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LiRa [457]
3 years ago
9

As the wavelength of a wave increases on the electromagnetic spectrum,

Physics
2 answers:
posledela3 years ago
8 0

Answer:

C

Explanation:

Waves can be measured using wavelength and frequency. The distance from one crest to the next is called a wavelength. The number of complete wavelengths in a given unit of time is called frequency. As a wavelength increases in size, its frequency, and energy decrease.

pishuonlain [190]3 years ago
4 0

Answer:

C

Explanation:

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Chicken Feet Or Butt Whole

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3 years ago
Distinguish between a converging lens and a diverging lens.
andriy [413]

Answer:A converging lens is thickest in the middle and causes parallel light rays to converge through the focal point on the opposite side of the lens. A diverging lens is thinner in the middle and causes parallel light rays to diverge away from the focal point on the same side of the lens.

Explanation:

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4 years ago
A skydiver has a mass of 65 kg.<br> Calculate the weight of the skydiver.
Vinil7 [7]
F=m*a
F=65 kg *9.8 m/s^2
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3 0
3 years ago
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
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