answer:

- the power rating of the resistance heater is 24139.5 W

- the inner surface temperature of the pipe at the exit is 96.34°C

Explanation:

Given the data in the question;

Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s        

The water is to be heated from 10°C to 80°C;

so Average or mean temperature [tex]T_{avg[/tex] will be;

[tex]T_{avg[/tex] = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C

Now, from the Table " Properties of Water " at average temperature;

at [tex]T_{avg[/tex] = 45°C

density p = 990.1 kg/m³

specific heat [tex]C_p[/tex] = 4180 J/kg-k

thermal conductivity k = 0.637 W/m-°C

Now, we determine the mass flow;

m" = pV"

we substitute

m" = 990.1 × 8.333 × 10⁻⁵

m" = 0.08250 kg/s

we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;

Q' = m"[tex]C_p[/tex]( T₂ - T₁ )

we substitute

Q' = (0.08250 × 4180 ) ( 80 - 10 )

Q' =  344.85 × 70

Q' = 24139.5 W

Hence, the power rating of the resistance heater is 24139.5 W

Next, we determine the average velocity of water in the tube;

[tex]V_{avg[/tex] = V" / [tex]A_c[/tex]

[tex]V_{avg[/tex] = V" / ( [tex]\frac{1}{4}[/tex]πD² )

given that;  flows through a 2-cm-internal-diameter; D = 0.02 m

we substitute

[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( [tex]\frac{1}{4}[/tex]π × (0.02)² )

[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )

[tex]V_{avg[/tex] =  0.265 m/s

Also, from table " saturated water property table "

At 45°C

viscosity μ = 0.596 × 10⁻³ kg/m-s

Prandtl number Pr = 3.91

Now, we determine the kinematic viscosity

v = μ / p

we substitute

v = ( 0.596 × 10⁻³ ) / 990.1

v = 6.01959 × 10⁻⁷ m²/s

so, Reynolds number in the flow region will be;

Re = ([tex]V_{avg[/tex] × D) / v

we substitute

Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)

Re = 8804.586

we can see that our Reynolds number (  8804.586 ) more than 2300 and less than 10,000.

Hydraulic and thermal entry length are equal in this flow region,

such that;

[tex]L_h[/tex] = [tex]L_t[/tex]

⇒ 10 × D = 10 × 0.02 = 0.2 m

we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.

So we the Nuddelt number

Nu = [tex]0.023Re^{0.8} Pr^{0.4[/tex]

Nu = 0.023 × [tex]8804.586^{0.8[/tex] × [tex]3.91^{0.4[/tex]

Nu = 56.8

Hence, the heat transfer coefficient h will be;

h = [tex]\frac{k}{D}[/tex] × Nu

we substitute

h = [tex]\frac{0.637}{0.02}[/tex] × 56.8

h = 31.85 × 56.8

h = 1809.1 W/m²-°C

Now, area of the heat transfer will be

A[tex]_s[/tex] = πDL

we substitute

A[tex]_s[/tex] = π × 0.02 × 13

A[tex]_s[/tex] = 0.8168 m²

Finally we determine the inner temperature of the pipe at exit. using the relation;

Q' = hA[tex]_s[/tex]( T₃ - T₂ )

we substitute

24139.5 = 1809.1 × 0.8168( T₃ - 10 )

24139.5 = 1477.67288( T₃ - 80 )

24139.5 = 1477.67288T₃ - 118213.8304

24139.5 + 118213.8304 = 1477.67288T₃

1477.67288T₃ = 142353.3304

T₃ = 142353.3304 / 1477.67288T

T₃ = 96.34°C

Therefore, the inner surface temperature of the pipe at the exit is 96.34°C

Answer:

Momentum(P) = 1336kgm/s

Explanation:

Momentum is the product of the mass and velocity, therefore Momentum(P) is = mass*velocity. since the mass is given 800kg, all we are left with is the displacement 500m and the time 300s which can be used to find the velocity. so velocity is the displacement divided by the time, ie 500/300 which will give us the velocity 1.67m\s. hence we multiply the mass and the velocity and we'll get 1336kgm/s.

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Force is defined as the rate of change of momentum. For an unchanging mass, this is equivalent to mass x acceleration. So, 1 N = 1 kg m s-2, or 1 kg m/s2.

Answer:

When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.

Explanation:

Answer:

When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.

Explanation:

pluto

Answer:

Speed = 328 m/s

Explanation:

Given the following data:

Wavelength = 4 m

Frequency = 82 Hz

To find the speed of the radio wave;

Speed = wavelength * frequency

Speed = 4 * 82

Speed = 328 m/s

Therefore, the radio wave is travelling at 328 meters per seconds.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.