Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine
= 62 MPa = 62 x 10^6 Pa
we also know that
= ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure
, we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
v = 57.2 m/s
Explanation:
The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:
v = s/t
where,
s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m
t = time taken to cover the distance = 2 h 14 min
Now, we convert it into minutes:
t = (2 h)(60 min/1 h) + 14 min
t = 120 min + 14 min = (134 min)(60 s/1 min)
t = 8040 s
Therefore, the value of velocity will be:
v = (4.6 x 10⁵ m)/8040 s
<u>v = 57.2 m/s</u>
Answer:

Explanation:
We are given that







We have to find the exit temperature.
By steady energy flow equation



Substitute the values




Answer:
a) θ₁ = 23.14 °
, b) θ₂ = 51.81 °
Explanation:
An address network is described by the expression
d sin θ = m λ
Where is the distance between lines, λ is the wavelength and m is the order of the spectrum
The distance between one lines, we can find used a rule of proportions
d = 1/600
d = 1.67 10⁻³ mm
d = 1-67 10⁻³ m
Let's calculate the angle
sin θ = m λ / d
θ = sin⁻¹ (m λ / d)
First order
θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₁ = sin⁻¹ (3.93 10⁻¹)
θ₁ = 23.14 °
Second order
θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₂ = sin⁻¹ (0.786)
θ₂ = 51.81 °
Answer:
The correct answer is "0.32 mL".
Explanation:
The given values are:
Density of gold bar,
d = 19.3 g/mL
Mass of gold bar,
m = 6.3 grams
Now,
The volume will be:
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 