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Pepsi [2]
2 years ago
11

Visible light travels at a speed 3.0 × 108 of m/s. If red light has a wavelength of 6.5 × 10–7 m, what the frequency of this lig

ht?
A) 2.2 × 10–15 HZ
B) 2.0 × 102 HZ
C) 3.0 × 108 HZ
D) 4.6 × 1014 HZ
Physics
2 answers:
WARRIOR [948]2 years ago
8 0

Answer on Edge is D.

Frequency = (3/6.5) x 10¹⁵ = 0.4615 x 10⁻¹⁵ = 4.615 x 10⁻14 second.

:)

gayaneshka [121]2 years ago
4 0

i think its d on edge

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The total energy of a 0.050 kg object travelling at 0.70 c is
quester [9]
Would presume the energy as kinetic energy.

E = (1/2)*mv²

But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s

Ke =  (1/2)*mv² = 0.5*0.05*(0.7*<span>3* 10⁸)</span>² = 1.1025 * 10¹⁵ Joules


There is no exact match from the options.
4 0
2 years ago
For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo
Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0
2 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
2 years ago
A helicopter is traveling at 86.0 km/h at an angle of 35° to the ground. What is the value of Ax? Round your answer to the neare
jasenka [17]
Answer: 70.5 km/h

Justification:

The question is not clearly stated but it seems you are asking for the x - component of the velocity of the helicopter.

You can find the x and y - components of the velocity using the trigonometric ratios sine and cosine.

The sine ratio relates the y-component and the velocity by:

sin(angle) = y-component of velocity / velocity

The cosine ratio related the x-component and the velocity by:

cos(angle) = x-component of velocity / velocity.

Since you have the angle and the velocity and are asked by the x-component of the velocity, you need to use the cosine ratio:

cos(35°)= x-component / 86.0 km/h

 => x -component = 86.0 km/h * cos(35°) = 70.5 km/h
6 0
2 years ago
Read 2 more answers
Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a
Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field F=-6i+5x^2j-5k

We need to calculate the flux

Using formula of flux

\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)

Put the value into the formula

\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz

\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz

\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}

\phi=682.6\ Wb

Hence, The flux is 682.6 Wb.

7 0
2 years ago
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