Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
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°
Answer:
The final temperature is T2= 5.35°C
Explanation:
Apply the Gay-lussacs's law we have
P1, initial pressure= 5.00 x 10^6 Pa
T1, initiation temperature= 25.°C
P2, final pressure= 1.07 x 10^6 Pa
T2, final temperature= ?
Cross multiplying and making T2 subject of formula we have
T2= 5.35°C