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4 years ago

Consider the following kinds of electromagnetic waves:

Physics

1 answer:

marysya [2.9K]4 years ago

7 0

Answer:

According to wavelengths in descending order: C, B, D, E and A

According to frequencies in descending order: A, E, D, B and C

According to speeds in vacuum: A =B =C =D =E

Explanation:

in the EM spectrum, radio waves have the longest wavelength while gamma rays have the shortest wavelength.

All EM waves travel with the same speed of $3*10^8m/s$ in a vacuum.

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Which statement best compares the accelerations of two objects in free fall?

gregori [183]

Answer:

D) The objects have the same acceleration. With no air resistance and the same gravity, the acceleration due to gravity should be the same for both objects.

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3 years ago

What is the RICE regimen for injuries?

amid [387]

Answer:

<h2>R.I.C.E. stands for rest, ice, compression, and elevation</h2>

Explanation:

I hope this helps

7 0

3 years ago

2. Is the original mixture homogeneous or heterogeneous? Why? .

myrzilka [38]

Answer:

There is no chemical combination of the substances in a mixture, so they retain their physical properties. There is the same composition throughout a homogeneous mixture. The structure of a heterogeneous mixture differs.

Explanation:

4 0

3 years ago

Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C

Degger [83]

Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h = - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

4 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago