The vector quantity with a magnitude equal to the product of the two equal charges with opposing polarity separated by a distance, d, is defined as the electric dipole moment associated with it.
The distance between the charges and a direction running along the line between the charges from the negative to the positive charge.
Two charges totaling 10.0 C and spaced 2.00 cm apart make up the dipole. The positive charge is further away from the line of charge than the negative charge, and the dipole's axis forms a 35.0° angle with the x-axis.
The dipole will be subject to a force that is Charges 2.0108C separated by a distance of 2.0103m make up an electric dipole. As seen in the picture, it is positioned so that the negative charge is 2.0 cm away from a long line charge with a linear charge density of 4.0104Cm1. The pressure on the Axis.
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(Figure 1) shows a thin liquid film bounded on the right side by a sliding wire that is attached to a spring with spring constant 0.50 N/m. The spring is stretched by 1.4 cm. Figure 1 of 1 < > 0.50 N/m 6.0 cm Mai W Part A What is the liquid's surface tension in mN/m? Express your answer in millinewtons per meter. || ΑΣφ ? 7= mN/m
The liquid's surface tension in mN/m is 58.33.
Surface tension is the tendency of the surface of a liquid at rest to contract into the smallest possible surface area. Surface tension allows objects denser than water, such as razor blades and insects, to float on the surface without being partially submerged.
Given:
Spring constant of the spring = k = 0.5 N/m
Distance by which the string is stretched = d = 1.4 cm = 0.014 m
Force of the spring on the wire = Fs
Fs = kd
Surface tension of the liquid = γ
Length of the sliding wire = L = 6 cm = 0.06 m
Surface tension force of the liquid on the wire = Ft
A liquid film has 2 surfaces therefore,
Ft = 2γL
The force of the spring on the wire and the surface tension force on the wire will be equal.
Fs = Ft
kd = 2γL
(0.5)(0.014) = 2γ(0.06)
0.007 = 0.12γ
γ = 0.05833 N/m
Converting the surface tension from Newton per meter to milli-Newton per meter, (1 N = 103 mN)
γ = (0.05833) x (103) mN/m
γ = 58.33 mN/m
Surface tension of the liquid = 58.33 mN/m
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when there is a net force exerted on an object, the magnitude of the acceleration of the object is [a] proportional to the magnitude of the net force
When there is a net force exerted on an object, the magnitude of the acceleration of the object is proportional to the magnitude of the net force and has a magnitude that is inversely proportional to the mass.--The statement is true.
Newton's second law of motion states that the acceleration of an object depends upon two variables i.e the net force which is acting upon the object and the mass of the object. The acceleration of an object is dependent directly upon the net force which is acting upon the object, and inversely upon the mass of the object. If the force acting upon an object is increased, the acceleration of the object is increased. If the mass of an object is increased, the acceleration of the object is decreased.
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The complete question is:
State whether a given statement is True or False.
When a net external force acts on an object of mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass.
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I have got the answers to these question but I need to find the working out pls if you are able to answer at least one please do
calm :), thanks mods
A uniform, rigid, thin board that is 2.4m long and weighs 200N is attached to a post by a pivot at point P and hangs over the edge of a building. A crate weighing 400N is attached to the right end of the board. The left end of the board, .8m from the pivot, is attached to a light vertical wire anchored at the other end to keep the board from rotating.
1. Draw and label vectors on a horizontal board to represent forces acting on the board (not components). Show each force vector originating at its point of application.
2. Calculate the magnitudes of the forces exerted on the board by the post and by the wire. If you need to draw anything other than what you have shown in part 1 to assist in your solution, use the space below. Do not add anything to the figure in part 1
3. The rotational inertia of the board alone about its center is (1/12)Ml2 , where M is the mass of the board and L is its length. Calculate the rotational inertia of the combined board-crate system about point P
4. Suppose that the wire breaks and the board begins to pivot about point P. Calculate each of the following
(a) The magnitude of the initial angular acceleration of the board-crate system.
(b) The magnitude of the initial linear acceleration of the left end of the board.
1. Diagram (Attached above)
2. By equilibrium of forces of vertical direction,
T + Wa +Wc = Rp
T + 200 + 400 = Rp
Rp - P = 600N ------------------(i)
Taking moment of forces about P (Clockwise positive),
Wb * 0.4m + Wc * (0.4 + 1.2m) – T * 0.8m = 0
200 * 0.4 + 400 * 1.6 – 0.8T = 0
0.8T = 80 + 640
T = 900N
Putting T in equation (i),
Rp - 900= 600
Rp = 1500N
So, Rp= 1500N, T = 900N, Wb = 200N, Wc = 400N
3. Moment of inertia about C.O.M. ‘O’ is Icom= 1/12 ML2
Moment of inertia about ‘P’ by parallel axis theorem
Ip = Icom + M * (OP)2
Ip= 1/12 ML2+ M * (0.4)2
For net MOI about P, mass of crate Mc at distance PC from P will also be taken in account.
Mc = 400 / 9.8 = 40.8Kg, PC = 1.6m
Mass of board M = 200/9.8 = 20.4 Kg
Length of board L = 2.4m
So, Net M.O.I. about P,
Ip net = Icom + M*(0.4)2 + Mc * (PC)2
Ip net= 1/12 * 20.4*(2.4)2 + 20.4*(0.4)2+ 40.8*(1.6)2
Ip net = 117.5 Kg m2
4.
a. As, the moment of inertia about point P, Ip net =117.5 Kg m2
Torque experienced when wire is cut
τ = Wb * 0.4 + Wb * (0.4 + 1.2)
τ = 200 * 0.4 + 400 *1.6
τ = 720 N-m
As τ = I α, where α is angular acceleration.
720 = 117.5 * α
α = 6.127 rad / s2
So, magnitude of initial angular acceleration is α = 6.127 rad / s2
b. Initial Acceleration (linear) of left most point A
aA = α * AP
aA= 6.127 * 0.8
aA= 4.9 m/s2
So, linear acceleration of point A is 4.9 m/s2
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What is the average speed of the dog if it’s goes 40 meter in 8 second
If 1 cm on the globe represents 555 km in the real world, how many kilometers would you have traveled in 16.8 cm? (round)
9324 km
On travelling 16.8 cm, the distance travelled in kilometres is 9324 kilometres.
As per the known fact, the ratio of map distance will be equal. So, we will write the two ratio and equate it. Let us assume the distance travelled in 16.8 cm be x. Representing the information of equation form -
1 : 555 = 16.8 : x
x = 16.8 × 555
Performing multiplication on Right Hand Side of the equation to find the value of x (Denominator of 1 can be ignored)
x = 9324 kilometres
Thus, the distance travelled in 9324 kilometres.
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A rectangular swimming pool has dimensions of 15 m x 30 m and is 3.5 m deep. When the pool is filled with fresh water, what is the force on the side of the pool due to the water?
The total force on the side of the pool due to the water is 2275 N/m^2 x (15 m x 3.5 m) = 1523125 N.
To find the force on the side of the pool due to the water, we first need to calculate the volume of the water in the pool. Since the pool has dimensions of 15 m x 30 m and is 3.5 m deep, its volume is 15 m x 30 m x 3.5 m = 1575 cubic meters. Next, we need to calculate the weight of the water in the pool. The weight of an object is equal to its mass multiplied by the acceleration due to gravity. The density of fresh water is about 1000 kg/m^3, so the mass of the water in the pool is 1575 m^3 x 1000 kg/m^3 = 1575000 kg. Since the acceleration due to gravity is approximately 9.8 m/s^2, the weight of the water in the pool is 1575000 kg x 9.8 m/s^2 = 15442500 N. Finally, we can calculate the force on the side of the pool due to the water. Since the force is distributed evenly over the surface of the side of the pool, the force on each square meter of the side of the pool is 15442500 N / (15 m x 3.5 m) = 2275 N/m^2. Therefore, the total force on the side of the pool due to the water is 2275 N/m^2 x (15 m x 3.5 m) = 1523125 N.
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which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
The majority of galaxies in the universe are elliptical galaxies, and galaxy collisions are frequent.
Mergers produce large elliptical galaxies. There is proof that galaxies develop through mergers and collisions. Star formation is triggered by galaxic collisions. All of the gas is transformed into stars before a disc can develop because the higher gas density produces stars more quickly. High redshift elliptical galaxies lack young, blue stars. Old red stars with erratic orbits in several planes are abundant in elliptical galaxies, which have a small amount of gas and dust. The largest galaxies that we are currently aware of are enormous elliptical galaxies. The majority of galaxies found now are small elliptical galaxies. Elliptical galaxies often have very little cold gas.
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complete question: Which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
A) Elliptical galaxies dominate the population in dense galaxy clusters.
B) Some ellipticals have stars and gas that rotate opposite to the rest of the galaxy.
C) Some elliptical galaxies are surrounded by shells of stars.
D) Computer simulations predict that the product of a galaxy collision is generally an elliptical galaxy.
E) Galaxy collisions are common and most galaxies in the universe are elliptical.
a toroid has a 50.8 cm inner diameter and a 53.7 cm outer diameter. it carries a 24.7 a current in its 604 coils.(figure 1) (figure 2). Determine the range of values for B inside the toroid. There should be two values, B(min) and B(max), in mT.
The range of values for B inside the toroid is: B(min) = 1.37 mT and B(max) = 3.37 mT
The range of values for B inside the toroid can be determined using the formula B = μ0nI/l, where μ0 is the permeability of free space, n is the number of turns, I is the current, and l is the average length of the toroid. The average length of the toroid is equal to the average diameter, which is (50.8 cm + 53.7 cm)/2 = 52.25 cm. Therefore, B = μ0nI/l = (4π×10^-7 H/m)(604 turns)(24.7 A)/(52.25 cm) = 0.0022 H/m. Since 1 mT = 10^-3 T, the range of values for B inside the toroid is 1.37 mT (B(min)) and 3.37 mT (B(max)).
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Mary weighs 525 N and she walks down a flight of stairs to a level 6.5 m below her starting point. What is the change in Mary’s potential energy? Answer in units of J.
The change in Mary’s potential energy is 3412.5 Joule.
What is potential energy?
Potential energy is a form of stored energy that is dependent on the relationship between different system components. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy. It is capable of performing more work when raised.
Given that: Mary weighs 525 N and she walks down a flight of stairs to a level 6.5 m below her starting point.
Hence, change in potential energy is = weight × change in height
= 525 N × 6.5 m
= 3412.5 Joule.
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Why do we think that Venus has so much more atmospheric gas than Earth?
Most of the gases that have been released from volcanoes on Earth later returned to the surface.
Venus has gained much more gas through outgassing than has Earth.
Earth has lost much more gas to thermal escape than has Venus.
Because of its lack of magnetic field, Venus has been able to gain gas through the impacts of solar wind particles, while Earth has not gained gas in this way.
We think Venus has more atmospheric gas than Earth because of its lack of magnetic field, Venus has been able to gain gas through the impacts of solar wind particles, while Earth has not gained gas in this way.
Venus's densest atmosphere of the four terrestrial planets consists of 96% carbon dioxide. Venus's surface atmospheric pressure is 92 times greater than Earth's. With an average surface temperature of 735 K (462 °C; 863 °F), Venus is the hottest planet in the Solar System. The planet has no carbon cycle that captures carbon in rocks and surface features, and no organic life that can sequester carbon in the form of biomass.
Venus may have had oceans, but those oceans are evaporating due to increasing temperatures caused by the continuous greenhouse effect. Much of the water may have photodissociated, and the solar wind has sent free hydrogen rage into space as a result of the lack of an internal magnetic field on Venus. The surface of Venus itself is deserted, dry, and punctuated with rock that is periodically renewed by volcanic activity.
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10 rooster head questions
Answer:
How does a rooster's head differ from that of a hen?What is the function of a rooster's crest on its head?Can a rooster's head and neck change color during mating season?How does a rooster use its head and beak during courtship displays?How does a rooster use its head to establish dominance in a flock?Can a rooster's head and beak be used as a weapon against predators?How does a rooster's head and beak help it to find food?How does a rooster's head and beak help it to eat and digest food?Can a rooster's head and beak be used for vocalizations, such as crowing?What are some common health issues that can affect a rooster's head and beak?Here are ten questions about the story entitled "cabeza de gallo":
Who are the main characters in the play?What is the central theme of the story?What kind of narrator tells the story?What was the rooster game about?Where is the place where most of the story takes place?Were the protagonist's predictions correct?What holiday was being celebrated?Which were knocked down as creatures struck in the abdomen?Who is the author of the work?What lesson or message does the story convey?¿What is a tale?A story is a narration, either oral or written, about a story that may have happened in real life, or is the product of the author's imagination.
The stories have the characteristic of having few characters, and are shorter than any common text.
¡Hope this helped!
Two pipes of identical diameter and material are connected in parallel. The length of pipe A is three times the length of pipe B. Assuming the flow is fully turbulent in both pipes and thus the friction factor is independent of the Reynolds number and disregarding minor losses, determine the ratio of the flow rates in the two pipes.
According to the question the ratio of the flow rates in the two pipes is [tex]\sqrt{3}[/tex] /1
What in physics is flow?Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As soon as unbalanced pressures are applied, this motion will persist.
Briefing:Lets take
Length of pipe B = L
Length of pipe A = 3 L
Discharge in pipe A = Q₁
Discharge in pipe B = Q₂
[tex]h_{f}[/tex] = (FLQ²)/12.1.[tex]d^{5}[/tex]
F=Friction factor, Q=Discharge,L=length
d=Diameter of pipe
here all only Q and L is varying and all other quantity is constant
So we can say that
LQ²= Constant
L₁Q₁²=L₂Q²₂
By putting the values
3LQ₁²=LQ²₂
Therefore
[tex]Q_{1}[/tex]/ [tex]Q_{2}[/tex] =[tex]\sqrt{3}[/tex]/ 1
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a 345 pf capacitor is charged to 145 v and then quickly connected to a 165 mh inductor. frequency of oscillation
A 165 mh inductor was attached right after that. The frequency of the oscillation is f=7957.74 Hz.
What is motion-induced oscillation?Periodic or oscillatory motion is defined as a motion that repeats itself. A restoring force or torque causes the object in this motion to oscillate about its equilibrium position.
For Class 8, what is an oscillation?Oscillation is a revolving motion between two states or locations. The side-to-side swing of a pendulum or the up-and-down motion of a spring with a weight are two examples of periodic motions that repeat themselves in a regular cycle and are considered oscillations.
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In shape and component stars, elliptical galaxies are most like our own
A. galactic center
B. spiral arms
C. nuclear bulge
D. halo
E. companions, the Magellanic Clouds
d
The correct option is D : Halo , In comparison to spiral galaxies, elliptical galaxies seem smooth, ellipsoidal, or spherical and have far less structure. Galaxy clusters are generally home to elliptical galaxies.
There are stars, planets, interstellar dust, and gases in elliptical, spiral, and irregular galaxies alike. The same gravitational forces that create other galaxies also create these three different sorts of galaxies.
Old stars, a tiny amount of gas, and dust can be found in elliptical galaxies. In contrast to open clusters, stars in these galaxies coalesce into globular clusters. Individual stars, nebula, and clusters are visible in some irregular galaxies but are invisible in other irregular galaxies.
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Student 1 brings a negatively charged rod with an insulating handle near to but not touching an uncharged metal sphere and finds that the objects attract. She wants the sphere to be repelled by the rod. Student 2, who is uncharged, suggests that this can be accomplished if he touches the sphere. Which of the following indicates whether this will work, and if so, gives a procedure that will cause the repulsion? (A) Yes, Student 2 could touch the sphere while the rod is held near it, because that would make the sphere negatively charged. (B) Yes, Student 2 could touch the sphere while the rod is kept far away, because that would make the sphere negatively charged. (C) Yes, Student 2 could touch the sphere while the rod is kept far away, because that would make the sphere positively charged. (D) No, there is no way that having Student 2 touch the sphere will work.
No, there is no way that having Student 2 touch the sphere will work. There will be no repulsion by touching the sphere whether the rod is kept near or away. No way cause the sphere negatively charged.
Charging an uncharged body by bringing near to it a charged body is called charging by induction. In the given case negatively charged rod is brought near to uncharged sphere, then all positive charges on it realigns towards the end of the rod by attraction.
When it is touched while keeping the rod closer, all negative charges are grounded passing through body. Thus sphere has now only positively charges. When rod is removed, the positive charges distributes uniformly ands sphere becomes positively charged.
In no way sphere will be negatively charged for repulsion to occur.
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The pressure of water vapor over ice is 3.88 mm Hg at – 2C and 4.58 mm Hg at 0 C. Estimate in J mol^(-1) the heat of vaporization of ice at -1 C.
The heat of vaporization of ice at -1°C is equal to 1.02 ×10⁵J/mol
What is the heat of vaporization?The Heat of Vaporization can be defined as the amount of heat that requires to be absorbed to vaporize a certain amount of liquid at a given temperature.
We know that the Clausius Clapeyron equation can be written as:
[tex]\displaystyle lln\frac{P_2}{P_1} =\frac{-\triangle H}{R}(\frac{T_1-T_2}{T_1T_2} )[/tex]
Given the initial pressure of water, P₁ = 3.88 mm Hg
The final pressure of the water, P₂ = 4.58 mm Hg
The initial temperature, T₁= - 1°C = 272 K
The final pressure of water, T₂ = 0°C = 273 K
[tex]\displaystyle ln\frac{4.58}{3.88} =\frac{-\triangle H}{8.314}(\frac{272-273}{272\times 273} )[/tex]
ΔH = 1.02 × 10⁵J/mol
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which of the following describes your ability to identify the sound of a musical instrument you have heard before
psychology
Answer: Recognition.
Explanation:
I'm pretty sure this is right !
An astronaut's pack weighs 19.0 N when she is on earth but only 3.84 N when she is at the surface of moon. Part A What is the acceleration due to gravity on this moon? Express your answer with the appropriate units.
The weight of an astronaut's pack is 19.0 N on earth and 3.84 N on moon. The acceleration due to gravity on the moon is 1.98 m/s²
According to the Newton's second law of motion, the relation between force acted on an object with its acceleration is given by:
F = ma
Where:
F = force acted on the object
m = mass
a = acceleration
Weight is another name for the force due to gravity. If an object is brought from the earth to the moon, its weight will be different because the gravitational accelerations are different. However, its mass remains the same.
m = F / a
In case a = g = acceleration due to gravity:
m = w/g
Let
g = acceleration due to gravity on the earth = 9.8 m/s²
g₂ = acceleration due to gravity on the moon.
Hence,
19/g = 3.84/ g₂
g₂ = 3.84/19 x 9.8 = 1.98 m/s²
Hence, the acceleration due to gravity on the moon is 1.98 m/s²
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the gravitational constant g was first measured accurately by henry cavendish in 1798. he used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.19 m apart. one of the spheres had a mass of 138 kg, while the mass of the other sphere was 0.93 kg.
The answer is
3.5*10^-8:1
Explanation:
F(gravitation) = G*M1M2/r^2
G = gravitational constant = 6.67*10^-11 Nm^2/kg^2
M1 = mass of the heavier sphere,
M2 = mass of the lighter sphere
r = their distance apart
Fg = 6.67*10^-11*188*0.93/(0.19)^2
Fg = 3.23*10^-7N
The ratio of the force of gravitation between them to the weight of the lighter sphere = 3.23*10^-7: (0.93*9.81) = 3.5*10^-8. : 1 by dividing both side by (0.93*9.81)
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difference between constant and
variable work
Work done on an object by a constant force is known as Constant Work, whereas the work done on an object by variable force is known as Variable Work.
In constant work or work done by constant force, the magnitude and direction of the force are constant or they do not change. Therefore, constant work is simply calculated by force acting on the object multiplied by displacement of the object.
But in case of variable work, things are not so easy. In this scenario, the force's magnitude and direction may alter at any point while the job is being done. The majority of the work we do on a daily basis is an instance of variable force work. The same calculation involves integration and is fairly difficult.
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two wires cross at a 400 without electrical contact. find the magnitude and direction of the magnetic field at point p. point p is 10 cm from the wire intersection and equally distant from both wires.
Both wire carrying current in same direction with the same magnitude A current-carrying wire produces a magnetic field and the Bio-Savart law enables us to calculate the magnitude and direction of this magnetic field. at any point where the magnetic field due to the segment of current-carrying wire is given by equation
[tex]B=\frac{\mu_0}{2 \pi} \frac{l}{r}[/tex]
where r is the distance between the wire and the point and I is the current of the wire. At point , the distance between it and the intersection is . So, its distance from the horizontal wire is B
[tex]r_1=(4 \mathrm{~cm}) \sin 75^{\circ}=3.86 \mathrm{~cm}[/tex]
This the same distance from the second wire as we are given
[tex]r_1=r_2=3.86 \mathrm{~cm}[/tex]
if we apply the right-hand rule, we find that the magnetic field at point of each wire has direction out of the magnetic field at point is the summation of both magnetic fields
[tex]B_1=\frac{\mu_0}{2 \pi} \frac{1}{r_2}+\frac{\mu_0}{2 \pi} \frac{1}{r_2}=2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \text { (2) }[/tex]
Now, we can use the values for to get the magnetic field at point by
[tex]B_1 & =2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \\& =2\left(\frac{\left(4 \pi \times 10^{-7} T \cdot m / A\right)(5 \mathrm{~A})}{2 \pi\left(3.86 \times 10^{-2} \mathrm{~m}\right)}\right) \\& =5.2 \times 10^{-5} \mathrm{~T}\end{aligned}[/tex]
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into . As both wires exert the same magnetic field at point is zero
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into the page. As both wires exert the same magnetic field at point is zero.
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Consider the broadcasting circuit for an AM radio station which broadcasts at a frequency of 1400 kHz. The free electrons in such a circuit are moving back and forth in simple harmonic motion. (a) How long does it take for the free electrons in this circuit to go back and forth once? Give your answer in microseconds. μs (b) Assuming the average speed of the free electrons is 100 μm/s, what is the range of motion of the electrons as they go back and forth in the wires of the circuit. (HINT: As a free electron goes back and forth, it travels its full range every half cycle.) Give your answer in nanometers. nm (c) What is the wavelength of the electromagnetic radio waves emitted by this broadcasting circuit? Give your answer in meters. m
a) Time period is given by 0.751 b) Displacement is 0.0751 nm c) The wavelength of the electromagnetic radio waves emitted is 225.56 m
T = 1/f
T = 1/(1330*10^3 Hz)
T = 751.88 ns
Time period T = 0.751 us
b) displacement is given by:
x = velocity * Time
x = 100*10^-6 m/sec*0.751*10^-6 sec
x = 0.0751 nm
c) lambda = c/f
lambda = 3*10^8/(1330*10^3 Hz)
lambda = 225.56 m
Displacement, commonly known as length or distance and denoted by the letters d or s, is a one-dimensional variable that represents the separation between two defined points. In the International System of Units (SI), the meter is the common unit of displacement (m). Transferring unfavorable emotions from one thing or person to another is referred to as displacement and is a defensive tactic. To "take out" their rage on a family member, for instance, a person can yell at them if they are upset with their job. Replace, supersede, and supplant are some frequent alternatives to the word "displace."
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How can a dilute solution of salt in water be made more concentrated?
Answer:
Heating it
Explanation:
Helps the salt dissolve faster and is more effective when heated up :)
The answer is A. by heating it
the luminosity of star a is 26 time bigger than luminosity of star b. star a is also known to be 3 times closer than star b
The luminosity of star a is 78 times bigger than star b.
The luminosity of star a is 78 times bigger than star b because the luminosity of star a is 26 times bigger than the luminosity of star b but star a is also 3 times closer than star b, so the difference in luminosity is amplified by a factor of 3. This means the luminosity of star a is 26 times 3, or 78 times bigger than star b.
The luminosity of a star is directly related to its distance from us. The closer a star is, the brighter it will appear. This is because the amount of light we receive from a star is inversely proportional to the square of its distance from us.
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a soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. the player's foot is in contact with the ball for 1.90 * 10^-3 s, and the force of the kick is given by f(t)
We set the derivative of F with respect to time equal to zero and solve for t since J=F avg t. The outcome is t=1.5103 s. It is clear that the force, F(t), reaches its maximum at t=0.0015s, with Fmax equal to 4500N.
How do you determine the force required to kick a ball?The formula Force (F) = Mass (M) x Acceleration (A), where the units are F (Newton) = Mass (kg) x Acceleration (m/s2), is used to determine the force of a kick.
What are the forces at work while kicking a soccer ball?The force exerted by the foot on the football during a kick is referred to as the action force. On the other side, the response force is the pressure the ball places on the foot. Because the action force and reaction force are of equal magnitude, no force is stronger than the other.
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The gauge pressure in a helium gas cylinder is initially 27atm . After many balloons have been blown up, the gauge pressure has decreased to 5atm .
One atm less than the absolute pressure is present in the gauge pressure. It is the variance between the absolute pressure inside the tank and the pressure outside. Therefor, the ratio is 0.185.
It is positive for pressures above atmospheric pressure and negative for pressures below atmospheric pressure; gauge pressure is the pressure as compared to atmospheric pressure. Any fluid that is not contained has its pressure increased by the atmospheric pressure. The pressure that exists inside the Earth's atmosphere is known as atmospheric pressure, also spelled barometric pressure (after the barometer). 101,325 Pa is the definition of the standard atmosphere, which is represented by the sign atm (1,013.25 hPa).
5/27 = 0.185 atm is the pressure ratio.
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in fig. 12-30, trying to get his car out of mud, a man ties one end of a rope around the front bumper and the other end tightly around a utility pole 18 m away. he then pushes sideways on the rope at its midpoint with a force of 550 n, displacing the center of the rope 0.30 m, but the car barely moves.what is the magnitude of the force on the car from the rope? (the rope stretches somewhat.)
The force on the car from the rope is 27496N.
The sum of forces in y direction is zero.
F = Tsinθ - Tsinθ =0
F - 2T SINθ = 0
T = F/2sinθ
Tan θ = .28/18 = 0.015
θ = [tex]tan^{-1}[/tex] 0.015 = 0.8593
T = 550 / 2sin (0.8593)
T = 27500 N
Fcar = Tcosθ
Fcar = Tcos[0.8593]
Fcar = 27496N
All the planets are maintained in their orbits around the sun by the force of gravity. Many of the large-scale structures in the Universe are caused by gravity because the gravitational attraction between the original gaseous stuff in the Universe allowed it to combine and form stars, which then condensed into galaxies. Although the reach of gravity is limitless, its effects diminish with increasing distance.
The general theory of relativity, which Albert Einstein proposed in 1915, is the theory that most accurately describes gravity. According to this theory, gravity is not a force but rather the curvature of spacetime brought on by an imbalance in the distribution of mass, which causes masses to move along geodesic arcs.
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Unpolarized light of Intensity Io is incident on a pair of (ideal) polarizers, as shown below. The direction of polarization of the first polarizer is vertical. The second polarizer has unknown orientation.For each part below, determine whether it is possible for the light that reaches the observer to have the given value of intensity.If so: Determine the direction of polarization of the second polarizer. Show all work.If not: Explain by not.a. No light reaches the observerb. Light of intensity (2*Io)/3 reaches the observer.c. Light of intesity Io/2 reaches the observer.d. Light of intensity Io/4 reaches the observer.
The un polarization of light at each state was explained and second polarizer is I2=I1=Io/2 and first polarizer is I2=Io/4=Io/2 cos^(q).
What is polarization?
Polarized light waves are defined as those light waves that only travel in one plane. The polarization of light is the process by which unpolarized light waves are converted to polarized light waves.
What is un polarization ?
Unpolarized light is a form of electromagnetic radiation that vibrates in more than one plane. Unpolarized light is the kind that comes from the sun, a desk lamp, or a candle flame.
a) When the pass axis of the two polarizers are perpendicular to each other.
b) Not possible, because after first polarizer intensity becomes Io/2.
from malus law after second polarizer I2=I1 cos^2(q)
I2=2Io/3, cos^2(q)=4/3, q=cos^-1(1.15) Which does not exist.
c) When the pass axis of the second polarizer is parallel to first i.e q=0
I2=I1=Io/2.
d) I2=Io/4=Io/2 cos^(q)
cos^(q)=1/2, q=45 degree w.r.t pass axis of the first polarizer.
Therefore, the un polarization of light at each state was explained and second polarizer is I2=I1=Io/2 and first polarizer is I2=Io/4=Io/2 cos^(q).
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An object attached to one end of a string moves in a circle at constant speed. Which of the following is correct?
the object is accelerating as it moves
the object's velocity is the same as its speed
the object does not require a force to keep its state of circular motion
if the string breaks, the object will keep its circular motion
If the string breaks, the object will move Radially away from the center of the cicl
The object is accelerating as it moves in a circle at constant speed. This is because acceleration is a measure of how an object's velocity changes over time, and the velocity of an object moving in a circle is constantly changing direction.
The object's velocity is not the same as its speed. Speed is a scalar quantity that refers to the distance an object travels per unit of time, while velocity is a vector quantity that refers to the speed and direction of an object's motion. The velocity of an object moving in a circle will vary as the object moves around the circle, even if its speed remains constant.
An object moving in a circle does require a force to keep its state of circular motion. This force is known as the centripetal force, and it is directed towards the center of the circle. The centripetal force is necessary to keep the object moving in a circular path, rather than following a straight-line path.
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