The electrostatic constant is also known as Coulomb's constant or the electric constant. Its unit is [tex]\rm \frac{Nm^2}{C^2}[/tex].
It is a fundamental physical constant that appears in Coulomb's law, which describes the interaction between electric charges. It's denoted by the symbol K.
Given information:
Unit of force (F) = Newton
Unit of charge (Q) = Coloumb
Unit of distance (r) = Meters
Now,
On substituting the values of given units:
[tex]\rm F = \frac{Kq_1q_2}{r^2}\\N =\frac{K C\times C}{m^2}\\N = \frac{K C^2}{m^2}\\K = \frac{Nm^2}{C^2}[/tex]
The unit of electrostatic constant (K) is [tex]\rm \frac{Nm^2}{C^2}[/tex]. This constant (K) plays a crucial role in understanding and calculating the forces and interactions between charged particles in the realm of electrostatics.
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Final answer:
The unit for the electrostatic constant k is Newton meters squared per Coulomb squared (N·m²/C²), and k is also known as Coulomb's constant with a value of approximately 8.99 × 10⁹ N·m²/C².
Explanation:
The unit for the electrostatic constant k in the formula F = k[tex]q_{1}[/tex][tex]q_{2}[/tex] /r² can be determined by rearranging the formula to solve for k, which gives us k = Fr² / [tex]q_{1}[/tex][tex]q_{2}[/tex] . Since the unit for the electrostatic force F is Newtons (N), the distance r is in meters (m), and the charge [tex]q_{1}[/tex]and [tex]q_{2}[/tex] are in Coulombs (C), the units for k would be Newton meters squared per Coulomb squared (N·m²/C²). The value of the electrostatic constant k is approximately 8.99 × 10⁹ N·m²/C², also known as Coulomb's constant.
Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its components. (d) The net displacement vector and its components.d1=6.00 mθ1=40°d2=8.00 mθ2=30°d3=5.00 mθ3=0°,where the last segment is parallel to the superimposed x axis. When we reach point c, what are the magnitude and angle of our net displacement →dnet from point i?
Answer:
18.3 m , 25.4°
Explanation:
d1 = 6 m, θ1 = 40°
d2 = 8 m, θ2 = 30°
d3 = 5 m, θ3 = 0°
Write the displacements in the vector form
[tex]d_{1}=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}[/tex]
[tex]d_{2}=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}[/tex]
[tex]d_{3}=5\widehat{i}[/tex]
The total displacement is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}[/tex]
[tex]d=16.53\widehat{i}+7.86\widehat{j}[/tex]
magnitude of resultant displacement is given by
[tex]d ={\sqrt{16.53^{2}+7.86^{2}}}=18.3 m[/tex]
d = 18.3 m
Let θ be the angle of resultant displacement with + x axis
[tex]tan\theta =\frac{7.86}{16.53}=0.4755[/tex]
θ = 25.4°
The field inside a charged parallel-plate capacitor is __________. zero directed from the negative to the positive plate parallel to the plates uniform
Answer:
Uniform
Directed from the negative to the positive plate
Explanation:
The field inside a charged parallel-plate capacitor is __________.
• zero
• directed from the negative to the positive plate
• parallel to the plates
• uniform
A parallel plate capacitor is made of two opposite charged plates.
As a universal rule the electric field is directed from the negative charge to the positive charge. So in the capacitor the field is directed from the negative plate to the positive plate
Also the electric field inside the capacitor is constant irrespective of the location.
[tex]E = \frac{\sigma }{ \epsilon_{0} \epsilon_{r} } \\[/tex]
Where
[tex]\sigma[/tex] is the surface charge density
[tex]\epsilon_{0}[/tex] is the permittivity of free space
[tex]\epsilon_{r}[/tex] is the relative permittivity of the material inside the plates
The field inside a charged parallel-plate capacitor is uniform, directed from the negative to the positive plate, and parallel to the plates, with magnitude E = σ / ε₀.
The field inside a charged parallel-plate capacitor is uniform, directed from the negative to the positive plate, and parallel to the plates. At points far from the edges of the plates, the electric field is perpendicular to the plates and is constant in both magnitude and direction, as described in Essential Knowledge 2.C.5. This uniform field can be used, for example, to produce uniform acceleration of charges between the plates, such as in the electron gun of a TV tube.
Note that the electric field outside the region between the two plates is essentially zero because the fields from the positive and negative plates point in opposite directions and cancel each other out, except near the edges of the plates. The magnitude of the electrical field in the space between the parallel plates can be expressed as E = σ / ε₀, where σ denotes the surface charge density on one plate and ε₀ is the permittivity of free space.
he distance from City A to City B is about 175175 mi by ferry. It takes 66 hr less to travel by the same ferry from City B to City C, a distance of about 7070 mi. What is the average rate of the fer
Answer:
The average speed will be 2.54 Km/h
Explanation:
Given that
A to B
Distance = 175175 m=175.175 KM
B to C
Distance = 7070 m=7.07 Km
Lets take time taken from B to C is t then A to B will be t+66
We know that
Distance = velocity x time
175.175 = V x (t+66) -------------1
7.070 = V x t -------------2
From equation 1 and 2
t=2.77 hr
Now by putting the values in equation 2
7.07 = V x 2.77
V = 2.54 Km/h
So the average speed will be 2.54 Km/h
You are in an egg drop contest which you win by breaking your egg the hardest. You drop your egg off of a 3 story building (12m above the ground). You can throw it up with speed 11m/s, or straight down with speed 11m/s. Write a way to break it.
Answer: Ok, first start writing our movement equation.
First our acceleration will be the gravity; so a= g
integrating we can obtain the velocity v= g*t + v₀
where v₀ depends on if we trhow the egg upsides or downsides, being the positive direction downsides.
after, the position will be r = [tex]\frac{gt^{2} }{2}[/tex] + v₀*t - 12m
where the -12m is because you are upside a building.
first case: straight down with speed 11m/s
so the velocity is positive
then, if we do [tex]\frac{10*t^{2} }{2}[/tex] + 11*t - 12m = 0 and with bashkara obtain the positive root for the time t=1.6 seconds
putting this on our velocity equation you will get v = 10*1.6 + 11 = 27m/s is the velocity of the egg when it hits the ground.
for the other case, our equation will have the form of:
[tex]\frac{10*t^{2} }{2}[/tex] - 11*t - 12m = 0
and the positive root of the time is: t = 3 seconds.
putting it in the velocity equation gives you v= 10*3 - 11 = 29m/s
which is bigger than the first case, so the egg hits the ground with more velocity, ence more energy.
Final answer:
Throwing the egg straight down with an initial speed of 11 m/s from a 12 m height will result in a higher impact speed, calculated using the kinematic equation for final velocity, which factors in both the initial speed and acceleration due to gravity.
Explanation:
To maximize the impact of the egg and break it the hardest in the egg drop contest, we need to consider the effect of throwing it downward with an initial speed. If the egg is thrown straight down with an initial velocity of 11 m/s from a 12 m high building, it will gain speed due to gravity in addition to the initial speed given. The final velocity just before impact can be calculated using the kinematic equation:
v2 = u2 + 2gh
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and h is the height.
Substituting the given values:
v2 = (11 m/s)2 + 2(9.8 m/s2)(12 m)
v2 = 121 m2/s2 + 235.2 m2/s2
v2 = 356.2 m2/s2
v = √356.2 m2/s2 ≈ 18.9 m/s
Therefore, throwing the egg straight down at 11 m/s will result in a higher impact speed upon hitting the ground compared to throwing it upwards or dropping it without initial velocity.
A bullet is fired straight up from a gun with a muzzle velocity of 136 m/s. Neglecting air resistance, what will be its displacement after 1.3 s? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Explanation:
Given that,
Speed with which the bullet is fired, u = 136 m/s
Time, t = 1.3 s
We need to find the displacement of the bullet after 1.3 seconds. It can be calculated using second equation of motion as :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Here, a = -g
[tex]s=ut-\dfrac{1}{2}gt^2[/tex]
[tex]s=136\times 1.3-\dfrac{1}{2}\times 9.8\times (1.3)^2[/tex]
s = 168.51 meters
So, the displacement of the bullet after 1.3 seconds is 168.51 meters. Hence, this is the required solution.
After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground. It is moving with a speed of 9 m/s when it reaches a maximum height of 7 m above the ground. What is the speed of the ball when it leaves Sarah's hand?
Answer:
13.82 m/s
Explanation:
You can see it in the pic.
Answer:
vo = 13.74 m/s
Explanation:
The goal is to find the initial velocity vo. According to the exercise, the maximum height that the ball will reach and the initial height will be respectively, yo = 1.5 m and ymax = 7 m, tmax will be equal to the time it takes for the ball to reach the maximum height. Having the equation:
vx = vo * cos∝o
clearing vo:
vo = vx/cos∝o
vy = vo * sin∝o + gtmax
0 = sin∝o * (vx/cos∝o) + gtmax
0 = vx * tan∝o + gtmax
gtmax = -vx * tan∝o
clearing tmax:
tmax = - (vx * tanao/g)
ymax = i + (vo * sin∝o) * tmax + (gtmax^2/2)
Replacing:
ymax = i - (vx^2 * tan^2∝o/2 * g)
clearing the angle a or:
∝o = arctan (((2 * g * (i - ymax)) / vx^2)^1/2)
substituting values:
∝o = arctan (((2 * (- 9.8) * (1.5 - 7)) / (9^2))) = 49.08 °
We find the initial speed with the following formula:
vo = vx/cos∝o = 9/cos49.08 ° = 13.74 m/s
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Use these two facts and the inverse square law for light to answer the questions.
Answer:
various parts have been answered
Explanation:
Inverse square for light is [tex]I_1r_1^2=I_2r_2^2[/tex]
initial distance from sun to earth is[tex]r_1=150\times10^6 [/tex]
and intensity or apparent brightness of sun is [tex]I_1=1300\ W/m^2[/tex]
a)
If distance from sun to earth is [tex]r_2=r_1/2=\frac{150\times10^6}{2}[/tex]
then apparent brightness is [tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(r_1/2)^2}=5200[/tex]
b)
If distance from sun to earth is [tex]r_2=2r_1[/tex]
then apparent brightness is [tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(2r_1)^2}=325\,W/m^2[/tex]
c)
If distance from sun to earth is [tex]r_2=7r_1[/tex]
then apparent brightness is
[tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(7r_1)^2}=26.5\,W/m^2[/tex]
Using the inverse square law, Earth's fraction of intercepted solar energy and the Sun's total power output are calculated based on the given distance and apparent solar brightness. The result shows Earth intercepts a minuscule part of the Sun's total energy output.
This question concerns the application of the inverse square law for light and the calculation of the Sun's total power output based on its apparent brightness from Earth. Given that Earth is approximately 150 million kilometers (or 1.5 × 1011 meters) away from the Sun and the apparent brightness of the Sun (also known as solar constant) at Earth is about 1300 watts/m2, we can determine several key aspects:
Fraction of the Sun’s energy striking Earth: Using Earth’s diameter of about 12,756 km (or 1.2756 × 107 meters), the cross-sectional area of Earth (which is a circular area) can be calculated as AEarth = π(1.2756 × 107/2)2 ≈ 1.28 × 1014 m2.
Total power output of the Sun: Applying the inverse square law, which states that the total power output P of the Sun is distributed over a sphere of radius equal to Earth’s orbit (1.5 × 1011 meters). The surface area of this sphere is A = 4π(1.5 × 1011)2 ≈ 2.83 × 1023 m2. The total power output (luminosity) L of the Sun can be calculated by multiplying this surface area by the apparent brightness (1300 W/m2), yielding L = (1300 W/m2) × (2.83 × 1023 m2) ≈ 3.68 × 1026 watts.
In essence, Earth intercepts a tiny fraction of the Sun’s total energy, owing to the vast difference in scales between the size of Earth and the distance to the Sun.
A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the acceleration of the truck as its trailer leaks sand at a constant rate through a hole in its bottom?
Answer:
Acceleration will increase.
Explanation:
The relation between force, mass and acceleration according to the Newton's second law of motion is given as:
F = ma
We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:
[tex]a=\frac{F}{m}[/tex]
Since, F is constant, the acceleration of the truck is inversely proportional to the mass.
There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.
Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.
So, the acceleration of the truck will increase.
As the truck loses mass due to leaking sand, while a constant driving force is maintained, the acceleration of the truck increases according to Newton's second law of motion.
Explanation:The question is about the changes in the acceleration of a truck as it loses mass due to leaking sand. From a physics perspective, the truck's acceleration will actually increase as it loses mass. This is because the force propelling the vehicle remains constant while its mass decreases. According to Newton's second law of motion, which states that force equals mass times acceleration (F=ma), if the force remains constant and mass decreases, the acceleration must increase to keep the equation balanced.
For example, with initial values, if the force propelling the truck is 30 newtons and the truck (with its sand) has a mass of 15 kilograms, the acceleration would be 2 m/s² (from F=ma, or 30N=15kg*a). Now suppose the sand leaks out and the truck’s mass drops to 10kg. If the driving force stays the same at 30N, when plugging these values back into F=ma, the new acceleration would be 3 m/s². Thus, the acceleration of the truck has increased as it has lost mass.
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A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the electric field at a point P that is directly above the 63.0 nC charge at a distance of 1.40 cm? Point P and the two charges are on the vertices of a right triangle.
Answer:
[tex]Ep_x = 288.97*10^3\frac{N}{C}[/tex]
[tex]Ep_y = 2770.6*10^3\frac{N}{C}[/tex]
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/r²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
r: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.
Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.
Known data
q₁ = 63 nC = 63×10⁻⁹ C
q₂ = -47 nC = -47×10⁻⁹ C
k = 8.99*10⁹ N×m²/C²
d₁ = 1.4cm = 1.4×10⁻² m
d₂ = 3.4cm = 3.4×10⁻² m
Calculation of r and β
[tex]r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m[/tex]
[tex]\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o[/tex]
Problem development
Ep: Total field at point P due to charges q₁ and q₂.
[tex]Ep = Ep_x i + Ep_y j[/tex]
Ep₁ₓ = 0
[tex]Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}[/tex]
[tex]Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}[/tex]
[tex]Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}[/tex]
Calculation of the electric field components at point P
[tex]Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}[/tex]
[tex]Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}[/tex]
While following the directions on a treasure map a pirate walks 37.0 m north and then turns and walks 8.5 m east what is the magnitude of his resultant displacement?
Answer: [tex]38\ m[/tex]
Explanation:
For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.
Then:
[tex]d^2=\triangle x^2+\triangle y^2[/tex]
You can observe that the square of the displacement is equal to the sum of the square of the horizontal displacement and the square of the vertical displacement.
Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:
[tex]\triangle x=37.0\ m\\\triangle y=8.5\ m[/tex]
Substituting values and solving for "d", you get:
[tex]d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m[/tex]
In a crash test, a 2,500 kg car hits a concrete barrier at 1.3 m/s2. If the car strikes the barrier with a force of 3,250 N, how much resistance force does the barrier provide? N This pair of forces is an example of Newton's law of motion.
Answer: 3250 N in the opposite direction
Explanation:
According to Newton's third law of motion:
"Each action has an equal and opposite reaction"
To understand it better:
When two bodies interact between them, appear equal forces and opposite senses in each of them. For example, if a body A exerts a force on another body B, it performs on A another force with the same magnitude and in the opposite direction.
Hence, this law states that for each force acting on a body, ther is a force of equal intensity but in the opposite direction on the body that produced it.
In this sense if the car hit the concrete barrier with a force of 3250 N, the concrete barrier will exert a force of -3250 N on the car (where the negative sign indicates the force is acting in the opposite direction).
Answer:3250 , third law
Explanation:
(1 point) A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6 ft/sec. At what rate is the tip of her shadow moving away from the light (ie. away from the top of the pole) when the girl is 45 ft away from the pole?
Answer:
[tex]\frac{dx}{dt}=7.14m/s[/tex]
Explanation:
As is showed at the figure annexed, we can solve this problem finding the relation between the girl displacement and the shadow displacement.
Relation the triangles (see figure annexed):
[tex]\frac{x}{H}=\frac{x-y}{h}\\x=\frac{H}{H-h}y[/tex]
We derive in order to find the speed of the shadow, because:
dx/dt: shadow's speed
dy/dt: girl's speed
[tex]\frac{dx}{dt} =\frac{25}{25-4}*6=7.14m/s[/tex]
At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 7.79 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?
Answer:
0.79 s
Explanation:
The initial velocity of the ball is, [tex]u=7.79 m/s[/tex].
And now we know that first equation of motion.
[tex]v=u+at[/tex]
Here, v is the final velocity, u is the initial velocity, t is time taken by the object, a is the acceleration.
Given that, final velocity is zero and ball is inb upward direction means opposite to acceleration due to gravity which means a=-g. and the value of g is [tex]9.8 m/s^{2}[/tex]
Therefore,
[tex]0=u-gt\\t=\frac{u}{g}\\ t=\frac{7.79}{9.8}\\ 0.79 s[/tex]
Therefore, it is the required minimum time that player must waiot to touch the ball.
While John is traveling along an interstate highway, he notices a 194-mile marker as he passes through town. Later John passes another mile marker, 104. (a) What is the distance between town and John's current location? (b) What is John's current position?
Answer:
a) 90 miles is the distance between town and John's current location.
b) John's current position is the 104 mile marker.
Explanation:
See in the picture.
A spring (k = 200 N/m) is suspended with its upper end supported from a ceiling. With the spring hanging in its equilibrium configuration, an object (mass = 2.0 kg) is attached to the lower end and released from rest. What is the speed of the object after it has fallen 4.0 cm?
To find the speed of the object after it has fallen 4.0 cm, we can use the principles of conservation of energy and Hooke's law. The potential energy stored in the spring when it is stretched by 4.0 cm is given by U = 1/2kx^2, where k is the spring constant and x is the displacement. The gravitational potential energy of the object when it is at a height of 4.0 cm can be calculated as mgx, where m is the mass and g is the acceleration due to gravity. Equating these two expressions for potential energy and solving for the speed of the object, we find v = sqrt((2gh - kx^2)/m), where h is the distance from the equilibrium position to the lowest point of the object. Plugging in the given values, we have v = sqrt((2 * 9.8 m/s^2 * 0.04 m - 200 N/m * 0.04 m^2) / 2 kg) = 1.6 m/s.
Explanation:To find the speed of the object after it has fallen 4.0 cm, we can use the principles of conservation of energy and Hooke's law. The potential energy stored in the spring when it is stretched by 4.0 cm is given by U = 1/2kx^2, where k is the spring constant and x is the displacement. The gravitational potential energy of the object when it is at a height of 4.0 cm can be calculated as mgx, where m is the mass and g is the acceleration due to gravity. Equating these two expressions for potential energy and solving for the speed of the object, we find v = sqrt((2gh - kx^2)/m), where h is the distance from the equilibrium position to the lowest point of the object. Plugging in the given values, we have v = sqrt((2 * 9.8 m/s^2 * 0.04 m - 200 N/m * 0.04 m^2) / 2 kg) = 1.6 m/s.
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The mass number of an atom of carbon (C) represents the total number of
A) protons in the atom.
B) neutrons and electrons in the atom.
C) protons and neutrons in the atom.
D) electrons in the atom.
Answer:
protons and neutrons.
Explanation:
protons and neutrons are the largest particles in an atom. Electrons are not included in the mass because they are too small to make a difference in the mass.
The mass number of an atom of carbon represents the total number of protons and neutrons in the atom. This is because the mass number, or atomic mass, is calculated by adding the number of protons and neutrons in an atom.
Explanation:The mass number of an atom of carbon (C) represents the total number of option C) protons and neutrons in the atom.
The mass number (also known as atomic mass) in an atom is calculated by adding the number of protons and neutrons, both of which are located in the nucleus of the atom. Electrons, on the other hand, contribute negligible mass to an atom and so are not counted towards the mass number. Accounting for only the protons would give us the atomic number (which is unique for each element), but the mass number includes both protons and neutrons.
For example, a Carbon-12 atom would have 6 protons and 6 neutrons as atomic mass is 12 (6 protons and 6 neutrons).
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A racehorse is running with a uniform speed of 69 km/hr along a straightaway. What is the time it takes for the horse to cover 400 meters?
21 seconds
2.1 minutes
0.35 hours
27.6 hours
Answer:
21 seconds
Explanation:
First, convert km/hr to m/s.
69 km/hr × (1000 m / km) × (1 hr / 3600 s) = 19.2 m/s
Distance = rate × time
400 m = 19.2 m/s × t
t = 20.9 s
Rounded to two significant figures, it takes the horse 21 seconds.
Answer:
The answer to your question is: 21 seconds
Explanation:
Data
speed = v = 69 km/h
distance = d = 400 m
time = t = ?
Formula
v = d / t
t = d / v
Process
Convert 400 m to km
1 km ----------------- 1000 m
x ------------------- 400 m
x = (400 x 1) / 1000
x = 0.4 km
Substitution
t = 0.4 / 69
t = 0.006 h
Covert time to minutes and seconds
60 min -------------------- 1 h
x ------------------- 0.006 h
x = (0.006 x 60) / 1
x = 0.36 min
Convert time to seconds
1 min ------------------ 60 s
0.36 min -------------- x
x = (0.36 x 60) / 1
x = 21.6 seconds ≈ 21 seconds
A charge of −3.77 µC is located at the origin, and a charge of −4.92 µC is located along the y axis at 2.86176 m. At what point along the y-axis is the electric field zero? The value of the Coulomb constant is 8.99 × 109 N · m2 /C 2 .
Answer:
1.336 m
Explanation:
q = - 3.77 micro C at y = 0 m
Q = - 4.92 micro C at y = 2.86176 m
Let the electric field is zero at P which is at a distance y from origin.
Electric field at P due to q is equal to the electric field at P due to Q.
[tex]E_{q}=E_{Q}[/tex]
[tex]\frac{kq}{OP^{2}}=\frac{kQ}{AP^{2}}[/tex]
[tex]\frac{3.77}{y^{2}}=\frac{4.92}{\left ( 2.86176-y \right )^{2}}[/tex]
[tex]\frac{\left ( 2.86176-y \right )}{y}}=1.142[/tex]
2.86176 - y = 1.142 y
2.142 y = 2.86176
y = 1.336 m
Thus, the electric field is zero at y = 1.336 m.
Distance of light travel the distance from moon can be found with the help of mirrors left on the moon by astronaut.A pluse of light is sent to the moon and returns to earth in 2.562s
Answer: 384,300,000 m
Explanation:
Speed [tex]S[/tex] is mathematically expressed as:
[tex]S=\frac{d}{t}[/tex]
Where:
[tex]S=c=3(10)^{8} m/s[/tex] is the speed of light
[tex]d[/tex] is the distance traveled
[tex]t=\frac{2.562 s}{2}=1.281 s[/tex] since [tex]2.562 s[/tex] is the time it takes to the pulse to go the moon and return.
Finding [tex]d[/tex]:
[tex]d=c t[/tex]
[tex]d=(3(10)^{8} m/s)(1.281 s)[/tex]
Finally:
[tex]d=384,300,000 m[/tex]
A flat disk, a solid sphere, and a hollow sphere each have the same mass m and radius r. The three objects are arranged so that an axis of rotation passes through the center of each object. The rotation axis is perpendicular to the plane of the flat disk. Which of the three objects has the largest rotational inertia?
a) The solid sphere and hollow sphere have the same rotational inertia and it is the largest.b) The hollow sphere has the largest rotational inertia.c) The solid sphere has the largest rotational inertia.d) The flat disk has the largest rotational inertia.e) The flat disk and hollow sphere have the same rotational inertia and it is the largest.
Answer:
b)
Explanation:
Rotational inertia of the sphere: [tex]\frac{2}{5}mr^2[/tex]
Rotational inertia of hollow sphere: [tex]\frac{2}{3}mr^2[/tex]
Rotational inertia of flat disk: [tex]\frac{1}{2}mr^2[/tex]
The largest value is 2/3 mr², therefore, hollow sphere's inertia is largest.
The hollow sphere has the largest rotational inertia among the flat disk, solid sphere, and itself because its mass is distributed farther from its axis of rotation.
The question asks which among a flat disk, a solid sphere, and a hollow sphere - each with the same mass and radius and an axis of rotation passing through their center - has the largest rotational inertia. To solve this, we consider the formula for the moment of inertia (rotational inertia) for each shape. A flat disk has a rotational inertia of ½ MR², the solid sphere has ¾ MR², and the moment of inertia for a hollow sphere is ¾ MR². Considering these formulas, the moment of inertia is directly influenced by how the mass is distributed relative to the axis of rotation. In objects where mass is distributed farther from the axis, like the hollow sphere, the rotational inertia increases. Therefore, the hollow sphere has the largest rotational inertia, making (b) The hollow sphere has the largest rotational inertia the correct answer.
Consider light of wavelengths 400 nm (UV), 550 nm (green, visible), and 900 nm (infrared). What is the energy associated with a 400 nm (UV) photon, a 550 nm (green, visible) photon, and a 900 nm (infrared) photon?
Answer:
Energy of UV light [tex]=4.95\times 10^{-19}j[/tex]
Energy of green light [tex]=3.6\times 10^{-19}j[/tex]
Energy of infrared light [tex]=2.2\times 10^{-19}j[/tex]
Explanation:
We have given the wavelength of UV light = 400 nm [tex]=400\times 10^{-9}m[/tex] , wavelength of green light = 550 nm and wavelength of infrared = 900 nm
Speed of light [tex]c=3\times 10^8m/sec[/tex]
Plank's constant [tex]h=6.6\times 10^{-34}[/tex]
Energy of the signal is given by [tex]E=h\nu =h\frac{c}{\lambda }[/tex]
So energy of UV light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{400\times 10^{-9}}=4.95\times 10^{-19}j[/tex]
Energy of green light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{550\times 10^{-9}}=3.6\times 10^{-19}j[/tex]
Energy of infrared light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{900\times 10^{-9}}=2.2\times 10^{-19}j[/tex]
What is the mass of a dog that weighs 382 N?(unit=kg)
Answer:
38.77 kg
Explanation:
The weight of an object is given by:
Weight = mg, g = acceleration due to gravity ( 9.8 m/s square)
and m is the mass.
So, in this case:
Weight of dog is 382 N, so, by putting the values of weight and g into the equation we will get the mass of dog.
382 = mass multiply with 9.8
so, the mass of the dog will be 38.77 kg
A tennis ball is dropped from 1.43 m above the
ground. It rebounds to a height of 0.945 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2. (Let
down be negative.)
Answer in units of m/s
(This I already know -5.29 but I just don't know how to continue and do the rest of the parts)
Part 2
With what velocity does it leave the ground?
Answer in units of m/s.
Part 3
If the tennis ball were in contact with the
ground for 0.00638 s, find the acceleration
given to the tennis ball by the ground.
Answer in units of m/s
2.
Answer:
Part 2: 4.30 m/s
Part 3: 1503 m/s^2
Explanation:
First you already had -5.29 m/s
Part 2
With what velocity does it leave the ground ?
it is the same process as question 1, but in this case the height is 0.945 m
[tex]v =\sqrt{2gh}[/tex]
[tex]v =\sqrt{2*9.8*0.945} \\v =4.30\ m/s[/tex]
Part 3
This is a simple formula of :
[tex]v_{f}-v_{i}=at \\where \\v_{f}: final\ speed \\v_{i}: initial\ speed \\a: acceleration\\t: time[/tex]
Final speed: is the speed the ball leaves the ground = 4.30 m/s
Initial speed: is the speed the ball hits the ground = -5.29 m/s
4.30 - (-5.29) = 0.00638a
a = 1503 m/s^2
A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum consists of a stationary 2.50- kg block of wood suspended by a wire of negligible mass. A 0.0100- kg bullet is fired into the block, and the block (with the bullet in it) swings to a maximum height of 0.650 m above the initial position (see class note for drawing). Find the speed with which the bullet is fired, assuming that air resistance is negligible.
Answer:
[tex]v_1 = 896.35 m/s[/tex]
Explanation:
As we know that bullet + pendulum system will move to the height of 0.650 m above the initial position
so here we can use energy conservation to find the speed just after the bullet hit the block
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.81)(0.650)}[/tex]
[tex]v = 3.57 m/s[/tex]
Now we can use momentum conservation to find the initial speed of the bullet
[tex]m_1v_1 = (m_1 + m_2)v[/tex]
[tex]0.0100 v_1 = (2.50 + 0.01)(3.57)[/tex]
[tex]v_1 = 896.35 m/s[/tex]
A pellet gun is fired straight downward from the edge of a cliff that is 12.7 m above the ground. The pellet strikes the ground with a speed of 27.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
Answer:
h₂ = 27.22m : height the pellet reaches
Explanation:
Pellet kinetics
Pellet moves with uniformly accelerated movement in
freefall
v f₁²=v₀²+2g*h₁ (formula 1) : ball movement down
v f₂²=v₀²-2g*h₂ (formula (formula 2) : ball movement up
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
g: acceleration due to gravity in m/s²
h :height in m
Known data
g=9.8 m/s²
h₁=12.7m
v f₁= 27.9 m/s
Initial velocity calculation (v₀)
We replace the data in formula 1:
27.9²=v₀²+2*9.8* 12.7
27.9²-2*9.8* 12.7=v₀²
529.49=v₀²
[tex]v_{o} =\sqrt{529.49}[/tex]
[tex]v_{o} =23.1\frac{m}{s}[/tex]
Calculation of the height that the pellet reaches when it goes up
Data:
[tex]v_{o} =23.1\frac{m}{s}[/tex]
v f₂=0 :When the pellet reaches the maximum height the final speed is zero
We replace the data in formula (2)
0²=23.1²-2*9.8*h₂
19.6h₂ =23.1²
h₂ =23.1²÷19.6
h₂ =27.22m
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.80 ✕ 105 m/s2 for 9.60 ✕ 10−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Answer:
Final velocity will be 556.8 m/sec
Explanation:
We have given the average rate of change of velocity that is [tex]a=5.8\times 10^5m/sec[/tex]
Time [tex]t=9.6\times 10^{-4}sec[/tex]
Initial velocity of bullet will be zero that is u=0
Now according to first law of motion
v=u+at, here v is final velocity, u is initial velocity and t is time
So [tex]v=0+5.8\times 10^5\times 9.6\times 10^{-4}=556.8m/sec[/tex]
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole
Answer:
electric field Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]
Explanation:
The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.
a) Let's write the electric field for each charge and the total field
E = k q /r
With k the Coulomb constant, q the charge and r the distance of the charge to the test point
Et = E1 + E2 + E3
E1 = k q / (x-a)²
E2 = k (-2q) / x²
E3 = k q / (x + a)²
Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]
The direction of the field is along the x axis
b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ where x << 1, for this we take factor like x from all the equations
Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]
We use binomial expansion
(1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1
(1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...
They replace in the total field and leaving only the first terms
Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]
Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]
Et = k q 2a²/x⁴
point charge
Et = k q 1/x²
Dipole
E = k q a/x³
The ________ proposes that the bodies of our solar system formed at essentially the same time from a rotating cloud of gases and dust. a. Plate Tectonics theory b. Nebular hypothesis c. Big Band theory d. Heliocentric theory
Answer:
B. Nebular Hypothesis
Explanation:
It is important to read the statement carefully so we can rule out options. The text is referring to solar system formation, it means we should discard all options which are not related to that. Big bang theory is a concept to explain the origin of the whole universe, not the solar system only, so it is discarded. Plate tectonics describes inner earth layers that are essential parts of the planet movemements like earthquakes and they are not related to solar system either. Discarded. Helioentric theory is the idea that defines the sun as the center of the solar system, but again it does not explain anything about its origin, so let me explain a litte in detail about the right choice: Nebular Hypothesis.
Nebular Hypothesis
This is the way astronomers accept the solar system was created as well as other other plantetary systems. Million of years after Big Bang, there were gases and small dust particles all over the universe and they started to join and concentrate into bigger and bigger amount of gases turning into enourmous clouds made of all the necessary elements to begin the formation of stars, planets and eventually the solar system.
You are on a train traveling east at speed of 18 m/s with respect to the ground. 1) If you walk east toward the front of the train, with a speed of 1.2 m/s with respect to the train, what is your velocity with respect to the ground?
Answer:
19.2 m/s
Explanation:
The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added
18 + 1.2 = 19.2
If you were walking backwards (west) your velocity with respect to the ground would be
18 - 1.2 = 16.8
Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air pressure and temperature at the nose of the shuttle are 1.2 atm and 300 K, respectively, what are the density and specific volume?
The density of the shuttle is 0.0491 mol/L, and the specific volume is 20.4 L/mol.
Explanation:To find the density, we can use the ideal gas law, which states that the density of an ideal gas is equal to its pressure divided by its temperature and the ideal gas constant. The formula for density (ρ) is: ρ = P / (R * T), where P is the pressure, R is the ideal gas constant, and T is the temperature in Kelvin. In this case, the pressure is 1.2 atm and the temperature is 300 K. Using the ideal gas constant for air (R = 0.0821 L * atm / (mol * K)), we can calculate the density:
ρ = 1.2 atm / (0.0821 L * atm / (mol * K) * 300 K) = 0.0491 mol / L
To find the specific volume, we can use the reciprocal of density. The specific volume (v) is equal to 1 / density. So, the specific volume for the given conditions is:
v = 1 / 0.0491 mol / L = 20.4 L / mol
Learn more about Density and Specific Volume here:https://brainly.com/question/13434300
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