Answer:
rolling friction is the lady in the forest
static friction is the man pushing the fridge
sliding friction is the snow kid
and fluid fraction is the bird I'm pretty sure
(a)
is the wavelength in air of such a sound wave.
(b)
is the wavelength of this wave in tissue.
<u>Explanation:</u>
Frequency and wavelength can be related by the equation,
Velocity = Wavelength x Frequency

where,
v - velocity of light for all EM (electromagnetic) waves in vacuum
Given:
f - 4.50 MHz = 
a) To find the wavelength in air
We know,
Speed of sound in air = 343 m/s
Apply given frequency and speed of sound in air, we get

b) If the speed of sound in tissue is 1500 m/s, find the wavelength of this wave in tissue
Speed of sound in tissue, v = 1500 m/s

Answer:
Period of oscillation = 1.33 seconds
Explanation:
The period of oscillation is given by:
T = 2π√[I/(MgL)]
for I = 2MR² and L = R,
Given: L = 0.22m = R
T = 2π√[2R/g]
T = 2 × 3.142 Sqrt[( 2 × 0.22)/ 9.8]
T = 6.284 Sqrt(0.44/9.8)
T = 6.284 Sqrt(0.0449)
T = 6.284 × 0.2119
T = 1.33 sec
Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:


I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
Answer:I have to say 56
Explanation: because it is going up by 8