how are characteristics of electromagnetic waves affected by medium

Answer:

Explanation:

The change must be 30 - - 40 which means it came in a 30 meters / second and went out in the opposite direction at 40 meters / second

The change is 70 m/sec.

You could show it to be - 70 meters per second as well. That's done by making the outgoing direction minus.

Delta v = vf - vi.

Now it depends on which way you define vf and vi.

Answer:

Explanation:

There are a total of 5 differen classification for the types of bodies of water that exist on Earth. These 5 would be the following: Oceans, seas, lakes, rivers, and canals. Oceans are the biggest bodies of water on Earth. If we ranked them by their size the ranking would be the following...

Pacific Ocean - being 155,556,651 sq km. in size

Atlantic Ocean - being 76,761,938 sq km. in size

Indian Ocean - being 68,555,923 sq km. in size

Antarctic / Southern Ocean - being 20,327,001 sq km. in size

Arctic Ocean - being 14,055,930 sq km. in size

If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. _ This is false statement.

When a particle is hurled obliquely near the surface of the earth, it travels along a curved path while accelerating continuously in the direction of the planet's center (we assume the particle stays close to the surface of the globe). Such a particle's motion is known as projectile motion, and its route is referred to as a projectile.

In projectile motion total energy is conserved. Hence, when a rock is thrown into the air, the increase in the height would increase the rock’s potential energy, and then the increase in the velocity as it falls to the ground would increase its kinetic energy.

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Answer:

this slow site thinks the answer is a link

Explanation:

this was a week ago so i dont know if u still need help

Answer:

reddish-orrange

Explanation:

please mark me as brainliest

a. The period of the child's motion is 3.171 seconds.

b. The frequency of the vibration is 0.315 Hz.

Since A child swings with a small amplitude on a playground with a 2.5m long chain.

a. So here the time period should be

[tex]2\times \ pi(L/g)^{1/2}\\\\2\times 3.14(2.5/9.81)^{1/2}[/tex]

T=3.171s

b) Now the frequency is

[tex]T=1\div f\\\\f=1\div 3.171[/tex]

f=0.315Hz

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describe how electrons create electricity

Atoms are made up of even smaller elements, called protons, electrons and neutrons. When electrical and magnetic forces move electrons from one atom to another, an electrical current is formed

Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity. ... In its balanced state, copper has 29 protons in its nucleus and an equal number of electrons orbiting around it.

Hope It Helps!

Answer:

c

Explanation:

because a drop of water is falling and that is gravitational potential energy into motion energy

Answer:

it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time.

Explanation:

the last one which is to increase the likelihood of accurate experiment results

Explanation:

Answer:

Advantages of mercury as a thermometric liquid.

-It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

-It does not wet (cling to the sides of) the tube.

-It has a high boiling point

-It expands uniformly (linear expansion) and responds quickly to temperature changes, hence is sensitive.

-It has a visible meniscus.

Disadvantages

-Mercury is very poisonous.

-its expansively is fairly low

-it is expensive

-It has a high freezing point therefore it cannot be used in places where the temperature gets very low.

Alcohol has a thermometric fluid

-Alcohol expands uniformly.

-It has a low freezing point (-115 degreecentigrade) therefore it is very suitable for place where the temperature gets very low.

-It has a large expansively

-It is an easily available cheap liquid, which is safe to use

Disadvantages of alcohol

-it wets the tube

-it has a low boiling point (cannot be used in places with high temperatures)

-it does not react quickly to changes in temperature

-It needs to be dyed, since it's colourless.

Disadvantages of water

-Water has high specific heat capacity. So it cannot be used for measuring small temperature differences.

- Water will wet the surface of the glass tube. It is a sticky substance.

- Water is transparent

Explanation:

Answer:

25 m/s

Explanation:

From the question,

Average speed = Total distance /total time.

S' = D/T........................... Equation 1

D = 40+60 = 100 Km = 100000 m.

T = t₁+t₂

t₁ = (40×3600/72) s = 2000 s

t₂ = 60000/30 = 2000 s

T = 2000+2000 = 4000 s.

SUbstitute the values of T and D into equation 1

S' = 100000/4000

S' = 25 m/s

Answer:

* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]

*E_ disk= 2kQ  [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]

Explanation:

Let's start by finding the electric field of the charged ring

in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.

            Eₓ = E cos θ          (1)

            E = k ∫  [tex]\frac{dq}{r^2}[/tex]

            cos θ = x / r

using the Pythagorean theorem

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

we substitute

            Eₓ = k ∫ [tex]\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }[/tex]

            Eₓ =  [tex]k \frac{x}{(c^2+y^2)^{3/2} }[/tex]   ∫ dq

             Eₓ = k \frac{x}{(c^2+y^2)^{3/2} }  Q

the ring's electric field

             E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]

Now let's find the electric field of the disk

The charge is distributed over the entire disk, so let's use the concept of charge density

              σ = [tex]\frac{dq}{dA}[/tex]

Let's approximate the disk as a group of rings, the width of each ring is dr, the area is

              dA = 2πr dr

we substitute

             σ = [tex]\frac{1}{2\pi r} \ \frac{dq}{dr}[/tex]

             dq = 2π σ r dr

we substitute in equation 1, where the electrioc field is of each ring

             Eₓ = [tex]k \int\limits^R_0 \ { \frac{x}{(x^2+r^2)^{3/2} } \ 2\pi \sigma \ r } \, dr[/tex]

if we use a change of variable

               dv = 2rdr

               v = r²

              Eₓ =  [tex]k x \pi \sigma \int\limits^a_b { \frac{1}{(x^2+v)^{3/2} } } \, dv[/tex]

we integrate

              Eₓ = k x π σ   [tex][ \frac{ (x^2 + r^2)^{-1/2} }{-1/2} ][/tex]

we value in the limits from r = 0 to r = R

              Eₓ = k π σ x  (-2) [ [tex]\frac{1}{ \sqrt{x^2+R^2} } - \frac{1}{x}[/tex]]

              Eₓ = 2π k  σ ([tex]1 - \frac{x}{(x^2 + R^2 ) ^{1/2} }[/tex]  )

             σ = Q/πR²

substitute

             Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )

             E_ disk= 2kQ  [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]

The two electric fields are

* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]

*E_ disk= 2kQ  [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]

we can see that the functional relationship of the two fields is different

Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.

When a tennis racket hits a ball, the ball exerts a force on the racket, and according to Newton's third law of motion, the racket exerts an equal and opposite force on the ball. This means that the ball pushes back on the racket in the opposite direction to which the racket struck the ball.

This principle is often referred to as "action-reaction" or "equal and opposite forces." When the racket collides with the ball, the force applied by the racket causes the ball to accelerate in the opposite direction, leading to its movement away from the racket.

Therefore, when a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.

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they are formed when hot

it's dependent on the rate of cooling of the melt, slow cooling allows large crystals to form, fast cooling yields small crystals. They cool too quickly to form crystals.

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Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

Answer:

Your answer is wood that is what they used

The primary energy source used by power plants to generate electricity is coal.

Power plants are plants which use fuels to generate electricity for use by homes and industries.

Power plants have different energy sources for their fuel.

The primary energy source used by power plants to generate electricity is coal.

Therefore, coal is a primary source of fuel for power plants.

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Answer:

Remember the relation:

Speed*Time = Distance.

Then first let's find the total distance that Hari drove.

First, he droves 2 hours at 60km/h, then the total distance traveled is:

Distance = 2h*60km/h = 120km

Then he droves 3 hours at 70km/h

The distance here is:

Distance = 3*70km/h = 210km

The total distance traveled is 120km + 210km = 330km

The total time of the travel is 2 hours  + 3 hours  = 5 hours.

The average speed is calculated as:

Av speed = (total distance)/(total time)

Av speed = (330km)/(5 h) = 66km/h

The average speed is 66km/h.

b) We know that the speed is 70km/h, but we want to write this in m/s.

Then we first know that 1 km = 1000m

then:

1 = (1000m/1km)

Then 70km/h = 70 km/h*(1000m/1km) = (70*1000) m/h = 70,000 m/h

Now we know that 1h = 3600 seconds, then:

1 = (1h/3600s)

So we can write:

70,000 m/h = (70,000 m/h)* (1h/3600s) = 70,000/3600 m/s = 0.019 m/s

and the distance is 135 km

Again, we can rewrite this as:

135km = 135km*(1000m/1km) = 135,000m

Then we want to solve:

(0.019m/s)*time = 135,000 m

We want to solve this for the time, then we get:

time = (135,000m)/(0.019m/s) = 7,105,263.2 seconds = 1.93 hours

Answer:

Explanation:

mass (m) = 65 kg

velocity (v) = 4.6 m/s

Kinetic energy (KE)

= 1/2 * m * v²

= 1/2 * 65 * 4.6²

=  687.7 J

hope it helps :)

Answer:

a. i) The pressure drop per unit length is 52,151.89 Pa

ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta

b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts

ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta

c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel

ii) The velocity at the center is approximately 2.04 m/s

Explanation:

The given diameter of the aorta, D = 2.5 cm = 0.025 m

The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s

Assumptions;

The blood flow is laminar

The blood is a Newtonian fluid

The viscosity of water ≈ 0.01 poise = 1 cp

a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;

[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]

Where;

Q = The flow rate = A·v

A = The cross sectional area

R = The radius = D/2

Δp/L = The pressure drop per unit length of the pipe

Therefore, we have;

[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]

The pressure drop per unit length ΔP/L = 52,151.89 Pa

ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;

∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa

Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;

[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175

Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta

b. i) The power, P = Q × ΔP

Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts

ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;

P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35

The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length

c. i) The velocity profile across the aorta is given as follows;

[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]

Where;

[tex]v_m[/tex] = The velocity at the center

We get;

[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]

The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s

ii) The velocity profile, v(r), is given by the following formula;

[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]

Therefore, we have;

[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]

The velocity profile of the pipe is created with Microsoft Excel

Answer:

the phenomenon of the interaction of electric currents or fields and magnetic fields.

Answer:

it would be 83% in a fatal crash.

Answer:

T.K.E = 750 Joules.

Explanation:

Given the following data;

Initial velocity, u = 5m/s

Final velocity, v = 10m/s

Mass of bicyclist = 50kg

Mass of bicycle = 10kg

Total mass, Tm = 50 + 10 = 60kg

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

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

Where;

K.E represents kinetic energy measured in Joules.

m represents mass measured in kilograms.

v represents velocity measured in metres per seconds square.

To find the total kinetic energy before accelerating simply means the kinetic energy due to the initial velocity and total mass;

[tex] T.K.E = \frac{1}{2}T_{m}U^{2}[/tex]

Substituting into the equation, we have;

[tex] T.K.E = \frac{1}{2}*60*5^{2}[/tex]

[tex] T.K.E = 30*25 [/tex]

T.K.E = 750 Joules.

Answer:

a) θ = 1.67 10⁻⁵ m,  b) Consequently we must affirm that the vehicle headlights cannot resolve.

Explanation:

a) To find the resolution of the camera we use the Rayleigh criterion for diffraction

            a sin θ = m λ

where m = 1 for the first zero of the slit

we must remember that the angles in the experiments are measured in radians and are very small

            sin θ = θ

we substitute

             θ = [tex]\frac{\lambda}{a}[/tex]

this expression is for a slit, in the case of circular openings the expression must be solved in polar coordinates giving

             θ = [tex]1.22 \ \frac{\lambda}{D}[/tex]

where D diameters of the opening

let's calculate

            θ = [tex]1.22 \ \frac{550 \ 10^{-9}}{ 40\ 10^{-3}}[/tex]

           θ = 1.67 10⁻⁵ m

b) let's use trigonometry to find the separation distance on earth

            tan θ = y / x

            y = x tan θ

let's calculate

remember that the angles must be in radians

                y = 250 10³ tan 1.67 10⁻0-5

                y = 4.18 m

as they indicate that the separation of the headlights is y = 1.6m,

we see that this separation is greater than the separation distance separation.

Consequently we must affirm that the vehicle headlights cannot resolve.

Answer:

A) Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]

B) Speed of Beth = 4v

Explanation:

Given - Consider two musicians, Alf and Beth. Beth is four times the

            distance from the inside of the curve as Alf.

To find - A) Knowing that If Beth travels a distance s during time Δt, how

                   far does Alf travel during the same amount of time.

              B) If Alf moves with speed v, what is Beth's speed?

Proof -

A)

As we know that

Speed = Distance / Time

⇒Distance = Speed ×Time

Now,

Given Speed of Alf = v

⇒Distance of Alf = vt

Also,

Distance of Beth = Speed of beth×time

Given that

Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time=

⇒Distance of Alf = Speed of Alf×time

                           =  [tex]\frac{1}{4}[/tex] Distance of Beth

⇒Distance of Alf = [tex]\frac{1}{4}s[/tex]

∴ we get

Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]

B)

We know the conversion of angular velocity

ω(Alf) = ω(Beth)

⇒V(Alf)/ r = V(Beth)/4r

⇒V(Alf) = V(Beth) / 4

⇒V(Beth) = 4 V(Alf)

As given, Alf moves with speed v

⇒V(Beth) = 4v

So, the correct option is - a.4v