Answer:
Choice a. [tex]70\; \rm N[/tex], assuming that the skating rink is level.
Explanation:
Net force in the horizontal directionThere are two horizontal forces acting on the boy:
The pull of his friend, andFrictions.The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.
The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.
[tex]\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N[/tex].
Net force in the vertical directionThe net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.
However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.
Net forceTherefore, the (combined) net force on this boy would be:
[tex]\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N[/tex].
The coil has a radius of 8 cm and 110 turns. While moving the magnet closer towards the coil, at a certain point in time, the voltage meter shows -310 mV. What is the magnitude of the rate of change of the flux through the coil?
Answer:
The magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s
Explanation:
Given;
radius of coil, r = 8 cm = 0.08 m
number of turns of the coil, N = 110 turns
the induced emf through the coil, E = -310mV
The induced emf through the coil is given by;
[tex]E = - N\frac{d \phi}{dt} \\\\\frac{d \phi}{dt} = \frac{E}{-N}[/tex]
Where;
dФ/dt is the magnitude of the rate of change of the flux through the coil
[tex]\frac{d\phi}{dt} = \frac{-0.31}{110} \\\\\frac{d\phi}{dt} =2.82*10^{-3} \ T.m^2/s[/tex]
Therefore, the magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s
The 60.0 g mass, attached to a light spring with a 40.0 N/m force constant, vibrates with an amplitude of 5.00 cm on a horizontal, frictionless plane. find (a) the total energy of the pulsating System, (b) the speed of mass when the displacement is 2.00 cm. When the displacement is 2.50 cm, (c) the kinetic energy and (d) the potential energy
Explanation:
(a) The total energy is the elastic potential energy at maximum displacement.
E = ½ kx²
E = ½ (40.0 N/m) (0.0500 m)²
E = 0.05 J
(b) At this displacement, there is both elastic potential energy and kinetic energy.
E = EE + KE
0.05 J = ½ kx² + ½ mv²
0.05 J = ½ (40.0 N/m) (0.0200 m)² + ½ (0.060 kg) v²
v = 1.18 m/s
(c) E = EE + KE
0.05 J = ½ kx² + ½ mv²
0.05 J = ½ (40.0 N/m) (0.0250 m)² + KE
KE = 0.0375 J
(d) EE = ½ kx²
EE = ½ (40.0 N/m) (0.0250 m)²
EE = 0.0125 J
A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is
Answer:
The force on the proton is 7.85 x 10⁻¹² N
Explanation:
Given;
kinetic energy of the proton, K.E = 2MeV = 2 x 10⁶ x 1.602 x 10⁻¹⁹ J
= 3.204 x 10⁻¹³ J
magnitude of the magnetic field, B = 2.5 T
The kinetic energy of the proton is given by;
[tex]K.E = \frac{1}{2} m v^2\\\\v^2 = \frac{2K.E}{m}\\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*3.204*10^{-13}}{1.67 *10^{-27}}} \\\\v = 1.959*10^7 \ m/s[/tex]
The force on the proton moving perpendicular to magnetic field is given by;
F = qvB
F = 1.602 x 10⁻¹⁹ x 1.959 x 10⁷ x 2.5
F = 7.85 x 10⁻¹² N
Therefore, the force on the proton is 7.85 x 10⁻¹² N
Select True or False for the following statements about Heisenberg's Uncertainty Principle.
A) It is not possible to measure simultaneously the x and z positions of a particle exactly.
B) It is possible to measure simultaneously the x and y momentum components of a particle exactly.
C) It is possible to measure simultaneously the y position and the y momentum component of a particle exactly.
Answer:
A and B are true C is false
Explanation:
Because it states that if we know everything about where a particle is located (the uncertainty of position is small), we know nothing about its momentum (the uncertainty of momentum is large), and vice versa.
So in A we can know the positions of two objects
In B we can measure the momentum at two different places
While in C we cannot measure both the position and momentum of y accurately
Which of the following is not an example of matter?
A. sand
B. heat from a fire,
C. helium in a balloon
D. fog paint on a canvas
Answer:
B. heat from fire
Explanation:
It is not an example of matter because heat is energy and not matter
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
A: What is the force exerted on the electron?
B: How strong is the electric field at the electron’s location?
C: How much work would be done on the electron if it was moved so that it’s .001m away from the sphere?
D: Now replace the electron with a positron (q=+1.602×10-19C). Explain in your own words how that would affect the results in parts A, B, and C.
Answer:
Explanation:
electric field at the location of electron
= 9 x 10⁹ x 7.2 / .03²
= 72 x 10¹² N/C
force on electron = electric field x charge on electron
= 72 x 10¹² x 1.6 x 10⁻¹⁹
= 115.2 x 10⁻⁷ N .
C )
work done = charge on electron x potential difference at two points
potential at .03 m
= 9 x 10⁹ x 7.2 / .03
= 2.16 x 10¹² V
potential at .001 m
= 9 x 10⁹ x 7.2 / .001
= 64.8 x 10¹² V
potential difference = (64.8 - 2.16 )x 10¹² V
= 62.64 x 10¹² V .
work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹
= 100.224 x 10⁻⁷ J .
D )
There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .
Work done in case of electron will be positive and work done in case of positron will be negative .
electric field due to charge will be same in both the cases .
The Kelvin temperature of the hot reservoir of an engine is twice that of the cold reservoir, and work done by the engine per cycle is 50 J.
Calculate:
(a) the efficiency of the engine,
(b) the heat absorbed per cycle, and
(c) the heat rejected per cycle.
Answer:
a) 50%
b) 100 J
c) 50 J
Explanation:
The cold temperature of the reservoir = [tex]T_{c}[/tex]
according to the problem, it is stated that the hot reservoir of an engine is twice that of the cold reservoir, therefore,
the hot temperature of the reservoir [tex]T_{h}[/tex] = [tex]2T_{c}[/tex]
The work done by the engine = 50 J
a) The max efficiency obtainable from a heat engine η = [tex]1 - \frac{T_{c} }{T_{h} }[/tex]
since [tex]T_{h}[/tex] = [tex]2T_{c}[/tex], the equation becomes
η = [tex]1 - \frac{T_{c} }{2T_{c} }[/tex] =
η = [tex]1 - \frac{1 }{2 }[/tex] = 0.5 = 50%
b) The heat absorbed per cycle will be gotten from
η = [tex]\frac{W}{Q}[/tex]
η is the efficiency of the system = 0.5
where W is the work done = 50 J
Q is the heat absorbed = ?
substituting, we have
0.5 = [tex]\frac{50}{Q}[/tex]
Q = 50/0.5 = 100 J
c) The heat rejected per cycle = 50% of the absorbed heat
==> 0.5 x 100 J = 50 J
Two monatomic ideal gases are in thermal equilibrium with each other. Gas A is composed of molecules with mass m while gas B is composed of molecules with mass 4m. The ratio of the average molecular kinetic energy KA / KB is
Answer: [tex]\frac{K_{A}}{K_{B}}[/tex] = 1
Explanation: Average molecular kinetic energy for monatomic idela gases is given by:
[tex]E=\frac{3}{2}RT[/tex]
where
R is gas constant of ideal gas
T is temperature
Which means kinetic energy depnends only on temperature.
Since gas A and gas B are at the same temperature, kinetic energy will be the same. Therefore:
[tex]\frac{K_{A}}{K_{B}}[/tex] = 1
Ratio of this average molecular kinetic energy is 1.
A catapult flings a stone at 16 m/s, giving it 1892 J of kinetic energy. What is the mass of the stone?
Calculate the mass of the stone.
Formula used:-Kinetic energy = 1/2 × m × v²
Solution:-We know that,
Kinetic energy = 1/2 × m × v²
⇒ 1892 = 1/2 × m × ( 16 )²
⇒ 1892 = 1/2 × m × 256
⇒ 1892 = 128m
⇒ m = 1892/128
⇒ m = 14.78 kg
A 817 kg car has four 8.91 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
0.0107
Explanation:
We know that
The rotational kinetic energy due to four wheel is
1/2ဃ²I x 4
So
1/4mR²(v/R)² = mv²
But kinetic energy along straight path of the car is 1/2mv²
=> 1/2( 817)v ²
Kc= 408.5v²
So The fraction of total kinetic energy that is due to rotation of the wheel about their axis
Is Kw/Kw+Kc
and Kw = 1/2* 8.91v²= 4.45v²
So 4.45v²/ 4.45v²+ 408.5v²
= 0.0107 as fraction of total kinetic energy
A flatbed truck is carrying a 20-kg crate up a sloping road. The coefficient of static friction between the crate and the bed is 0.40, and the coefficient of kinetic friction is 0.30. What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place
Answer:
The angle is [tex]\theta =21.8 ^o[/tex]
Explanation:
From the question we are told that
The mass of the crate is [tex]m_c = 20 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.40[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.30[/tex]
Generally for the the crate not to slip , the static frictional must be equal to the force driving the truck
i.e
[tex]F_f = F[/tex]
Now since we are considering a slope that static frictional force is mathematically represented as
[tex]F_f = mg * cos(\theta) * \mu_s[/tex]
While the force driving the truck is mathematically represented as
[tex]F = mg * sin (\theta )[/tex]
Here mg is the weight of the crate so
So
[tex]mg * cos (\theta ) \mu_s = mg * sin (\theta )[/tex]
=> [tex]\frac{sin (\theta )}{cos (\theta)} = \mu_s[/tex]
=> [tex]\theta = tan ^{-1} [\mu_s ][/tex]
=> [tex]\theta = tan ^{-1} [0.40 ][/tex]
=> [tex]\theta =21.8 ^o[/tex]
A diagram show an illustration is on the first uploaded image
A runner jumps off the ground at a speed of 16m/s .At what angle did he jumped from the ground if he lands 8m away?
Answer:
128 degrees
Explanation:
speed divided by distance travelled
the angle he jumped from the ground is 8.92°
The question above is a projectile motion problem, and we can solve it using the formula of range.
⇒ Formula:
R = u²sin2∅/g................... Equation 1⇒ Where:
R = Range of the runneru = initial velocityg = acceleration due to gravity∅ = angle to the horizontalFrom the question,
⇒ Given:
R = 8 mu = 16 m/sg = constant = 9.8 m/s²⇒Substitute these values into equation 1
8 = 16²sin(2∅)/9.8⇒ Solve for ∅
8×9.8 = 16²sin2∅78.4 = 256sin2∅sin2∅ = 78.4/256sin2∅ = 0.306252∅ = sin⁻¹(0.30625)2∅ = 17.83∅ = 17.83/2∅ = 8.92°Hence the angle he jumped from the ground is 8.92°
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Assuming two hypothetical maps that each cover a standard 8.5 by 11-inch sheet of paper, the larger-scale map would cover a larger geographic area than the smaller-scale map.
a) true
b) false
Answer:
b) false
Explanation:
The scale of a map is the ratio of a distance on the map to the corresponding distance on the ground. Scaling allow us to capture a large geographical area on a reduced platform while still retaining the relative sizes and positioning of places on the map to their real life sizes and positioning. If both maps cover a standard 8.5 by 11-inch sheet of paper, then the map with the smaller ratio will have the bigger geographical area.
To understand better, let us assume two geographical areas A and B. A is bigger than B. If we were to put them both on the same area of map paper, then we'll have to scale up the smaller geographical area B so as to fit into the map paper. This means that the geographical area with the smaller area B will have the larger scale on the map.
g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all planets orbiting the sun, assuming circular orbits?
Explanation:
Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :
[tex]T^2=\dfrac{4\pi ^2}{GM}a^3[/tex]
Here,
G is gravitational constant
M is mass of sun
It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.
If you wish to detect details of the size of atoms (about 1 ✕ 10−10 m) with electromagnetic radiation, it must have a wavelength of about this size.
(a) What is its frequency?
(b) What type of electromagnetic radiation might this be?
Answer:
a) 3×10^18 Hz
b) The electromagnetic wave is an x-ray
Explanation:
We know that the speed of an electromagnetic wave is given by;
c= λf
Where;
c= speed of electromagnetic waves = 3×10^8 ms-1
f= frequency of electromagnetic waves= the unknown
λ = wavelength of the electromagnetic wave= 1 ×10^-10 m
Hence;
f = c/λ= 3×10^8/ 1×10^-10
f= 3×10^18 Hz
b) The electromagnetic wave is an x-ray.
(a) The frequency of the electromagnetic radiation is [tex]3\times 10^{18} \ Hz[/tex]
(b) The electromagnetic radiation is ultra violet ray.
The given parameters;
distance of the atom, λ = 1 x 10⁻¹⁰ mThe frequency of the electromagnetic radiation is calculated as follows;
[tex]c =f \lambda \\\\f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^8}{1\times 10^{-10}} \\\\f = 3\times 10^{18} \ Hz[/tex]
The frequency range of ultra violet ray is between 10¹⁴ to 10¹⁸ m.
Thus, the electromagnetic radiation is ultra violet ray.
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Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.
Answer:
She will speed up because air resistance will decrease and be less than gravity.
Explanation:
While Luz was falling initially, she was experiencing a gravitational force downwards, and air resistance that arise from the drag force on her body
Her gravitational force downwards is constant, and she fall down with a net force of
[tex]F_{net}[/tex] = [tex]F_{g}[/tex] - [tex]F_{d}[/tex]
where [tex]F_{net}[/tex] is the net force on Luz downwards
[tex]F_{g}[/tex] is the gravitational force on Luz
[tex]F_{d}[/tex] is the drag force on Luz
The drag force on on Luz is proportional to her attack surface area. When Luz changes her body position from her frontal area, parallel to the ground to falling with her feet first, she reduces the area available for drag force from her whole frontal area to just about a little more than the areas of the sole of her feet. This action reduces the drag force due to air resistance on her body.
Answer:
The correct answer to the question is the third statement.
Explanation:
She will speed up because air resistance will decrease and be less than gravity. Changing position during skydiving is one of the factors affecting the speed. Since the cross section area is smaller than the first position, she experiences lesser air resistance, which causes her to speed up.
Allowing all but one allows for accurate results
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)
Answer:
r = 5.08 m
Explanation:
The electric force of attraction or repulsion is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.
So, the force from the proton is balanced by the mass of the electron.
[tex]\dfrac{kq_pq_e}{r^2}=mg[/tex]
r is distance
[tex]r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m[/tex]
So, proton have to be at a distance of 5.08 meters above the electron.
When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:
Answer:
The comoving distance and the proper distance scale
Explanation:
The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.
A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is doubled, approximately what will the pendulum's new period be?
A: T/2
B: T
C: sqrt(2)*T
D: 2T
Answer:
B. T
Explanation:
This is because the mass of the Bob does no relate to the period as given by the relationship
T = 2π(√L/g) so double mass is still T
Answer:
The correct option is (B).
Explanation:
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is the length of the pendulum
g is acceleration due to gravity
It is very clear that the time period is independent of the mass of the bob. If the mass of the bob is doubled, the time period of the pendulum remains the same i.e. T. Hence, the correct option is (B).
Pls answer 9 Through 12
Answer:
physics,chemistry,biology,astronomy and earht sciences
principles of science are integrity of knowledge honesty,collegiality(cooperation between colleagues) objectivity and openness
examples of science are actually branches of science so those are biology,mathematics,chemistry and physics etc
science is importent to provide our basic needs and to improve our living standard.It makes our life much easier by providing different technologies.It helps in the diagnosing and treatment of a disease
Which of the following elements tend to be radioactiv
Check all that apply.
A. Ne
B. Am
C. U
D. K
When you squeeze together the coils of a spring and then release them, you are creating a ____ wave? transverse compressional water seismic
Answer:
Hey there!
Because you are compressing the spring, these are compressional waves.
Let me know if this helps :)
Gauss’s law applies to:_____________
a. lines.
b. flat surfaces.
c. spheres only.
d. closed surfaces.
Answer:
D. Closed Surfaces
The gaussian surface could be a sphere and can be applied to flat surfaces, but it's always a closed surface.
If we use to construct the latches on the windows and doors, then the magnetism will keep thee latches secure.
a. True
b. False
Answer:
a. True
Explanation:
I'll assume the question is about magnetic latches and locks.
Magnetic door locks use an electromagnetic force to stop doors from opening, so they are ideal for security. There are two main types of electric locking devices. Locking devices can either be a fail-secure locking device that remains locked when power is lost, or a fail-safe locking device that is unlocked when de-energized. An electromagnetic lock creates a magnetic field when energized or powered up, this causes an electromagnet and armature plate to become attracted to each other strongly enough to keep a door from opening.
the lenses in a students eyes have arefractive power of 52. 0 diopters when she is able to focus on the board if the distance between the ey lens and the retina is 2.00 cm find how
Answer:
p = 49.95 cm
This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row
Explanation:
In medicine it is very common to express the potential visual corrections that is
P = 1 / f
where P is the power and f the focal length in meters
In this exercises give the power, let's find the focal length
f = 1 / p
f = 1/52
f = 0.01923 m = 1.923 cm
For geometrical optics calculations the most used equation is the constructor equation
1 / f = 1 / p + 1 / q
Where p and q are the distance to the object and the image, respectively
To be able to see an object clearly, its image must be on the retina,
q = 2.00 cm
find the distance to the object
1 / p = 1 / f - 1 / q
1 / p = 1 / 1,923 - 1/2.00
1 / p = 0.02002
p = 49.95 cm
This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.30 m before stopping. How far does the lighter fragment slide
Answer:
the distance d traveled by the lighter fragment is 58.1 m.
Explanation:
mass of the lighter fragment = m
the lighter fragment traveled a distance = ?
mass of the heavier fragment = 7m
the distance covered by the heavier fragment = 8.30 m
The two particles will be given the same amount of energy from the explosion. This energy is used to do work by the two fragments.
work done by heavier fragment w = mgd
where m is the mass
g is acceleration due to gravity
d is the distance traveled.
substituting, the work done by the heavier fragment is
w = 7m x g x 8.3 = 58.1mg
The same way, the lighter fragment does work of
w = mgd
equating the two work done since they are given the same amount of energy from the explosion, we have
58.1mg = mgd
mg cancels out, we have
the distance d traveled by the lighter fragment d = 58.1 m
A boat sailing against the current experience a negative acceleration of 11 m/s^2. If the boat's initial velocity is 44m/s upstream, how long until it comes to a stop?
Answer:
4 s
Explanation:
Given:
v₀ = 44 m/s
v = 0 m/s
a = -11 m/s²
Find: t
v = at + v₀
0 m/s = (-11 m/s²) t + 44 m/s
t = 4 s
If the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.
What is Acceleration?Acceleration may be defined as the rate of change of velocity with respect to time. It is a vector quantity as it has both magnitude and direction. It is also the second derivative of position and the first derivative of velocity with respect to time.
According to the question,
v₀ = 44 m/s
v = 0 m/s
a = -11 m/s²
Now, you have to calculate the time, t.
So, you calculate the time, t with the help of the following formula:
v = at + v₀0 m/s = (-11 m/s²) t + 44 m/s
t = 4 s.
Therefore, if the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.
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(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
exposed. What force is required to immerse it completely under water?
(Density of wood = 600 kg/m3, water = 1000 kg/m3)
[8]
Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.
A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comfortable internal pressure of 101 kilopascals. How deep can it dive in the ocean before it would risk collapsing from the pressure
Answer:
The depth will be equal to 6141.96 m
Explanation:
pressure on the submarine [tex]P_{sea}[/tex] = 62 MPa = 62 x 10^6 Pa
we also know that [tex]P_{sea}[/tex] = 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
[tex]P_{sea}[/tex] = 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine [tex]P_{g}[/tex] = 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 [tex]P_{sea}[/tex], we have
62 x 10^6 = 10094.49h
depth h = 6141.96 m