Answer:
7.1 m
Explanation:
Given:
Distance traveled by the student in the first attempt = 
Distance traveled by the student in the second attempt = 
So, the maximum distance that the student could travel in this attempt = 
So, the maximum distance that the student could travel in this attempt = 
Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.
So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.
And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.
Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.
Well gravity is 9.8m/s so i would assume 19.6m/2 is the velocity
By definition, the refractive index is
n = c/v
where c = 3 x 10⁸ m/s, the speed of light in vacuum
v = the speed of light in the medium (the liquid).
The frequency of the light source is
f = (3 x 10⁸ m/s)/(495 x 10⁻⁹ m) = 6.0606 x 10¹⁴ Hz
Because the wavelength in the liquid is 434 nm = 434 x 10⁻⁹ m,
v = (6.0606 x 10¹⁴ 1/s)*(434 x 10⁻⁹ m) = 2.6303 x 10⁸ m/s
The refractive index is (3 x 10⁸)/(2.6303 x 10⁸) = 1.1406
Answer: a. 1.14
Answer:
60 kg
Explanation:
The man's weight is equal to the weight of the water he displaces.
mg = ρVg
m = ρV
m = (1000 kg/m³) (3 m × 2 m × 0.01 m)
m = 60 kg
The man's mass is 60 kg.