What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s?

You want to put a new CD in your car stereo.While driving at 45 mi/h,you look away for 2.0s.What distance (in miles) do you driv

KengaRu [80]

Answer:

You drive 0.025 miles blind (or 132 ft)

Explanation:

(distance) = (speed) x (time)

(distance blind) = (speed) x (time looking for CD)

(distance blind) = (45 mi/hr) x (2 s) x (1/3600 hr/s) = 0.025 miles = 132 ft

3 0

2 years ago

Which band of the electromagnetic spectrum has the lowest frequency

Murrr4er [49]

Answer:

Radio waves

Explanation:

Electromagnetic waves are produced by the oscillations of electric and magnetic field. They are transverse waves, which means that the oscillations occur in a direction perpendicular to the direction of propagation of the wave, and they are the only type of waves that can travel through a vacuum.

Electromagnetic waves are classified into 7 different types, depending on their frequencies. From lowest to highest frequencies, we have:

Radio waves

Microwaves

Infrared

Visible light

Ultraviolet

X-rays

Gamma rays

Radio waves are the electromagnetic waves with lowest frequency. They are used, for examples, for satellites, telecommunication, broadcasting.

5 0

2 years ago

A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time inte

serious [3.7K]

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

3 0

2 years ago

hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body

bagirrra123 [75]

Answer:

(a)

$\lambda _{m}=9.332 \times 10^{-6}m$

(b)

$\lambda _{m}=1.632 \times 10^{-6}m$

(c) $\lambda _{m}=4.988 \times 10^{-7}m$

Explanation:

According to the Wein's displacement law

$\lambda _{m}\times T = b$

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{310}$

$\lambda _{m}=9.332 \times 10^{-6}m$

(b) T = 1500°C = 1500 + 273 = 1773 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}$

$\lambda _{m}=1.632 \times 10^{-6}m$

(c) T = 5800 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}$

$\lambda _{m}=4.988 \times 10^{-7}m$

5 0

2 years ago

Dynamic stretching serves 2 purposes warming up and controlling BMI true or false

Ket [755]

Uhm i think false but i’m not sure

8 0

2 years ago