The surface area to volume ratio declines with cell expansion. This is due to the fact that as a cell expands, its volume (internal contents) develops more quickly than its surface area (its cell membrane). Because the cell membrane needs to have enough surface area for the transfer of materials into and out of the cell, this is an issue for cells. The cell must go through mitotic cell division or cease to function if its volume grows to the point where not enough materials can be carried into and out of it. Cells are so tiny because of this.
A cell's surface area to volume ratio will decrease as its surface area will decrease keeping the volume constant. A cell with a smaller surface area to volume ratio has less plasma membrane and more volume.
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the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.
Three-dimensional shape the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.
What do you mean by three-dimensional shape?A solid figure, object, or shape with three dimensions—length, breadth, and height—is known as a three-dimensional shape in geometry. Three-dimensional shapes have height, which is equivalent to thickness or depth, unlike two-dimensional shapes. These objects are sometimes referred to as 3-D forms because three dimensions can also be written as 3-D. Every 3-D shape takes up space, which is quantified by volume. For instance, The fundamental 3-dimensional shapes we encounter every day include a cube, rectangular prism, sphere, cone, and cylinder. We can see 3-D shapes all around us. A square prism can be seen in a book and a box, a cube in a Rubik's Cube and a die, etc.
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ECONOMICS GRADE 10 CASE STUDY TOPIC: South African growth and development: Mining and industry Manufacturing and services.
The correct answe is South Africa - Economic Growth and Development:
After the formal end of the previous apartheid system two decades ago, South Africa can now boast having one of the richest economies in Africa and a well-functioning democracy. It is the largest economy in Africa, but it also has deeply ingrained structural issues that limit its ability to expand and flourish.
One of the largest economies on the African continent is that of South Africa. However, despite a period of rapid development from 2003 to 2007, its real GDP average yearly growth rate between 2001 and 2010 has been very weak and unquestionably significantly below the African average. A number of African nations have had substantially faster growth rates, which has improved a number of development-related indices.
By the standards of the recent growth records of several Euro Zone nations, South Africa's development is hardly sluggish! One crucial aspect of the economy is that South Africa has achieved relatively modest progress in meeting a number of important development targets and some of the Millennium Development Goals, but her economy does not appear to have achieved the "take-off" required to kick start significant development progress, especially against the backdrop of her deep social problems.
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If the mass of a planet is 3. 10 1024 kg, and its radius is 2. 00 106 m, what is the magnitude of the gravitational field, g, on the planet's surface?
The gravitational field strength is 51.6925 N/kg.
We need to know about the gravitational field to solve this problem. The gravitational field is the area where affected by gravitational force. The magnitude of the gravitational field can be calculated by this equation
g = G . m / r²
where g is the gravitational field, G is gravitational constant (6.67 x 10¯¹¹ Nm²/kg²), m is the mass of the planet and r is the radius of the planet
From the question above, we know that
m = 3.10 x 10²⁴ kg
r = 2.00 x 10⁶ m
By substituting the given parameter, we get
g = G . m / r²
g = 6.67 x 10¯¹¹ . 3.10 x 10²⁴/ (2.00 x 10⁶)²
g = 51.6925 N/kg
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What happens to the interference pattern that results from the diffraction of electrons passing through two closely spaced slits if the rate of electrons going through the slits is decreased to one electron per hour?.
The interference pattern remains the same if the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process.
What is Interference?This is referred to as a condition which occurs when two waves meet and combine while traveling along the same medium such as air, water etc.
The interference pattern is the same if the the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process. The decrease in the rate of electrons doesn't affect the pattern of the waves being a constructive or destructive one.
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a negative charge is placed at the center of a ring of uniform positive charge. what is the motion (if any) of the charge? what if the charge were placed at a point on the axis of the ring other than the center?
The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
if the charge were placed at a point on the axis of the ring other than the center the electric field then there is no motion.
electric field - It is a force produced by any charge near its surrounding.
If the charge were placed at a point on the axis of the ring other than the center then there is no motion because at the center of the ring E.F ( Electric field is zero.
electric filter the point on the axis of uniformly charged ring depend on the distance of the point from the center of the ring.
.
E.F = 0 because each half cancels each other . the negative charge will be in equilibrium as every part of the ring is uniformly attracted by it.
The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
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Explain grounding in your own words
Badminton game played with one player per side. (identification)
(Physical education)
Answer:
Badminton game played with one player per side is true and Badminton game played with two players per side is also true.
Explanation:
What is the active volcanoes
A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0. 20 kg, and the robot has a mass of 0. 30 kg. The robot pushes on the doll with a force of 0. 30 n. The magnitude of the acceleration of the robot is.
The robot is moving with an acceleration of 1.0 m/s.
What do you understand by magnitude?Magnitude is simply "distance or quantity" in the context of physics. In terms of motion, it shows the absolute or relative size, direction, or movement of an object. It is used to describe something's size or scope. In physics, the term "magnitude" often refers to a size or amount.
Scalar values serve as units of measurement, and scalar multiplication serves as the definition of magnitude. A quantity's magnitude in a particular unit multiplied by that unit equals the original quantity. This is true for all tensor types, including vectors and real-number tensors.
The given parameters;
mass of the toy doll, m = 0.2 kg
mass of the robot, M = 0.3 kg
force applied by the robot, F = 0.3 N
The magnitude of the acceleration of the robot is calculated from Newton's second law of motion;
F = ma
where;
a is the acceleration of the robot
Thus, the magnitude of the acceleration of the robot is 1 m/s²
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The area of a typical eardrum is about 5.00 × 10⁻⁵m² .(a) Calculate the average sound power incident on an eardrum at the threshold of pain, which corresponds to an intensity of 1.00 W/m² .
The sound power incident on the eardrum at the threshold of pain is,
[tex]5 \times 10^{-5} W[/tex]
What do you mean by the intensity?In physics, the power transferred per unit area is known as the intensity or flux of radiant energy, where the area is measured on a plane perpendicular to the direction of the energy's propagation. Watts per square meter (W/m2) and kilograms per square meter (kg/s3) are the units used in the SI system. With waves like acoustic waves (sound) or electromagnetic waves like light or radio waves, intensity is most usually employed to describe the average power transfer across one period of the wave. Other situations where energy is exchanged can also be described in terms of intensity. One could, for instance, figure out how much kinetic energy each drop of water from a sprinkler is carrying.
Power = intensity x area
Given:
Area, A = [tex]5 \times 10^{-5} \;m^{2}[/tex]
Intensity, I = [tex]1\;W/m^{2}[/tex]
We know that,
Sound Power, P = [tex]I \times A[/tex]
P = [tex]1 \times 5 \times 10^{-5}[/tex]
P = [tex]5 \times 10^{-5}\;W[/tex]
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Suppose you are transported to a planet with twice the mass of earth, but the same radius of earth. Your weight would __________ by a factor of __________.
Answer:تاتا البال،ًـلاللبببلا
لولاليثقغاايصسضشطذدزلب
Explanation:ابرلرتهىغرلب
GP Submarine A travels horizontally at 11.0 m/s through ocean water. It emits a sonar signal of frequency f = 5.27× 10³Hzin the forward direction. Submarine B is in front of submarine A and traveling at 3.00 m/s relative to the water in the same direction as submarine A. A crewman in submarine B uses his equipment to detect the sound waves ("pings") from submarine A. We wish to determine what is heard by the crewman in submarine B. (b) In Equation 17.19 , should the sign of v(S)be positive or negative?
GP Submarine A travels horizontally at 11.0 m/s through ocean water. It emits a sonar signal of frequency f = 5.27× 10³Hzin the forward direction. Submarine B is in front of submarine A and traveling at 3.00 m/s relative to the water in the same direction as submarine A. A crewman in submarine B uses his equipment to detect the sound waves ("pings") from submarine A. We wish to determine what is heard by the crewman in submarine B. (b) In Equation 17.19, the sign of v(S) should be positive.
What is frequency in physics?Recurrence is the quantity of events of a rehashing occasion for each unit of time. It is additionally once in a while alluded to as fleeting recurrence to underscore the difference to spatial recurrence, and customary recurrence to stress the differentiation to precise recurrence. Recurrence is communicated in units of hertz (Hz) which is comparable to one (occasion) each second. The comparing time frame is the time term of one cycle in a rehashing occasion, so the period is the corresponding of the recurrence. Recurrence is estimated in units called hertz (Hz). At the point when one wave passes in a single second its recurrence is 1 wave each second or 1 Hertz. Recurrence and period are really alternate extremes. While period is estimated in seconds per cycle, recurrence is estimated in cycles each second. Think about wave with a time of 2 seconds.
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determine the car’s instantaneous velocity at t=5s
Answer:
In the given graph,
Velocity is constant between 4 to 8 which is 2m/s
So, at t=5s
Instantaneous velocity=2m/s
Explanation:
hope this help u thank u
A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric flux through this area (b) when the electric. field is parallel to the surface.
The value of electric flux will be zero.
For a uniform electric field passing through a plane surface,
electric flux(ϕ) = EAcosθ
where,
θ is the angle between the electric field and the normal to the surface
E is the electric field
A is surface area
(a)
The electric field is perpendicular to the surface,
so,
θ=0
electric flux(ϕ) = EAcosθ
electric flux(ϕ) =(6.20×10⁵N/C)(3.20m²)cos0
electric flux(ϕ) =1.98×10⁶Vm
(b)
The electric field is parallel to the surface
θ=90°
so,
cosθ=0 and the flux is zero For a uniform electric field passing through a plane surface,
therefore, electric flux(ϕ) =0
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In the electron-transport chain, as electrons move along a series of carriers, they release energy that is used to do what?.
In the electron-transport chain, as electrons move along a series of carriers, they release energy that is used to pump protons across a membrane.
In plants and animals, the electron transport chain points to the consistent streaming of electrons across cells where electron donors donate electrons and electron acceptors accept electrons through an oxidation-reduction reaction.
As a result of the oxidation-reduction reaction, an electrochemical gradient is created and ATPs are generated for energy purposes.
Moreover, the chain comprises 4 key protein complex molecules that are transferred across the plasma membrane cell.
This cellular process takes place in mitochondria and is responsible for chemical reactions such as photosynthesis and cellular respiration. The four principle steps of this process are Krebs' cycle, glycolysis, oxidative phosphorylation, and pyruvate oxidation.
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a block weighing 10 lbf and having dimensions 10 in. on each edge is pulled up an inclined surface on which there is a film of sae 10w oil at 100c f. if the speed of the block is 2 ft/s and the oil film is 0.001 in. thick, find the force required to pull the block. assume the velocity distribution in the oil film is linear. the surface is inclined at an angle of 25° from the horizontal.
The force required to pull the block will be 13.89 kN.
What is the force?Let μ be the viscosity of oil, y be thickness, v be the velocity of the block, and A be the base area of the block. Then the force is given as,
F = μAv / y
An inclined surface with a coating of sae 10w oil at 100°C is dragged up by a block that weighs 10 lbf and measures 10 in on each edge. if the oil coating is 0.001 in. thick and the block is moving at a speed of 2 ft/s. Assume that the oil film's velocity distribution is linear. A 25° angle from the horizontal separates the surface from the horizontal.
The diagram is given below. Then the force is given as,
F - 10 sin 25° = 10 × (10 / 12)² × 2 / 0.001
F = 4.226 + 13888.89
F = 13893.1 N
F = 13.89 kN
13.89 kN of force will be required for pulling the block.
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A meteor is falling to earth and enters the Earth’s atmosphere at 0=−435 ŷ, if it experiences an acceleration due to Earth’s gravity of =−9.8 2ŷ, how what will its velocity be after t= 13s? (assume all motion is vertical, which means on the y-axis. You will use ŷ for up, and - ŷ for down)
At 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s
We know that,
v = u + at
where,
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
Given that,
u = - 435 m / s
a = - 9.8 m / s²
t = 13 s
v = - 435 - ( 9.8 * 13 )
v = - 562.4 m / s
Velocity is the rate of change of distance with respect to time. It is denoted by V.
Therefore, at 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s
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6 A gun fires a shell at an angle of elevation of 30⁰ with a velocity of 2 x 10³ ms. What are the horizontal and vertical components of the velocity? What is the range of the shell? How high will the hall rise?
1. The horizontal component of the velocity is 1.73×10³ m/s
2. The vertical component of the velocity is 1×10³ m/s
3. The range of the shell is 3.53×10⁵ m
4. The maximum height reached is 1.53×10⁵ m
1. How to determine the horizontal component of the velocity
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Horizontal velocity =?Horizontal velocity = u × Cosθ
Horizontal velocity = 2×10³ × Cos30
Horizontal velocity = 1.73×10³ m/s
2. How to determine the vertical component of the velocity
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Vertical velocity =?Vertical velocity = u × Cosθ
Vertical velocity = 2×10³ × Sin30
Vertical velocity = 1×10³ m/s
3. How to determin the range
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Acceleration due to gravity (g) = 9.8 m/s²Range (R) =?R = u²Sine(2θ) / g
R = (2×10³)² × Sine (2×30) / 9.8
R = 3.53×10⁵ m
4. How to determine the maximum height
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Acceleration due to gravity (g) = 9.8 m/s²Maximum height (H) =?H = u²Sine²θ / 2g
H = [(2×10³)² × (Sine 30)²] / (2 × 9.8)
H = 1.53×10⁵ m
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Calculate the centripetal force exerted on the earth by the sun. Assume that the period of revolution for the earth is 365. 25 days, the average distance is 1. 5 × 108 km and the earth’s mass is 6 × 1024 kg.
The centripetal force exerted on the earth by the sun is 3.56775 × 10²² N.
Mass of the Earth, m = 6 × 10²⁴ kg
The average distance between the Sun and the Earth is, r = 1.5 × 10⁸ km . r = 1.5 × 10¹¹ m
The path of the Earth around the Sun is circular, with a circumference:
c = 2πr
c = 2 × 3.1416 × 1.5 × 10¹¹ m
c = 9.42478 × 10¹¹ m
The Earth takes time of T = 365.25 days to complete one revolution around the Sun, that is to travel the distance of c.
T = (365.25 days) × (24hr/1day) × (60min/1hr) × (60s/1min)
T = 3.15576 × 10 ⁷s.
Thus the velocity of the Earth is,
v = c/T
v = (9.42478 × 10¹¹ m) / (3.15576 × 10 ⁷s)
v = 2.98653 × 10⁴ m/s
The centripetal force exerted on the Earth by the Sun is given by,
F = 2mv/r
F = 2× (6 × 10²⁴ kg) × (2.98653 × 10⁴ m/s ) /(1.5 × 10¹¹ m)
F = 3.56775 × 10²² N
Therefore, the centripetal force exerted on the earth by the sun is 3.56775 × 10²² N
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which describe the study of spectroscopy? select the two correct answers.(1 point) interaction of light and atoms interaction of light and atoms emission and absorption of light emission and absorption of light the number of galaxies the number of galaxies reflection of light by earth
Spectroscopy is the study of the absorption and emission of light and other radiation by matter.
How can we conclude the statement?Spectroscopy is the interaction of lights and atoms and the study of the emission and absorption of light.
Spectroscopy can be defined as the study of light absorption as well as its emission. It is also uses to study of structure of atoms.
It offers techniques that has to do with how a sample responds to the radiation of light.
Its applications are:
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A thin, square, conducting plate 50.0cm on a side lies in the x y plane. A total charge of 4.00× 10⁸ is placed on the plate. Find (a) the charge density on each face of the plate
The calculated charge density on each plate is 8*10⁻⁸ C/m²
The charge density is a measurement of the amount of electric charge present per unit of surface area, body volume, or field. How much charge is held in a specific field is determined by the charge density. It is possible to calculate charge density in terms of length, area, or volume.
L = 50cm = 0.5m for side length. Q = 4*10⁻⁸C, charge on the plate
= Q/A for surface charge density
Each component's surface charge density is then equal to half of the plate's overall charge density. Thus,
σ(face) = 1/2σ
(face = Q/2A)
(face = Q/2L2) Now that we have Q and L, we can plug in.
σ(face) = 4*10⁻⁸/ 2*0.5²
σ(face) = 4*10⁻⁸ / 0.5
(face) = 8 * 10 ⁻⁸ C/m²
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An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.800 c relative to the mother ship. As measured on the Earth, the spaceship is 0.200 ly from the Earth when the landing craft is launched.(d) If the landing craft has a mass of 4.00 × 10⁵ kg , what is its kinetic energy as measured in the Earth reference frame?
The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
What is its kinetic energy as measured in the Earth reference frame?It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.
[tex]V_x'=0.8c\\V=0.6c\\m=4*10^5kg[/tex]
Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,[tex]KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}[/tex]
So, to [tex]V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016[/tex]find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. We have an expression from Lorents transformation for relativistic law of addition of velocities as,[tex]V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V[/tex]
Substituting values, we get,[tex]V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c[/tex]
Thus, the KE will be,[tex]KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J[/tex]
Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
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100 points,
The National Park Service sometimes creates controlled burns to mitigate wildfires. How does a controlled burn limit the spread of wildfires?(1 point)
A controlled burn is used to make a path that helps a spreading wildfire arrive at a source of water.
A controlled burn temporarily shuts down parks so people won’t start campfires.
A controlled burn helps to remove plants and vegetation around buildings so a wildfire won’t destroy them.
A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
What is a controlled burns?The term controlled burn refers to setting up an area in which the fire is controlled in order to avoid wild fires. These are deliberately set up in order to avoid the bush from burning down.
Let us recall that a wild fire is able to blaze across a large causing damage to a buildings as well as life and other properties in the way of the fire and could cause huge looses including loss of habitat.
Thus, the National Park Service sometimes creates controlled burns to mitigate wildfires because a controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
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Study the graph below, it represents a motion with 3 distinct parts. Pay particular attention to label on the y-axis. Using at least 1 complete sentence, describe what is happening during each part of the motion. [your answer must include words and phrases that describe motion. Suggested examples are things like accelerating, constant velocity, backward, forward, at rest, stop]
We describe the motion in the graph as
In the first 2 seconds, the motion of the object is that it at rest. From t = 2 seconds to t = 4 seconds, the motion of the object is that it accelerating. For t = 4 seconds to t = 6 seconds, the motion of the object is that it is moving with constant velocityWhat is motion?Motion is the change in position of an object with time.
How to describe the motion in the graph?Since on the graph
the y - axis is represented by velocity and the x - axis is represented by time,Motion from t = 0 to t = 2 secondsFrom the graph, in the first 2 seconds, the motion of the object is described as the object is at rest.
Motion from t = 2 to t = 4 secondsFrom the time , t = 2 seconds to t = 4 seconds, we see that the graph has a positive slope. During this time interval, from t = 2 seconds to t = 4 seconds, the object is accelerating.
So, the motion of the object is described as the object is accelerating
Motion from t = 4 to t = 6 secondsFor the time interval t = 4 seconds to t = 6 seconds, we see that the velocity is a straight line parallel to the x(time) axis. This implies that the velocity of the object is constant for t = 4 seconds to t = 6 seconds. So, for t = 4 seconds to t = 6 seconds, the motion of the object is described as object is moving with constant velocity
So, we describe the motion in the graph as
In the first 2 seconds, the motion of the object is that it is at rest. From t = 2 seconds to t = 4 seconds, the motion of the object is that it is accelerating. For t = 4 seconds to t = 6 seconds, the motion of the object is that it is moving with constant velocityLearn more about motion graph here:
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what is the magnitude (in n/c) and direction of an electric field that exerts a 2 x 10-5 n upward force on a -1.2 µc charge? 1µc
|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.
We have a Charge particle present inside a Electric field.
We have to determine the magnitude (in n/c) and direction of an electric field that exerts a 2 x [tex]10^{-5}[/tex] Newton.
What is Electric Field Intensity?The electric field intensity at a point R meters away from a charge Q is the amount of force experienced by the test charge ([tex]$q_{o}[/tex]) of unit magnitude at a distance R meters away the Charge Q. Mathematically -
E = [tex]$\frac{F}{q}[/tex]
According to the question, we have -
F = 2 x [tex]10^{-5}[/tex] Newton
|Q| = - 1.2 [tex]\mu[/tex]C = - 1.2 x [tex]10^{-6}[/tex] [tex]\mu[/tex]C = 1.2 x [tex]10^{-6}[/tex] [tex]\mu[/tex]C
Therefore -
E = [tex]$\frac{2\times 10^{-5} }{1.2\times 10^{-6} }= \frac{100}{6} = 16.67\;N/C[/tex]
Therefore -
|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.
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A cubical surface surrounds a point charge q . Describe what happens to the total flux through the surface if (d) the charge is moved to another location inside the surface
Gauss law states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.
Mathematically,
Φ=Q ϵo
Where;
Q is enclosed charge
ϵo is the permittivity of the free space
If the charge is moved to another location inside the same cube, the flux in the electric field remains the same. This is because as long as the charge remains within the plane of the electric field, the flux is calculated in all directions and thus the flux remains unchanged. It only changes when the charge is moved outside the cube.
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A reconnaissance plane flies 615 km away from
its base at 918 m/s, then flies back to its base
at 1377 m/s.
What is its average speed?
Answer in units of m/s.
The average speed of the plane is 3,967.7km/h or 1,102.15m/s.
Speed is a scalar number in kinematics that quantifies how quickly an item is moving. Since the plane's speed might fluctuate, we refer to the plane's average speed when describing its rate of motion. The whole distance travelled is divided by the total amount of time required to go that distance to arrive at this pace.
[tex]speed = \frac{distance}{time}[/tex]
For, [tex]time (t_{1} ) = \frac{615}{918}[/tex]
as 1m/s = 3.6km/h , therefore,
[tex]time (t_{1} ) = \frac{615}{918*3.6}[/tex]
[tex]time (t_{1} ) = 0.186 hours[/tex]
For, [tex]time(t_{2} ) = \frac{615}{1377}[/tex]
as 1m/s = 3.6km/h , therefore,
[tex]time (t_{2} ) = \frac{615}{1377*3.6}[/tex]
[tex]time (t_{2} ) = 0.124 hours[/tex]
Thus, the average speed of plane is,
[tex]s = \frac{615+615}{0.124+0.186}[/tex]
s = 3967.7km/h
or, [tex]\frac{3967.7}{3.6}[/tex] m/s = 1,102.15m/s
Therefore, the average speed of the plane is 3,967.7km/h or 1,102.15m/s.
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To launch a 100 kg human so that he leaves a cannon moving at a speed of 4 m/s, you need a spring with an appropriate spring constant. This spring will be compressed 2. 0 m from its natural length to launch the person. Which spring constant do you need?.
To launch a 100 kg human so that he leaves a cannon moving at a speed of 4 m/s, you need a spring with an appropriate spring constant. This spring will be compressed 2. 0 m from its natural length to launch the person.400N/m spring constant is needed.
What is spring constant k?The spring constant, k, is a proportional constant. It gauges how firm the spring is. When a spring is compressed or extended to a length that differs by an amount x from its equilibrium length, it produces a force F = -kx that pushes it back towards its equilibrium position.
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f a ball is thrown into the air with an initial velocity of 50 ft/s, its height in feet after t seconds is given by y
The instantaneous velocity of the ball is 18ft/s.
What is velocity?
The pace at which an object's position changes as perceived from a particular point of view and as measured by a particular unit of time (for example, 60 km/h northbound) is defined as its velocity. Velocity is a fundamental concept in kinematics, the branch of classical mechanics that deals with the motion of bodies.
In order to be defined, the physical vector quantity known as velocity needs to have both a magnitude and a direction. Speed is a coherent derived unit whose quantity is measured in metres per second (m/s or m/s1) in the SI. Speed is the scalar absolute value (magnitude) of velocity (metric system).
Calculations:
at t=1s, the height is y=50(1)-16(12)=34ft.
at t=1.1s, the height is y=50(1.1)-16(1.12)=35.64ft
Between these two heights, the change in time is 0.1s. The change in height is 35.64ft-34ft=1.64ft.
Therefore [tex]V_{avg}=\frac{del y}{del t} = \frac{1.64ft}{0.1s} =16.4ft/s[/tex]
at t=1.01s, the height is y=50(1.01)-16(1.012)=34.1784ft
between t=1.01s and t=1s the change in time is 0.01s and the change in height is 0.1784ft.[tex]V_{avg}=\frac{0.1784ft}{0.01s} =17.84ft/s[/tex]
at t=1.001, the height is y=50(1.001)-16(1.0012)=34.017984ft
between t=1s and t=1.001s, the change in time is 0.001s and the change in height is 0.017984ft
[tex]V_{avg} =\frac{0.017984ft}{0.001s}=17.984ft/s[/tex]
It appears that as the time interval approaches 0, the average velocity approaches 18ft/s. Therefore it is safe to assume the instantaneous velocity at t=1s is 18ft/s.
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An electric field of magnitude 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350m wide and 0.700m long (c) if the plane contains the y axis and its normal makes an angle of 40.0° with the x axis.
The value of electric flux for mentioned plane will be 657N⋅m²/C at 40° with the x axis.
For a uniform electric field passing through a plane surface,
electric flux(ϕ)=EAcosθ
where,
θ is the angle between the electric field and the normal to the surface,
E is the electric field
A is surface area
(a) The electric field is perpendicular to the surface so θ=0
=(3,50×10³ N/C)[(0.350m)(0.700m)]cos0
=858N⋅m²/C
(b)
The electric field is parallel to the surface θ=90
, so cosθ=0
therefore,
the flux is zero.
(c) For the mentioned plane,
electric flux(ϕ)=(3.50×10³N/C)[(0.350m)(0.700m)]cos40.0
=657N⋅m²/C
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