Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
I’m sorry if i took up a lot of space, hope this is a valid approximate answer
Answer:
0.557 s
Explanation:
Given:
v₀ = 5.46 m/s
v = 0 m/s
a = -9.8 m/s²
Find: t
v = at + v₀
0 m/s = (-9.8 m/s²) t + 5.46 m/s
t = 0.557 s
The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.
Answer:
dt/dx = -0.373702
dt/dy = -1.121107
Explanation:
Given data
T(x, y) = 54/(7 + x² + y²)
to find out
rate of change of temperature with respect to distance
solution
we know function
T(x, y) = 54 /( 7 + x² + y²)
so derivative it x and y direction i.e
dt/dx = -54× 2x / (7 +x² + y²)² .........................1
dt/dy = -54× 2y / (7 + x² + y²)² .........................2
now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2
dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²
dt/dx = -0.373702
and
dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²
dt/dy = -1.121107
