Answer:
The answer is D. Four covalent bounds
What is the value of 4p − 2, when p = 8? 16 24 30 34
Answer:
30
Explanation:
Plug in p to get 4(8) - 2
32 - 2 = 30
How are gases and liquids different from each other? A. Gases can change volume, and liquids cannot. O B. Liquids can change volume, and gases cannot. C. Gases can change shape, and liquids cannot. D. Liquids can change shape, and gases cannot.
Answer:
A.gasses can change volume, and liquids cannot
Explanation:
if you pour a glass of water you don't have to worry about it expanding such that it overflows. however if you air up a ballon you may notice that it becomes lager the warmer it gets and smaller the cooler it gets
Paragraph/Comprehension type questions.
A body weighs 500gf in air and 300gf when completely immersed in water
36. Find the apparent loss in weight of the body.
1)500gf
2)300gf
3)200gf
4)800gf
37. Find the buoyant force acting on the body
1)500gf
2)300gf
3)200gf
4)800gf
Answer:
1>500gf
1>300gf
its answer
1-Two objects a distance apart are experiencing 40 N of force. How much force wouldthere be if you DOUBLED BOTH of their masses??
We know that
• The applied force is 40 N.
,• Their masses are being doubled.
Following Newton's Second Law, the mass is directly proportional to the applied force, which means if we double the mass, then the force will be double too.
\(F=ma\to F=2ma\to F=2(ma)\)However, if we double both masses at the same time, then the force between will be four times stronger. So, F = 4*40N = 160N.
Therefore, the force between the masses will be 160 N.May someone please give me a hand?
circuit diagram
this is a circuit diagram
According to the principal of superposition,
Answer:
the resultant wave is the algebraic sum of all the waves reaching that particular point at a given time.
Explanation:
imagine two or three waves reaching a particular particle x at the same time. The particle will vibrate those waves and give out or transmit a resultant wave which is the algebraic sum of the incoming two waves. If both the waves have the same amplitude and phase, the resultant wave will be amplified. However if the waves have the same amplitude and equal but opposite phase then the resultant wave will be a straight line
Jupiter's Great Red Spot is
1.slowly shrinking
2.a large storm
3.smaller than Saturn's spot
4.a large area of liquid hydrogen
Answer:
2 : A large storm
Explanation: I hope this helps
Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experience the same magnitude of acceleration down an inclined plane on Earth. Neglecting friction and air resistance, which angle should the inclined plane have with respect to the horizontal ? (in deg)
Answer:
Angle of incline is 20.2978°
Explanation:
Given that;
Gravitational acceleration on a planet a = 3.4 m/s²
Gravitational acceleration on Earth g = 9.8 m/s²
Angle of incline = ∅
Mass of the stone = m
Force on the stone along the incline will be;
F = mgSin∅
F = ma
The stone has the same acceleration as that of the gravitational acceleration on the planet.
so
ma = mgSin∅
a = gSin∅
Sin∅ = a / g
we substitute
Sin∅ = (3.4 m/s²) / (9.8 m/s²)
Sin∅ = 0.3469
∅ = Sin⁻¹( 0.3469 )
∅ = 20.2978°
Therefore, Angle of incline is 20.2978°
no i dont have one soory
Answer:
u dont have a question
Explanation:
?? bc i would gladly help u
Write a one or two summary paragraph discussing this experiment and the results. Use the following questions and topics to help guide the content of your paragraph.
According to your data, was your hypothesis correct? (Be sure to refer to your data when answering this question.)
Summarize any difficulties or problems you had in performing the experiment that might have affected the results. Describe how you might change the procedure to avoid these problems.
List at least two real world examples that apply the findings of this experiment. (Hint: An example of this type was given in the introduction to this project.)
The first hypothesis in determining density experiment was right, but the second was wrong. The density of the penny will determine its mass-to-volume ratio; if it is higher than that of the nickel and quarter.
Explain real-world examples that apply the findings of this experiment.The student had no significant problems during the experiment. Just because the penny was smaller than a nickel, he assumed it would have a lower density. This experiment must be used by people in other situations, such as while attending the Huncle Berry Competition. This is because in order to play the game, every machine there uses the density of a coin.
Pay phones may be used back then as well. Pay phones only accepted quarters, therefore to verify that it is a quarter, it looks at the density. Additionally, some piggy banks count the coins based on density; nevertheless, this might occasionally result in some minor errors.
The second hypothesis, if the penny has a lower density than the nickel and quarter, then it will have a lower mass-to-volume ratio.
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Say an impulse is applied opposite the go-kart's direction of travel. What happens to
the go-kart if its momentum + impulse = 0?
The go kart stops comes to a stop.
The go kart slows down but keeps moving.
The go kart speeds up.
There is no change in the speed of the go kart.
If the impulse is strong enough and lasts for a sufficient amount of time, the go-kart will eventually come to a stop.
Option A is correct.
What is meant by impulse?impulse is described as the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector quantity.
If the force is insufficient to stop the go-kart entirely, it will slow down but continue to move. The force and duration of the impulse, along with the mass and speed of the go-kart, will all affect how much deceleration occurs.
Given that momentum plus impulse equals zero, the go-kart's change in momentum as a result of the impulse will be equal in amount but will move in the opposite direction of its original momentum.
As a result, the go-kart's final momentum will be zero, suggesting that it has either stopped or is travelling very slowly.
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A car is heading North at 18 m/s.
The driver flicks the wheel, and within 3 seconds he has the car headed South at 12
m/s.
What is the acceleration?
Answer:
10 m/s south
Explanation:
From 18 m/s North to 12 m/s South is a total change of 30 m/s towards the South.
So, Change in velocity = 30 m/s South
Time = 3 s
30/3=10 so 10 m/s South
Plane-polarized light is incident on a single polarizing disk with the direction of E0 parallel to the direction of the transmission axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of 1.22
Malus's law allows us to find the result for which is the angle of the polarizer for a reduction of the intensity is:
θ = 34.9º
Malus's law establishes a mathematical relationship for the light transmitted by a polarizer is:
I = I₀ cos² θ
where I and I₀ are the transmitted and incident intensities, respectively, θ is the angle between the two polarizers.
They indicate that the incident light is polarized, so we do not need the first polarized to create the polarized light, ask what the polarized angle should be for the intensity to be reduced by a factor.
\(\frac{I_o}{I } = 1.22 \\\frac{I}{Io} = 0.81967\)
Let's use Malus's law
i / Io = cos² θ
cos θ = \(\sqrt{\frac{I}{I_o} }\)
θ = cos⁻¹1 0.81967
θ = 34.9º
In conclusion, using Malus's law we can find the angle of the polarized for a reduction of the intensity is:
θ = 34.9º
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Electric field lines are used to represent the vector electric field around point charges and charged objects. Which of the following statements are true about electric field lines. Select ALL that apply.
Select all that apply
A. Electric field lines cannot cross.
B. Lines of electric field only originate from positive charges.
C. Field lines point in the direction of the force the electric field creates on an electron.
D. The strength of the electric field is greater in regions where the field lines are closer together.
E. In an electric-field-line drawing with many point charges, the number of field lines originating or terminating on each charge is proportional to the charge. That is, bigger charges have proportionally more field lines. F. The true strength of an electric field at any point can be determined from an electric field representation.
Electric field lines are a powerful tool to understand and visualize electric fields. They help to represent the direction and magnitude of the electric field at various points around a charged object.
The following statements are true about electric field lines:
A. Electric field lines cannot cross: This is because at the point where two field lines cross, there would be two directions for the electric field, which is impossible. Hence, the lines do not cross, and this is one of the fundamental characteristics of electric field lines.
B. Lines of electric field only originate from positive charges: Electric field lines originate from positive charges and terminate at negative charges. This is because positive charges repel positive charges and attract negative charges. Therefore, the electric field lines originating from a positive charge terminate at a negative charge.
C. Field lines point in the direction of the force the electric field creates on an electron: Electric field lines point in the direction of the force that would be experienced by a positive charge placed at any point in the field. Electrons, being negatively charged, would experience a force in the opposite direction to the electric field.
D. The strength of the electric field is greater in regions where the field lines are closer together: The density of field lines indicates the strength of the electric field. The closer the lines are, the stronger the field at that point.
E. In an electric-field-line drawing with many point charges, the number of field lines originating or terminating on each charge is proportional to the charge. That is, bigger charges have proportionally more field lines: The number of field lines originating or terminating on each charge is directly proportional to the magnitude of the charge.
F. The true strength of an electric field at any point can be determined from an electric field representation: The strength of the electric field at a point can be determined by the density of electric field lines at that point. However, the actual strength of the field would require quantitative measurements using instruments such as a voltmeter or an electrometer.
In conclusion, electric field lines are an essential tool in understanding the behavior of electric fields. They provide a visual representation of the electric field, its direction, and its strength at various points in space.
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Multiply.
(2x + 4)(x - 4)
Answer:
(2x + 4)(x - 4)=2x^2-4x-16
During batting practice, the ballplayer swings his bat with a velocity of 15
m/s. If the bat is 0.8 m long, what is the centripetal acceleration at the end of
the bat? (Ignore the length of the ballplayer's arm.)
O A. 240.6 m/s2
B. 281.2 m/s2
C. 187.5 m/s2
O D. 326.9 m/s2
what is a shargaff rule
According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.
Who is Chargaff ?Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.
Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.
He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.
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The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2
. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2.2 s. To bring the wheel to rest, the mechanic gently applies the disk brakes, which squeeze pads against a metal disk connected to the wheel. The pads touch the disk 7.1 cm from the axle, and the wheel slows down and stops in 1.2 s.
What is the magnitude of the friction force on the disk?
the magnitude of the friction force on the disk is approximately 1.16 N.
We can use the conservation of energy to find the friction force on the disk. The initial kinetic energy of the wheel is equal to the work done by the friction force on the disk:
K_i = W_friction
The initial kinetic energy of the wheel can be found from its moment of inertia and angular velocity:
K_i = (1/2) I \(ω^2\)
where I is the moment of inertia, ω is the angular velocity, and the factor of 1/2 comes from the rotational kinetic energy formula.
The final kinetic energy of the wheel is zero, since it comes to a stop. The work done by the friction force can be found from the distance over which it acts:
W_friction = F_friction d
where F_friction is the friction force and d is the distance over which the pads act on the disk. We can find the distance from the angular displacement of the wheel:
θ = ω t
where θ is the angle through which the wheel rotates, t is the time for the wheel to come to a stop, and the factor of 1/2π converts from rotations to radians. The distance over which the pads act is then:
d = r θ = 0.071 m × (5/2π) ≈ 0.562 m
Now we can put everything together:
K_i = W_friction
(1/2) I \(ω^2\) = F_friction d
We can solve for the friction force:
F_friction = (1/2) I \(ω^2\) / d
Plugging in the given values:
F_friction = (1/2) ×\(0.097 kg⋅m^2\) × \((5/2.2π rad/s)^2\) / 0.562 m ≈ 1.16 N
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what is the pressure of a tank of uniform cross sectional area 4.0m2 when the tank is filled with water a depth of 6m when given that 1 atm=1.013 x 10^5pa density of water=1000kgm-3 g=9.8m/s2
The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)
Pressure (P) = Density (ρ) × g × Depth (h)
Given: Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Depth (h) = 6 m
Using the given values, one can calculate the pressure:
Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure
= 58800 kg·m⁻¹·s⁻²
Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:
Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)
Pressure = 580.124 atm
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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
A set of 500-g masses is placed one at a time on a digital balance during quality control testing. The mass readings are 397 g, 401 g, and 403 g. Describe the accuracy and precision of the scale. (1 point)
not accurate and not precise
not accurate and not precise
both accurate and precise
both accurate and precise
accurate but not precise
accurate but not precise
precise but not accurate
Considering the definition of precision and accuracy, the mass readings of the digital balance are accurate but not precise.
Definition of precision and accuracyPrecision as the proximity between the indications or measured values of the same measurand, obtained in repeated measurements, under specified conditions.
Accuracy is defined as the closeness between the measured value and the "true" value of the measurand.
In other words, accuracy is how close a measurement is to the true value, while precision is how close the values of several measurements are to a point.
Precision and accuracy are independent of each other. Thus, the results in the values of a measurement can be precise and not exact (and vice versa).
Accuracy and precision in this caseA set of 500-g masses is placed one at a time on a digital balance during quality control testing. The mass readings are 397 g, 401 g, and 403 g.
In this case, the measurement is accurate, since the results of each individual measurement are quite similar. But the measurements are not exact (not precise) because the results are far from the real value.
In summary, the mass readings of the digital balance are accurate but not precise.
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1.) Consider the graph shown to the right. The two lines denote the motion of two different objects which move in the x direction only.
2.) Five vectors, A, B, C, D, E are shown at the right. Which of the combinations below has the largest magnitude ?
Graph is shown in the picture, please help!
The change in position of an object over time is known as velocity. The thing in question is traveling correctly. Vector c has the largest magnitude of all the vectors. And as demonstrated, a half-arrow can be used to represent this vector.
What are some types of objects?Everything that is readily apparent or palpable and has a stable form. An object of study in medicine is anything, anyone, or anything to which action or thinking is focused.
What is a kid-friendly object?A person or thing that the verb directs its action toward is called an object. A noun, pronoun, or noun phrase that is impacted by the action verb is referred to as an object. A sentence is also finished by it.
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Two students were climbing stairs at school. Student 1 has a weight of 700 N. Student 2 has a weight of 650 N. How much power would each student have if each took 6 s to climb a 4 m tall flight of stairs?
Student 1 would have 4200 W, and Student 2 would have 3900 W of power.
Student 1 would have 116 W, and Student 2 would have 108 W of power.
Student 1 would have 2800 W, and Student 2 would have 2600 W of power.
Student 1 would have 467 W, and Student 2 would have 433 W of power.
Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.
From the question,
We are to calculate the power each student would have to climb the flight of stairs.
Power can be calculated using the formula
\(P = \frac{F \times d}{t}\)
Where
P is Power
F is the force
d is the distance
and t is the time
NOTE: The weight of the students represent the force
For student 1F = 700 N
d = 4 m
t = 6 s
∴ \(P = \frac{700 \times 4}{6}\)
P = 467 W
For student 2
F = 650 N
d = 4 m
t = 6 s
∴ \(P = \frac{650 \times 4}{6}\)
P = 433 W
Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.
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A small object of mass 3.80g and charge −18.0μC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?
The electric field will have a magnitude of 2.07 x \(10^{3}\) N/C and the electric field will be directed downwards.
How do you determine the size of an electric field?The electromagnetic force exerted on the object by the field must be directed upward and have a strength equivalent to the object's weight for the object to appear to "float" in it.
Thus, \(F_{e} = qE = mg\)
The magnitude of the electric field is,
E = mg/ IqI = (3.80 x \(10^{-3} Kg\))(9.80 m/s²)/1.80 x \(10^{-6}\) C = 2.07 x \(10^{3}\) N/C.
How do you tell which way the electric field is going?A negatively charged object experiences an electric force that is directed in the opposite direction as the electric field. Because it must be directed downward, the electric force.
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What is the kinetic energy of a car with a mass of 1200 kg and is traveling at 40 m/s?
Answer:
9,600,000 JExplanation:
The kinetic energy of an object can be found by using the formula
\(k = \frac{1}{2} m {v}^{2} \\ \)
m is the mass
v is the velocity
From the question we have
\(k = \frac{1}{2} \times 1200 \times {40}^{2} \\ = 600 \times 1600\)
We have the final answer as
9,600,000 JHope this helps you
The Sun radiates energy at a rate of about 4×1026W. At what rate is the mass decreasing?
4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.
The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.
To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.
Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.
ΔE = 4×10^26 W
c = 3×10^8 m/s (speed of light)
Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².
Substituting the values, we have:
Δm = (4×10^26 W) / (3×10^8 m/s)²
Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.
This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.
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A ball of mass 3 kg is released at the top of a track, as shown in the image below. The top of the ramp is 1 metre higher than
the lower, horizontal section of the track. This horizontal section is 5 metres long.
Assumptions:
• There is no loss of energy due to friction or air resistance.
• Gravitational acceleration is 9.8 m/s?
3 kg
1 m
5 m
(a) [1 MARK] What type of energy does the ball have when it is being held at the top of the track?
potential energy
(b) How much work must be done to hold the ball at the top of the track?
• [0.5 MARK] Write the amount of work done (number only) here: 0
[0.5 MARKI Write the unit for the work done here (symbol, not words):
.
(C) (1 MARK] What is the velocity (in m/s) of the ball when it is at the start of the horizontal section of the track? (Write the
number only to two decimal places. Do not include the unit.)
(d) [1 MARK] What is the velocity (in m/s) of the ball when it is at the end of the horizontal section of the track? (Write the
number only to two decimal places. Do not include the unit.)
Answer:
potential energy
Explanation:
Suppose that a magnetic field of 1 T goes out of the plane of the page. A straight wire carries a current 1 A from right to left. Find the direction of force acting on the wire.Group of answer choicesO out of the page into the page left,O in the plane of the page right, O in the plane of the page up, O in the plane of the page
The required direction of magnetic field is observed to be directed up, in the plane of the page. Correct option is C.
The magnetic field is given as B = 1 T.
Th current through the wire as I = 1 A.
The right-hand rule specifies the magnetic force's orientation. It states that if the index finger is pointing in the direction of a magnetic field and the thumb is pointing in the direction of current flow, the middle finger will point in the direction of the magnetic force when the index, thumb, and middle digits are held perpendicular to one another.
By using this formula, we can see that the middle finger is pointing upward towards the page's horizontal plane.
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A horizontal pipe has varying cross-section, with one section having a diameter of 5.5cm and another section having a diameter of 4.6 cm. When a fluid of density 3.38 g-cm-3 flows through the pipe, it has a flow speed of 715 cm-s-1 and pressure of 3.45 x 105 Pa in the section with diameter 5.5 cm.A) Calculate the flow speed (in m-s-1) of the fluid in the section with diameter 4.6 cm.B) What is the pressure (in × 105 Pa) in the section with diameter 4.6 cm?
Given data:
* The given section with the diameter d_1 = 5.5 cm.
* The flow speed of the fluid in the section with the diameter d_1 is v_1 = 715 cm/s.
* The diameter of the other section is d_2 = 4.6 cm
* The density of the fluid is,
\(d=3.38gcm^{-3}\)* The pressure of fluid in the section with diameter d_1 is,
\(P_1=3.45\times10^5\text{ Pa}\)Solution:
(A). The flow rate of the fluid remains constant, thus,
\(A_1v_1=A_2v_2\)where A_1 is the area of the section with diameter d_1, A_2 is the area of the other section with diameter d_2, and v_2 is the velocity of the fluid in section with diameter d_2,
The area of the section with diameter d_1 is,
\(\begin{gathered} A_1=\pi\times(\frac{d_1}{2})^2 \\ A_1=\frac{\pi d^2_1}{4}_{} \end{gathered}\)Substituting the known values,
\(\begin{gathered} A_1=\frac{\pi\times(5.5)^2^{}}{4} \\ A_1=23.76cm^2 \end{gathered}\)The area of the other section of pipe with diameter d_2 is,
\(\begin{gathered} A_2=\frac{\pi d^2_2}{4} \\ A_2=\frac{\pi\times4.6^2}{4} \\ A_2=16.62cm^2 \end{gathered}\)Thus, the velocity of the fluid from the other section with diameter d_2 is,
\(\begin{gathered} A_1\times v_1=A_2\times v_2 \\ 23.76cm^2\times715\times10^{-2}ms^{-1}=16.62cm^2\times v_2 \\ 23.76cm^2\times7.15ms^{-1}=16.62cm^2\times v_2 \\ v_2=\frac{23.76\times7.15}{16.62}ms^{-1} \end{gathered}\)By simplifying,
\(\begin{gathered} v_2=\frac{169.9}{16.62}ms^{-1} \\ v_2=10.22ms^{-1} \end{gathered}\)Thus, the velocity of the fluid from the other section with a diameter of 4.6 cm is 10.22 m/s.
(B). From Bernoulli's principle, the relation between the pressure and velocity on each section is,
\(P_1+\frac{1}{2}d_{}v^2_1=P_2+\frac{1}{2}dv^2_2\)where d is the density of the fluid, P_1 is the pressure of fluid in the section with diameter d_1 and p_2 is the pressure of the fluid in the section with the diameter d_2,
Substituting the known values,
\(3.45\times10^5\times\frac{1}{2}\times3.38\times10^{3^{}}\times(715\times10^{-2})^2=P_2\times\frac{1}{2}\times3.38\times10^{3^{}}\times(10.22)^2\)By simplifying,
\(\begin{gathered} 3.45\times10^5\times\frac{1}{2}\times3.38\times10^{3^{}}\times(7.15)^2=P_2\times\frac{1}{2}\times3.38\times10^{3^{}}\times(10.22)^2 \\ 298.07\times10^8=P_2\times176.5\times10^3 \\ P_2=\frac{298.07\times10^8}{176.5\times10^3} \\ P_2=1.69\times10^5\text{ Pa} \end{gathered}\)Thus, the pressure of fluid in the section with a diameter of 4.6 cm is,
\(\text{1}.69\times10^5\text{ Pa}\)50POINTS PLEASE HELP SPACE QUESTION
Explain how and why seasons vary in the Northern and Southern Hemispheres.
The seasons in the Northern Hemisphere are the opposite of those in the Southern Hemisphere. Seasons occur because Earth is tilted on its axis relative to the orbital plane, the invisible, flat disc where most objects in the solar system orbit the sun.