Conserving fuel is important because burning fuel adversely affects the environment.
How burning fuel affect our environment?When fossil fuels are burned, they release large amounts of carbon dioxide which is a greenhouse gas into the surrounding environment. Greenhouse gases trap heat in the atmosphere that is responsible for global warming. Due to this burning of fuel, a large number of glaciers start melting which leads to drowning of coastal areas and large number of floods.
So we can conclude that Conserving fuel is important because burning fuel adversely affects the environment.
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Oliver repeats his run five times.
Give two reasons why.
Answer:
•So he stays healthy.
•He has energy and time to do it.
A constant force applied to object A causes it to accelerate at 5 m/s². The same force applied to object B causes an accelera-
tion of 3 m/s². Applied to object C, it causes an acceleration of
8 m/s².
a. Which object has the largest mass?
b. Which object has the smallest mass?
c. What is the ratio of mass A to mass B (mA/MB)?
(i) The largest mass will be of the object B.
(ii) The smallest mass will be of the object C.
(iii) The ratio of the mass of object A and b will be 3:5.
According to the Newton’s 2 law for the same force applied acceleration and mass will be in the inverse ratio .it means the more the acceleration the less will be the mass of the object for the same force applied. As the acceleration of the object B is less so its mass will be more and vice versa. And the ratio of the masses will be the inverse ratio of the acceleration of object.
Hence, the ratio will become inverse ratio of acceleration of both the object that is 3:5.
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M A uniformly charged ring of radius 10.0cm has a total charge of 75.0μC . Find the electric field on the axis of the ring at from the center of the ring.(b) 5.00 cm,
The electric field on the axis of the uniformly charged ring at the center of the ring at 5.00cm is 2.41×〖10〗^7 N/C
Where a positive charge (q) is given, always note that the charges point away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,
E=(K_e×b)/(r^3×Q)
Where
E is Electric field
Ke is electric field constant
a is radius of the charge q from the center of the circle
Q is electric charge = 75.0μC
The radius of the charge is a = 10cm= 0.1m
Distance from the electric field = 5cm = 0.05m
Ke is a constant given as 8.99×〖10〗^(9 ) Nm^2/c^2
E=(8.99×〖10〗^(9 ) Nm^2/c^2×0.05)/(〖0.1〗^3×75×〖10〗^(-6) )
E=2.41×〖10〗^7N/C
Therefore, the electric field away from the center of the ring is 2.41×〖10〗^7N/C away from the center of the ring
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A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)
Answer:
48.9 m/s
Explanation:
I am ignoring air resistance in this:
horizontal velocity stays constant, so we need to find vertical velocity
v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)
v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s
the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s
Answer:
V = 49.2 m/s
Explanation:
Given:
Vₓ = 47.4 m/s
t = 1.23 s
___________
V - ?
Vy = Voy + g·t = 0 + 9.8·1.23 ≈ 12,1 m/s
V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s
The tensile stress in a thick copper bar is 99.5 % of its elastic breaking point of 13.0× 10¹⁰ N/m² . If a 500-Hz sound wave is transmitted through the material, (b) What is the maximum speed of the elements of copper at this moment?
The speed of the wave in the rod is
[tex]The speed of the wave in the rod is $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$[/tex]
5044m/s
What is sound wave?
A sound wave is the pattern of disruption brought on by the movement of energy moving through a medium as it propagates away from the source of the sound (such as air, water, or any other liquid or solid substance). Pressure waves are produced when an item vibrates, such as a ringing phone, and these waves are known as sound waves. The surrounding medium's particles are disturbed by the pressure wave, and those particles disrupt the particles next to them, and so on. Like ocean waves, the disturbance's pattern causes outward movement in all directions. Usually in all directions and with decreased intensity as it gets further away from the source, the wave transmits the sound energy through the medium.
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how many times higher could an astronaut jump on the moon than on earth if his takeoff speed is the same in both locations (gravitational acceleration on the moon is about 1/6 of size 12{g} {} on earth)?
An astronaut could 6 times higher on the moon than on earth if his takeoff speed is the same in both locations.
What was it like for astronauts on the lunar surface?From lunar circle, space explorers guided cameras through the window of their space apparatus toward catch photographs of the moon's surface. The nearest look we've had at the moon came from the sendoff of NASA's Apollo program during the 1960s. Somewhere in the range of 1967 and 1972, a progression of missions handled the primary men on the moon. At the point when the following space explorer to arrive at the moon strolls on the lunar surface in 2024, she'll confront radiation levels multiple times higher than on Earth. While Apollo mission space travelers. The prompt surface was extremely fine, as best we could see peering down from 15 feet. Off somewhere out there was an exceptionally clear skyline, perhaps with a rock.
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the universe originated from the explosion and expansion of all matter and energy. this statement is an example of .
The Big Bang Theory is an example of the universe originated from the explosion and expansion of all matter and energy
The Big Bang Theory is the way that astronomers explain how the universe began as a tiny, dense, fireball that exploded 13.8 billion years ago. It includes Albert Einstein's general theory of relativity along with standard theories of fundamental particles.
The theory was born of the statement that other galaxies are moving away from our own at countless speed in all directions, as if they had all been forced by an ancient explosive force.
What is the solar system?The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.
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Me.Henderson Porsche accelerates from 0 to 60 mi/hr in 4 seconds. It’s acceleration is?
The acceleration of Mr. Henderson Porsche is 54545.45 mi/hr².
Velocity is the directional speed of an object, body, or particle in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.
We know that acceleration of a body is Change in Velocity per unit time.
[tex]a=\frac{v-u}{t}[/tex]
Here ,
v is the Final Velocity,
u is the Initial Velocity and;
t is the time taken.
In the given question ,
The Porsche accelerated from 0 to 60 miles per hour in four seconds.
u = 0 mi/hr
v = 60 mi/hr
t = 4 sec
[tex]t=\frac{4}{3600}{~}hr\\[/tex]
[tex]t=\frac{1}{900} {~}hr[/tex]
t = 0.0011 hr
Then,
[tex]a=\frac{v-u}{t}[/tex]
[tex]a=\frac{60-0}{0.0011}[/tex]
[tex]a=54545.45 {~}mi/hr^{2}[/tex]
Hence, the acceleration of the Porsche is 54545.45 mi/hr².
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Ricardo came up with three different designs for paper airplanes. He would like to find out which design flies the farthest. Ricardo plans to fly each paper airplane five times. What step should ricardo take after he flies each plane?.
Ricardo came up with the three different paper airplane designs. Ricardo plans to fly each paper airplane five times to see which design flies the furthest. Ricardo should calculate the aircraft's flight distance after each flight and record it.
How can one determine a distance?Two of the most popular methods in agricultural surveying involve taking measurements of angles and distances with simple equipment. even when the equipment is simple. A high level of precision can be reached with perseverance and practice.
The primary method for measuring distance is (1) pace. The mileage counter. (3) taping or chaining four phases and (6) methods for electronic distance measurement, optical range finders (EDM).
It will be helpful to first examine the common units of distance (displacement). Common English units are as follows:
A foot is made up of 12 inches (in)
Yd (yard) equals 3 ft (ft)
The length of one rod is 16.5 feet and 5.5 yards (yd)
5280 feet, 1760 yards, and 320 rods make up a mile.
The units cancellation method is very helpful when converting between different units of measurement.
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Assume two identical sinusoidal waves are moving through the same medium in the same direction. Under what condition will the amplitude of the resultant wave be greater than either of the two original waves?(a) in all cases(b) only if the waves have no difference in phase (c) only if the phase difference is less than 90°(d) only if the phase difference is less than 120° (e) only if the phase difference is less than 180°
The correct options are B and C
Let us assume the amplitude of the two sinusoidal waves to be A1 and A2
Since, it is mentioned that the waves are identical, therefore let us assume A1=A2= a
The resultant wave of any two waves is given by the formula:
Resultant wave , R = [tex]\sqrt{A1^2+A2^2+2A1A2cos\alpha }[/tex]
where A1 and A2 are amplitudes of wave 1 and 2 and [tex]\alpha[/tex] is the angle between the waves
Now as per the formula, it is the the factor 2cos[tex]\alpha[/tex] that will decide that amplitude is greater than original or not
Hence we know that maximum value of cos [tex]\alpha[/tex] is when [tex]\alpha[/tex] is 0 degrees
Now it is known that value of cos[tex]\alpha[/tex] varies from 1 to 0 for o degrees to π/2
and the value varies from 0 to -1 for angle π/2 to π angle
Case 1: when [tex]\alpha[/tex]=0 degree
Resultant wave R = [tex]\sqrt{A1^2+A2^2+2cos0A1A2}= \sqrt{2A^2+2A^2}=\sqrt{4A^2} =2A[/tex]
Case 2: When [tex]\alpha[/tex] is less than 90
let us consider [tex]\alpha = 45 degree[/tex]
Resultant wave R =
[tex]\sqrt{A1^2+A2^2+2cos45A1A2}= \sqrt{2A^2+2A^2/2}=\sqrt{3A^2} =\sqrt{3} A[/tex]
case 3:
When [tex]\alpha[/tex] is less than 120
[tex]\sqrt{A1^2+A2^2+2cos120A1A2}= \sqrt{2A^2-2A^2/2}=A[/tex]
case 4: when [tex]\alpha[/tex] is less than 180 degrees
the value will be 0
Hence only option b and d are correct
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In an ultrahigh vacuum system (with typical pressures lower than 10⁻⁷ pascal), the pressure is measured to be 1.00 ×10⁻¹° torr (where 1 torr = 133 Pa). Assuming the temperature is 300 K , find the number of molecules in a volume of 1.00m³ .
The number of molecules in the given volume is 5.33 x 10⁻¹² moles.
What is the number of molecules of the gas?The number of molecules of the gas is calculated as follows;
PV = nRT
n = PV/RT
where;
P is the pressure of the gasV is the volume of the gasT is temperature of the gasn is number of molecules of the gas1 torr = 133 Pa
1 x 10⁻¹⁰ torr = ?
= 1.33 x 10⁻⁸ Pa
n = (1.33 x 10⁻⁸ x 1) / (8.314 x 300)
n = 5.33 x 10⁻¹² moles
Thus, the number of molecules in the given volume is 5.33 x 10⁻¹² moles.
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what does it mean when an emission line of helium from a galaxy is shifted toward red?(1 point) the galaxy contains a small percent of hydrogen. the galaxy contains a small percent of hydrogen. the galaxy contains a large percent of hydrogen. the galaxy contains a large percent of hydrogen. the galaxy is moving toward the viewer. the galaxy is moving toward the viewer. the galaxy is moving away from the viewer.
When an emission line of helium from a galaxy is shifted toward red, it means that the galaxy is moving away from the viewer and is denoted as option D.
What is a Galaxy?This refers to the systems of stars and interstellar matter that make up the universe which are gravitationally bound together and are made up of four main components which are:
The disk.The halo.The central bulge.The center or black hole.When an emission line of helium from a galaxy is shifted toward red, there is a displacement to a longer wavelengths and this alteration is referred to as a doppler shift.
Moving away from the view means that the galaxy will be viewed in the red spectrum of the visible light which is a part of the electromagnetic spectrum.
This is therefore the reason why option D was chosen as the most appropriate choice.
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Compared with a 1-kg block of solid iron, a 2-kg block of solid iron has the same select one:
a. mass.
b. volume.
c. weight.
d. all of these
e. none of these
The content of matter in a body is called mass. Weight can be defined as the amount of force required to accelerate an object with mass, while volume is the space occupied by the mass.
On comparing with 1-kg block of solid iron, a 2-kg block of solid iron, mass is different for both the blocks.
So (a) mass cannot be the correct option.
According to this formula ,[tex]Density = \frac{Mass}{Volume}[/tex] , Density for both the blocks would be same, so volume will be different for both the blocks because of the different masses.
So (b) Volume cannot be the correct option.
Weight is mass dependent therefore
(c) weight cannot be the correct option
Since (a), (b), (c), are incorrect so (d) can not be the correct answer
Therefore on Comparing a 1-kg block of solid iron, with a 2-kg block of solid iron, (e) none of these would be same.
Correct option is (e)
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Four toy cars are moving at a constant speed until they experience an unbalanced force of 12 n each. Which toy car would have an acceleration of 3 m/s2 ?.
The car's acceleration would be for a mass of 4 kg is 3 m/s²
Given the information below:
12 Newtons of force.
Increasing speed Equals
By using Newton's Second Law of Motion, we can determine the mass of the car:
According to Newton's second law of motion, force is equal to the rate at which momentum changes. Force is defined as mass times acceleration for a constant mass.
force= mass* acceleration
This formula provides Newton's Second Law of Motion mathematically;
12= mass*3
mass=12/ 3
mass=4 kilogram
By changing the parameters in the formula, we obtain;
= 4 kilos of mass.
As a result, the vehicle with a mass of 4 kg would accelerate at 3 m/s².
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A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationship between the direction of the velocity and acceleration vectors? assume air resistance is negligible.
The dot product of velocity vector and acceleration vector at maximum height is equal to zero.
[tex]$\overrightarrow{v}.\overrightarrow{a}= 0[/tex]
We have a football player who kicked a football in order to do a goal.
We have to determine the relationship between the direction of the velocity and acceleration vectors when the football reaches its maximum height.
What is Projectile Motion ?Projectile motion is the two - dimensional motion of an object thrown or projected into the air such that it moves under the influence of acceleration of gravity.
According to the question -
Assume that the velocity with which the projectile was fired is u m/s.
Therefore, it can be resolved into two components as -
[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + u sinθ [tex]a_{y}[/tex].
Now, when the projectile reaches the maximum height, it will no longer cover any vertical height but it will keep moving in the horizontal direction. Therefore - the vertical component of the velocity will become zero. Therefore -
[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + 0 = u cosθ [tex]a_{x}[/tex]
Now, refer to the figure attached for reference -
At the maximum height, the velocity vector is in the horizontal direction and the vector for the acceleration due to gravity is vertically downwards.
Therefore - the acceleration vector and velocity vector will have an angle of 90° between them. Assume -
Velocity at maximum height = v = u cosθ [tex]a_{x}[/tex]
and acceleration = a = g = 9.8 m/[tex]s^{2}[/tex]
Therefore -
The dot product of velocity vector and acceleration vector is equal to zero.
[tex]\overrightarrow{v}.\overrightarrow{a}= 0[/tex]
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Two spheres are made of the same metal and have the same radius, but one is hollow and the other is solid. The spheres are taken through the same temperature increase. Which sphere expands more?.
The expansion of both will be the same. Then the correct option is C.
What is thermal expansion?Thermal expansion, which often excludes phase transitions, is the propensity of matter to alter its form, area, volume, and concentration in response to a variation in heat.
The same metal is used to create two identically sized spheres, but only one is solid and the other is hollow. The same rate of warming is applied to the spheres.
The formula for the thermal expansion is given as,
[tex]\rm \alpha _L = \dfrac{1}{L} \cdot \dfrac{dL}{dT}[/tex]
[tex]\alpha_{L}[/tex] = thermal expansion
L = particular length measurement
dL = change in length
dT = change in temperature
If the material of the sphere is the same, then the coefficient of the expansion of the sphere is the same, then the expansion of both will be the same.
Both will expand at the same rate. Then, C is the right answer.
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help! i dont now the answer. you will get 15 points if you answer
Answer:
i think a or b is the answer
Explanation:
i also don't know exactly
If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?
Answer:
[tex]{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}[/tex]
A balloon clings to a wall after it is negatively charged by rubbing.(a) Does that occur because the wall is positively charged?
A balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.
Is opposite charge attract each other?Yes, opposite charges attract each other. When a positive charge and a negative charge interact with each other, their forces act from the direction of positive to the direction of negative charge. As a result opposite charges attract each other while on the other hand, similar charges repel each other because their forces move in the opposite direction so they repel each other.
So we can conclude that a balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.
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A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude.(ii) Next assume the end at which reflection occurs is free to slide up and down. Now what is the net displacement at a point on the string where two pulses are crossing? Choose your answer from the same possibilities as in part (i).
(ii) The net displacement at a point on the string where the pulses cross is 0.2 m.
The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.
A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.
At the post, the pulses are reflected and return along the string without losing any of their amplitude.
Now, let's say the ends are free.
There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.
Now, since A = 0.1 m
Then, 2A = 2(0.1) = 0.2 m
As a result, the net displacement at the string's intersection of two pulses is 0.2 m.
The correct option is (c).
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How quickly would a 60kg object accelerate if the person applied a 500N force?
Answer:
8.33 m/s^2
Explanation:
The equation for force is Force = mass * acceleration. The force is 500 N, and the mass is 60 kg, so substituting those into the equation you get that 500 = 60 * acceleration. Divide 500 by 60 to get the acceleration, which is 8.33 m/s^2.
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a 3.5 kg block is pushed along a horizontal floor by a force : f of magnitude 15 n at an angle ???? 40° with the horizontal (fig. 6-19). the coefficient of ki- netic friction between the block and the floor is 0.25. calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.
The frictional force is 11 N and the block’s acceleration is 0.14 m/s².
What is the coefficient of friction?It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.
For part (a):
Taking x component:
F(x) = F cosθ
Taking y component:
F(y) = -F sinθ
Applying Newton’s second law to the y-axis:
F(N) = 15 sin40 + (3.5)(9.8)
F(N) = 44 N
coefficient U = 0.25
f(k) = 11 N
For part (b):
Applying Newton’s second law to the x-axis:
a = ((15) cos40 - 11)/3.5
a = 0.14 m/s²
Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².
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explain the way to get wind energy for our use
Answer:
YOUR ANSWER IS HERE....
Wind turbines work on a simple principle: instead of using electricity to make wind—like a fan—wind turbines use wind to make electricity. Wind turns the propeller-like blades of a turbine around a rotor, which spins a generator, which creates electricity.USES OF THE WIND ENERGY.generating electricity.generating electricity.milling grain.generating electricity.milling grain.pumping water.generating electricity.milling grain.pumping water.powering cargo ships (via kites)generating electricity.milling grain.pumping water.powering cargo ships (via kites)reducing carbon footprint.generating electricity.milling grain.pumping water.powering cargo ships (via kites)reducing carbon footprint.sailing.generating electricity.milling grain.pumping water.powering cargo ships (via kites)reducing carbon footprint.sailing.windsurfing.generating electricity.milling grain.pumping water.powering cargo ships (via kites)reducing carbon footprint.sailing.windsurfing.land surfing.HOPE IT HELPS! MARK AS BRAINLIAST.A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field(a) 0 cm
The magnitude of the electric field is 0 N/C.
We choose a Gaussian sphere of radius r > R concentric with the solid sphere.
Apply Gauss's law from Equation (1):
[tex]\Phi_{E} = \oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]
Since the electric field vector E is parallel to the area vector dA of the curved part the dot product can be the simple product EdA:
[tex]\oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]
since the electric field is constant we can put E out of the integral:
[tex]E \oint dA = \frac{q_{in}}{\in_{0} }[/tex]
[tex]EA = \frac{q_{in}}{\in_{0} }[/tex]
Substitute for A from Equation (2) and for [tex]q_{in}[/tex] from Equation (3)
recognizing that the enclosed charge by the Gaussian sphere is less than Q:
[tex]E(4\pi r^{2}) = \frac{pVgaussian}{\in_{0} }[/tex]
Substitute for V gaussian (volume of the Gaussian surface) from Equation (4):
[tex]E(4\pi r^{2}) = \frac{\rho (\frac{4}{3} \pie r^{2})} {\in_{0} }[/tex]
[tex]E = \frac{\rho r}{3\in_{0} }[/tex]
Where p is the charge per unit volume of the solid sphere which is equal to the total charge Q of the sphere divided by the volume V of the sphere :
[tex]E = \frac{(\frac{Q}{V} )r}{3\in_{0} }[/tex]
[tex]E = \frac{Qr}{(\frac{4}{3} \pi R^{3})3\in_{0} }[/tex]
[tex]E = \frac{Qr}{4\pi \in_{0}R^{3} }[/tex]
Substitute numerical values:
[tex]E = \frac{(26\times10^{-6})(0)}{4\pi(8.8542 \times10^{-12})(0.4)^{3}}[/tex]
[tex]E = 0 N/C[/tex]
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M A particle with positive charge q = 3.20 ×10⁻¹⁹C moves with a velocity →v = (2i + 3j - k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle (in unit-vector notation), taking B = (2i + 4j + k) T and E =(4i - j - 2k ) V ,
The Magnitude of force on the charged particle is -
F = (12.8 x 10⁻¹⁹C)i - (3.2 x 10⁻¹⁹C)j - (6.4 x 10⁻¹⁹C)k
We have a positively charged particle that moves through a region where both a uniform magnetic field and a uniform electric field exist.
We have to determine the total force on the moving particle.
What is Lorentz Force?The force exerted on a charged particle (q) moving with velocity (v) through an Electric field (E) and Magnetic field (B) is called Lorentz force. Mathematically -
|F| = qE + qv x B
According to the question, we have -
q = 3.20 × 10⁻¹⁹C
v = (2i + 3j - k) m/s
B = (2i + 4j + k) T
E = (4i - j - 2k) V
Now, using the formula to calculate Lorentz force -
F = q(4i - j - 2k) + q(2i + 3j - k) x (2i + 4j + k)
F = q(4i - j - 2k) + (2qi + 3qj - qk) x ((2i + 4j + k)
F = q(4i - j - 2k) + 0
F = q(4i - j - 2k)
F = (12.8 x 10⁻¹⁹C)i - (3.2 x 10⁻¹⁹C)j - (6.4 x 10⁻¹⁹C)k
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____?_______ is the muscular partition which between chest and abdominal cavity.
Answer: The diaphragm
Explanation:
The diaphragm is a thin dome-shaped muscle which separates the thoracic cavity (lungs and heart) from the abdominal cavity (intestines, stomach, liver, etc.)
What happens when you push or pull a water tube
The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.
It is very hard when pulling water and if a person need to consistently pull air out of the tube before water makes it to the pump, it is very difficult and some pumps may or cannot do it.
What is the pulling about?In regards to pushing, when a person is pushing above the Center of Gravity, one is pushing the said object into the ground and thus is increasing the rate of friction.
If a person is pulling you are said to be pulling above the Center of Gravity and lowering friction.
So, The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.
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a water tank in the shape of a hemispherical bowl of radius 5 m5 m is filled with water to a depth of 2 m.2 m. how much work is required to pump all the water over the top of the tank? (the density of water is 1000 kg/m31000 kg/m3 ). (use symbolic notation and fractions where needed.) W
The work required to pump all the water over the top of the tank of radius 5 m and depth 2 m is 5.12 x 10^6 J.
The work required to pump the water over the top of the tank is equal to the gravitational potential energy of the water inside the tank.
The potential energy is given by the following equation,
U= mgh
Here, m is the mass of water present inside the tank.
We know that, Density( Mass/Volume
Hence, m= Density x volume
So, the equation becomes
U= ρ x V x g x h
Where V is the volume of hemispherical tank= 2/3π x r^3
Putting the given values in the above equation,
U= 1000 x 2/3 x 3.14 x 53 x 9.8 x 2 = 5.12 x 10^6 J
Hence, the work required to pump all the water over the top of a tank of radius 5 m and a depth of 2 m is 5.12 x 10^6 J.
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A speeding car passes a highway patrol checkpoint, then decelerates at a constant rate. After 5 s, the car is 225 m from the checkpoint, and its speed is then 30 m/s. What was the car’s velocity when it passed the checkpoint?.
The car’s velocity when it passed the checkpoint was of: 60 m/s
The formula and procedure we will use to solve this exercise is:
vi = [(2 *x)/ t] - vf
Where:
x = distancet = timevi = initial velocityvf = final velocityInformation about the problem:
x = 225mt = 5 svf= 30 m/svi = ?Applying the initial velocity formula we have:
vi = [(2 *x)/ t] - vf
vi = [(2 * 225m)/5 s ] - 30 m/s
vi = [(450 m)/5 s ] - 30 m/s
vi = 90 m/s - 30 m/s
vi = 60 m/s
What is velocity?It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).
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which si unit for legnth is greater than a meter a. decimeter b. centimeter c. kilometer d. milimeter
Option(c) is Kilometer is correct. Kilometer unit for length is greater than a meter.
What do you mean by SI units?Worldwide adoption of the International System of Units (SI) as a system of measurement units is widespread. For convenience, this updated version of the metric system is based on the number 10. The metric prefixes or the SI prefixes are a collection of prefixes that have been established. According on the prefixes, the unit is either a multiple or a fraction of base ten. It enables the zeros of very small or very high numbers to be reduced, for example, from 7,500,000 Joules to 1 nanometer and from 0.000000001 meter to 7.5 Megajoules, respectively. Additionally, some SI prefixes have a group of symbols before the unit symbol.
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