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3 years ago

Describe the frequency and wavelength range of radio waves

Physics

1 answer:

sammy [17]3 years ago

3 0

Answer:

Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light. They have frequencies from 300 GHz to as low as 3 kHz, and corresponding wavelengths from 1 millimeter to 100 kilometers.

Explanation:

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A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev

Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²

a= 4×π²×2×3²

a=710.6 m/s²

5 0

3 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

3 years ago

Based on Newton’s 3rd law if you were to push on the wall with a force of 100 N, how much force would the wall push back towards

rusak2 [61]

The answer is 100N. Look up the definition of Newton's third law.

3 0

3 years ago

17.a car is traveling at a constant velocity of 60 km/h for 4 hours means

Setler [38]

<span>Assuming the car is travelling in the same direction for the entire hour, the acceleration is zero.</span>

7 0

3 years ago

A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that

irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

$h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)$

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0

3 years ago