Answer:
Explanation:
See the file attached .
b ) Range of projectile
= u²sin2θ / g
= 42² sin32 x 2 / g
= 42² sin64 / 9.8
= 161.8 m
c )
Max height = u² sin²32 / 2 g
= 42² sin²32 / 2x 9.8
= 25.27 m .
Answer:
Explanation:
We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.
The velocity of sound is found in:
v = 331.5 + .606T
We need to find that first in order to fill it into the frequency equation which is
where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:
v = 331.5 + .606(25) and
v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:
v = 347 m/s
Filling that into the frequency equation:
and
so
Answer:
θ_c = 36.87°
Explanation:
Index of refraction for index medium; n_i = 2
Index of refraction for Refractive medium; n_r = 1.2
Formula to find the critical angle is given;
n_i(sin θ_c) = n_r(sin 90)
Where θ_c is critical angle.
Thus;
2 × (sin θ_c) = 1.2 × 1
(sin θ_c) = 1.2/2
(sin θ_c) = 0.6
θ_c = sin^(-1) 0.6
θ_c = 36.87°
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Answer:
They will be moving away from each other at the same speed.
Explanation:
The final speed may be identical to the initial speed if the collision is perfectly elastic. The final speed will be zero if the collision is perfectly in-elastic.
In all likelihood, the final speed will be slightly slower than the initial speed.