Answer:
c heats up cancer cells and weakens them.
Answer:
I think it is C
Explanation:
because
please help!!!
When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways? O An open circuit is being changed into a closed circuit. A closed circuit is being changed into an open circuit. O A parallel circuit is being changed into a series circuit. A series circuit is being changed into a parallel circuit.
Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
What would cause surface ocean water to have a higher salt content?
A.
Surface ocean water will have a higher salt content from the melting of sea ice
B.
Surface ocean water will have a higher salt content from low rates of evaporation and high rates of precipitation.
C.
Surface ocean water will have a higher salt content from water flowing out of a river into the ocean
D.
Surface ocean water will have a higher salt content from high rates of evaporation and low rates of precipitation
Answer:
d I think? not sure I don't know much abt the ocean
The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.26 cm. At the same moment, both particles are released.
Required:
Determine the distance from the positive plate at which the two pass each other.
Answer:
The distance from the positive plate at which the two pass each other is 0.0023 cm.
Explanation:
We need to find the acceleration of each particle first. Let's use the electric force equation.
[tex]F=Eq[/tex]
[tex]ma=Eq[/tex]
For the proton
[tex]m_{p}a_{p}=Eq_{p}[/tex]
[tex]a_{p}=\frac{Eq_{p}}{m_{p}}[/tex]
[tex]a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}[/tex]
[tex]a_{p}=6.19*10^{10}\: m/s^{2}[/tex]
For the electron
[tex]m_{e}a_{e}=Eq_{e}[/tex]
[tex]a_{e}=\frac{Eq_{e}}{m_{e}}[/tex]
[tex]a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}[/tex]
[tex]a_{e}=1.14*10^{14}\: m/s^{2}[/tex]
Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.
[tex]x_{p}+x_{e}=0.0426[/tex]
Both of them have an initial speed equal to zero. So we have:
[tex]\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426[/tex]
[tex]t^{2}(a_{p}+a_{e})=2*0.0426[/tex]
[tex]t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}[/tex]
[tex]t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}[/tex]
[tex]t=2.73*10^{-8}\: s[/tex]
With this time we can find the distance from the positive plate (x(p)).
[tex]x_{p}=\frac{1}{2}a_{p}t^{2}[/tex]
[tex]x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}[/tex]
[tex]x_{p}=0.0023\: cm[/tex]
Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.
I hope it helps you!
when air mass is caught between two cold fronts the result is a _______ front.
Answer choices
A.occluded
B.warm
C.cold
D.stationary
How did she change the circuit?
In a movie, the heroine saves the day by cutting a wire
before the time runs out on a timer.
O She changed a short circuit to an open circuit.
O She changed an open circuit to a closed circuit.
O She changed a closed circuit to an open circuit.
O She changed a closed circuit to a short circuit.
Answer:
She changed a closed circuit to an open circuit
Explanation:
This means that she has changed from a short circuit to an open circuit. Then the correct option is C.
What is electric current?The stream of positive charges which flow from the positive terminal to the negative terminal of the battery attached in a circuit.
An interrupted consistency in an electrical circuit prevents current from flowing, which is what the term "OPEN CIRCUIT" refers to.
A closed circuit is comparable to a road that uses a bridge to cross a stream.
An electrical circuit in a gadget with a current rating than a typical circuit, particularly one created by unintentional component engagement and subsequent current misdirection.
In a movie, the heroine saves the day by cutting a wire before the time runs out on a timer.
This means that she has changed from a short circuit to an open circuit.
Then the correct option is C.
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A 5.1 g bullet is fired into a 2.3 kg ballistic pendulum. The bullet emerges from the block with a speed of 221 m/s, and the block rises to a maximum height of 7 cm . Find the initial speed of the bullet. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:
748.62 m /s.
Explanation:
mass of bullet m = .0051 kg .
mass of block M = 2.3 kg
block rises to a height of .07 m so velocity of block after collision
V = √ 2 gh
=√ (2 x 9.8 x .07 )
= 1.17 m /s
velocity of bullet after collision v = 221 m /s
Now we shall apply law of conservation of momentum to find out the velocity of bullet before collision .
Let it be Vx . then
5.1 x 10⁻³ x Vx + 0 = 5.1 x 10⁻³ x 221 + 2.3 x 1.17
= 1.127 + 2.691 = 3.818
Vx = 748.62 m /s
I need help please will mark brainliest
Answer:
c
Explanation:
A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall building. The rocket's engine produces a horizontal acceleration of (1.60 m/s^3)t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, downward. Air resistance can be neglected.
Required:
What horizontal distance does the rocket travel before reaching the ground?
Answer:
Explanation:
Consider downward displacement first .
downward displacement h = 38 m .
downward acceleration = g
downward initial velocity u = 0
h = ut + 1/2 g t ²
38 = 0 + 9.8 t ²
t = 1.97 s
During this period , there will be horizontal displacement with initial velocity u = 10 m /s and acceleration a = 1.6 m /s²
s = ut + 1/2 a t²
= 10 x 1.97 + .5 x 1.6 x 1.97²
= 19.7 + 3.10
= 22.8 m
An ideal spring is lying horizontally on a frictionless surface. One end of the spring is attached to a wall. The other end is attached to a moveable block that has a mass of 5 kg. The block is pulled so that the spring stretches from its equilibrium position by 0.65 m. Then the block is released (from rest), and as a result the system oscillates with a frequency of 0.40 Hz (that's 0.40 rev/sec) Find:
a) the acceleratiuon of the block when the spring is stretched by 0.28 m.
b) the maximum force magnitude exerted by the spring on the block.
c) the oscillation frequency of a 2.5 kg blcok under the same circumstances (i.e. with the same spring and initial displacement).
Answer:
a) a = - 1.76 m / s², b) F = 20.5 N, c) w = 3.55 rad / s
Explanation:
a) a simple harmonic motion is described by the expression
x = A cos (wt + Ф)
in this case they give us the frequency
w = 2π f
w = 2π 0.40
w = 2.51 rad / s
as the maximum elongation is 0.65 m this corresponds to the amplitude of the movement
A = 0.65 m
to find the phase constant (Ф) we use the initial condition that for t = 0 v = 0 and x = A, we substitute
A = A cos (0+ Ф)
cos Ф = 1
Ф = 0
the resulting equation is
x = 0.65 cos (2.51 t)
Let's find the time it takes to get to x = 0.28 m
0.28 = 0.65 cos 2.51 t
2.51 t = cos-1 (0.28 / 0.65)
remember angles are in radians
t = 1.1254 / 2.51
t = 0.448 s
the acceleration is
[tex]a = \frac{dv}{dt} = \frac{d^2x}{dt^2}[/tex]
a = -A w² cos wt
we subtitle
a = - 0.65 2.51² cos (2.51 0.448)
a = - 1.76 m / s²
b) the maximum acceleration occurs when the cosine is ±1
a = A w²
a = 0.65 2.51²
a = 4.10 m / s²
Let's use Newton's second law
F = m a
F = 5 4.1
F = 20.5 N
c) The angular velocity is given by
w² = k / m
let's find the spring constant
k = m w²
k = 5 2.51²
k = 31.5 N / m
therefore if the block is exchanged for another with mass m'= 2.5 kg
w = √(31.5 / 2.5)
w = 3.55 rad / s
A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.203 m/s2 in the x direction. After 56.7 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answer:
Explanation:
Initial velocity in y direction Vy = 22.3 m /s
initial acceleration in x direction ax = .203 m /s ²
time of acceleration t = 56.7 s
final velocity in x direction
v = u + a t
Vx = 0 + .203 x 56.7 = 11.51 m /s
Final velocity in y direction will remain same as initial velocity in y direction = 22.3 m /s because there is no acceleration in y direction .
Magnitude of final velocity
= √ ( Vx² + Vy²)
= √ (22.3² + 11.51² )
= √ ( 497.29 + 132.48)
= 25.1 m /s
Direction of final velocity from y direction be Ф
TanФ = Vx / Vy = 11.51 / 22.3 = .516
Ф = 27.3° .
Compare and contrast climate and weather
What happens to the force attraction of the distance two objects is increased?
Answer:
Explanation:
The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.
As the distance between the sun and earth decreases, the speed of the planet
a
increases
b
decreases
c
stays the same
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.400 m and the length of the copper section is 0.800 m . Each segment has cross-sectional area 0.00700 m2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings.
(a) What is the temperature of the point where the brass and copper segments are joined?
(b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?
Answer:
a) 36°
b) 0.109 kg
Explanation:
Heat flows from brass to copper with the brass having its temperature
Length of brass = 0.4
Length of copper = 0.8
Temperature of = 36.15
See attachment for calculation
The temperature at the joint is 36.15°C
The amount of ice melted is 1.086 kg
The rate of transfer of thermal energy,
H = Q/t = KAΔT/L
where, K is the thermal conductivity of the substance, A is cross-sectional area, ΔT is temperature difference at the ends and L is the length
As given in the question,
the length of the brass section [tex]L_{1}[/tex] = 0.4 m
it's thermal conductivity [tex]K_{b}[/tex] = 109 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 373K
the length of the copper section [tex]L_{2}[/tex] = 0.8 m
it's thermal conductivity [tex]K_{c}[/tex] = 385 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 273K
cross-sectional area of both the substance is same A = 0.007 [tex]m^{2}[/tex]
Let the temperature at the joint be T
The rate of heat flow must be constant across the whole length of the setup.
Hence at the joint,
[tex]\frac{K_{b}A(T_{1}-T) }{L_{1} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ 109*A*(373-T)}{0.4} =\frac{385*A(T-273)}{0.8}[/tex] ⇒ T=309.15 K
⇒ T = 36.15°C is the temperature at the joint.
Now we have to calculate the equivalent thermal conductivity K of the setup in order to calculate the amount of heat transfer.
considering equivalent thermal conductivity K throughout the setup we can form the following equation to calculate its value
[tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ K*A*(100)}{1.2} =\frac{385*A(36.15)}{0.8}[/tex]
⇒ K = 208.76 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the amount of heat transferred at the copper end in ice-water mixture in 5 minutes(300 seconds) :
Q = [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} }[/tex] × t = [tex]\frac{208.76*0.007*100}{1.2}[/tex] × 300 = 36533 J
latent heat of fusion of ice [tex]L_{f}[/tex] = 33600 J/kg
[tex]Q=mL_{f}[/tex]
[tex]m=\frac{Q}{L_{f} }[/tex]
[tex]m=\frac{36533}{33600}[/tex] ⇒ m = 1.086 kg of ice is melted in 5 minutes
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In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m. Calculate the bulk modulus of the liquid.
Answer:
The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Explanation:
Given;
density of the liquid, ρ = 1500 kg/m³
frequency of the wave, F = 410 Hz
wavelength of the sound, λ = 7.80 m
The speed of the wave is calculated as;
v = Fλ
v = 410 x 7.8
v = 3,198 m/s
The bulk modulus of the liquid is calculated as;
[tex]V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2[/tex]
Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = [tex]\frac{1}{2}[/tex][tex]m_{p}[/tex]v²
v = √( 2K / [tex]m_{p}[/tex] )
lets relate the cross-sectional area A of the beam to its diameter D;
A = [tex]\frac{1}{4}[/tex]πD²
now, we substitute for v and A
n = I / [tex]\frac{1}{4}[/tex]πeD² ×√( 2K / [tex]m_{p}[/tex] )
n = 4I/π eD² × √([tex]m_{p}[/tex] / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Two masses are being pulled up a 30.0-degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s2 and the velocity is down the incline. The force is applied to a 200-kg mass and a string connects the 200-kg mass to a 150-kg mass. The coefficient of kinetic friction is 0.200. The force F is
Answer:
Explanation:
The acceleration up the incline is 1.00 m/s²
Net force acting on two masses = total mass x acceleration
= 350 x 1 = 350 N
weight acting down the plane = m g sinФ
= 350 x 9.8 x sin30 = 1715 N
Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction
= 350 x 9.8 x cos30 x .2 = 594N
Net force acting on total mass
= F - 1715 - 594 = 350 , where F is required force
F = 2659 N .
A 45945990 prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism
Answer:
Explanation:
Angle of incidence at the longer face = 45⁰ ( see the figure in the attached file )
For total reflection from longer face
i = C where C is critical angle .
And relation between critical angle and refractive index is as follows .
μ = 1 / sinC
= 1 / sin45
= √2
= 1.414 .
What is the car's acceleration from 0 to 1 second?
A. 8 mph/s
B. 20 mph/s
C. 60 mph/s
D. 10 mph/s
As the distance between the sun and earth decreases, the force of gravity
a
Increases
b
decreases
c
stays the same
Answer:
B Decrease
Explanation:
ik im right cuz i looked up the answer
The open-circuit voltage of a car battery is measured to be 12 V. During engine startup, the battery delivers 10 A to the starter motor, and the battery voltage drops to 11.7 V. Draw the Thévenin equivalent circuit for the battery. How much power does the battery deliver to the starter motor?
Answer:
- the Thevenin equivalent circuit for the battery is uploaded below
- the battery delivered 117 watts of power to the starter motor
Explanation:
Given the data in the equation
diagram of the Thevenin equivalent circuit for the battery is uploaded below.
Current I = 10 A
Voltage 1 = 12 V
voltage 2 = 11.7 v
R = (V1 - V2) / I
R = (12-11.7)/10
R = 0.3 / 10
R = 0.03Ω
Thevenin equivalent circuit
[tex]R_{L}[/tex] = V2 / I = 11.7 / 10
[tex]R_{L}[/tex] = 1.17Ω
so, power delivered to the starter motor will be;
p = (V2)² / [tex]R_{L}[/tex]
P = ( 11.7 V )² / 1.17Ω
p = 136.89 / 1.17
p = 117 watts
Therefore, the battery delivered 117 watts of power to the starter motor
a childs weight is 331 N. what is the childs mass in kg?
Answer:
the child's mass is 33.1 kg
how long it take a train to cover 630km having a speed of 30 km/hr
Answer:
21 hours
Explanation:
well 30 x 20 = 600 than 21 = 630
Explain how you could use iron filings and a piece of paper to help reveal the effect of a magnetic field.
Answer:
you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet
Explanation:
The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of the horizontal velocity (which is constant) and the amount of time the pumpkin is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 14 m/s, and fire the pumpkin several times while varying the angle between the cannon and the horizontal.
Required:
For which angle is the range a maximum (with the initial speed held constant)?
Answer:
Explanation:
For range o a projectile , the formula is as follows
R = u² sin2Ф / g where u is initial velocity of throw , Ф is angle of throw and g is acceleration due to gravity .
Here u = 14 m /s
R = 14² sin2Ф / 9.8
R = 20 sin2Ф
Now R will have maximum value when sin2Ф has maximum value .
Maximum value of sin2Ф = 1
sin2Ф = 1 = sin 90°
Ф = 45°
So when throw is aimed at 45° , range will be maximum .
What is the average speed of a cheetah that sprints 100 m in 4s? How about if it sprints 50 m in 2 s
Speed = (distance covered) / (time took)
A characteristic of a nebula is that it-
Answer:
Center of solar system
Explanation:
Answer: b
Explanation:
What Is a Sound Wave? Learning Goal: To understand the nature of a sound wave, including its properties: frequency wavelength, loudness, pitch, and timbre. Sound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to understand the physical nature of a sound wave and its properties to correct common misconceptions about sound propagation Most generally, a sound wave is a longitudinal wave that propagates in a medium (ie, air) The particles in the medium oscillate back and forth along the direction of motion of the wave. This displacement of the particles generates a sequence of compressions and rarefactions of the medium Thus, a sound wave can also be described in terms of pressure variations that travel through the medium. The pressure fluctuates at the same frequency with which the particles positions oscillate When the human ear perceives sound. It recognizes a series of pressure fluctuations rather than displacements of individual air particles. Part 1 Figure 1 of 2 > Fi MA length Part A Based on the information presented in the introduction of this problem, what is a sound wave? Propagation of sound particles that are offerent from the particles that comprise the medium Propagation of energy that does not require a medium Propagation of pressure fluctuations in a medium Propagation of energy that passes through empty spaces between the partides that com Submit Request Anst Part B Complete previous parts) Part hall to the other? Does air play a role in the propagation View Available Hints) SUITE Part D The graphs shown in (Figure 1) represent pressure variation versus time recorded by Enter the letters of all the correct answers in alphabetical order.
Answer:
A) Propagation of pressure fluctuations in a medium
B) air is the medium in which the wave is transported,
Explanation:
Part A.
A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.
The most correct answer is:
* Propagation of pressure fluctuations in a medium
Part b
air is the medium in which the wave is transported, otherwise it cannot propagate
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?
Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
keli learned that an air mass is a very large body of air with similar temperature humidity and pressure and the air mass are constantly in motion she knows that you're messing depending on the temperature and moisture content tent of region where they form she looked up more information about what makes them move what are the major causes for moving & Masten North America choose two that apply.
Answer choices
A. changing humidity
B. low temperature
C. jet storm
D. prevailing westerlies
Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.
What is meant by air mass?An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.
Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.
Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.
Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.
Therefore, the correct answer is option C. jet storm.
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