Answer : The correct option is (d) 2.73 m
Explanation :
By the 2nd equation of motion,
where,
s = distance or height = ?
u = initial velocity = 3.0 m/s
t = time = 0.5 s
a = acceleration due to gravity =
Now put all the given values in the above equation, we get:
Therefore, the correct option is (d) 2.73 m
Answer:
The correct option is;
C. 1,715 m
Explanation:
We are given the information from the group of teen at the City edge
Time of arrival of explosion sound = 5 s after sighting
Time of sighting explosion = 5 s before hearing the boom
Speed of sound in air ≈ 343 m/s
Speed of light = 299,792 km/s
Therefore, distance covered by sound in 5 seconds is given by the following equation;
Hence Distance = 343 m/s × 5 s = 1715 m
To check, we compare the time it would take for the light to cover 1715 m
That is which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.
Therefore, the distance of the students from the factory is approximately 1,715 m
Answer:
Explanation:
Let i be the angle of incidence and r be the angle of refraction .
From the figure
Tan ( 90 - i ) = 2.5 / 8
cot i = 2.5 / 8
Tan i = 8 / 2.5 = 3.2
i = 72.65°
From snell's law
sini / sin r = refractive index
sin 72.65 / sinr = 1.333
sin r = .9545 / 1.333
= .72
r = 46⁰
From the figure
Tan r = d / 4
Tan 46 = d /4
d = 4 x Tan 46
= 4 x 1.0355
=4.14 m .
