An icy object moving through space in a highly eccentric orbit is called a

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

NikAS [45]

Answer:

a) 68.943 m

b) 41.846 m

c) 80.648 m

Explanation:

Given:

Delay in time for spectator A, t₁ = 0.201 s

Delay in time for spectator B, t₂ = 0.122 s

the delay in sound heard is the due to the distance being traveled by the sound from the kicker to the spectator

thus,

a) Distance of the kicker from A,

d₁ = speed of sound × time taken

d₁ = 343 m/s × 0.201 s = 68.943 m

b) Distance of the kicker from B,

d₂ = speed of sound × time taken

d₂ = 343 m/s × 0.122 = 41.846 m

c) Since the angle between the two spectators for the player is 90°

thus, a right angles triangle is formed.

where, the distance between the spectators is the hypotenuses (s) of the so formed triangle

Therefore,

s² = d₁² + d₂²

on substituting the values, we get

s² = 68.943² + 41.846²

or

s² = 6504.22

or

s = √6504.22

or

s = 80.648 m

hence, the distance between the spectators is 80.648 m

3 0

3 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$