Answer:
Kinetic energy
Solids 1, liquids 2, gases 3
Temperature
Solids 1, liquids 2, gases 3
Density
Solids 3, liquids 2, gases 1
Explanation:
Most of elements can be in the three different states at different temperatures, solid in lower temperatures then liquid and gas in higher temperatures.
When it's gaseous it is more expanded, it's volume is bigger then it's density is lower. As the molecules are more separated they can move more easily.
On the opposite hand in solid state, it's more compressed, the volume is smaller and the density is higher. The molecules are more compressed an they can't move easily.
The acceleration of a cart rolling down a ramp depends on __________. The acceleration of a cart rolling down a ramp depends on __________. both the angle of the ramp and the length of the ramp the length of the ramp the angle of the ramp neither the angle of the ramp or the length of the ramp
Answer:angle of ramp
Explanation:
The acceleration of a cart rolling down a ramp depends on the angle of the ramp and not on the length of the ramp.
Let the car have wheel in form of thin cylinder, therefore its acceleration is given by
[tex]a=\frac{gsin\theta }{1+\frac{I}{mr^2}}[/tex]
where
a=acceleration
g=acceleration due to gravity
I=moment of inertia of the wheel
m=mass of wheel
r=radius of wheel
[tex]\theta [/tex]=angle made by ramp with the horizontal
In above term there is no sign of length of ramp thus it is independent of it.
The acceleration of a cart rolling down a ramp is primarily influenced by the angle of the incline and gravitational acceleration, but in practical scenarios, friction and other factors also play a role. The final velocity as the cart leaves the ramp is a function of the acceleration and the distance over which it acts.
Explanation:The acceleration of a cart rolling down a ramp depends on several factors, including the angle of the incline and the presence of friction. When considering a cart on a frictionless inclined plane, the acceleration only depends on the angle of the ramp and gravitational acceleration. However, in real-world scenarios, factors such as friction, the mass of the cart, and the initial velocity may also play a role. Acceleration is directly related to the sine of the angle of the incline. Furthermore, the final velocity of the cart as it leaves the ramp would depend on both the acceleration of the cart down the ramp and the distance over which the acceleration acts. In the absence of air resistance, all objects slide down a frictionless incline with the same acceleration if the angle is the same. Also, linear and angular accelerations are directly proportional, with the radius of wheels also affecting the acceleration.
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
A cylindrical insulated wire of diameter 2.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center?
The magnitude of the magnetic field at the center of the solenoid is 2π x 10^-5 Tesla, found using the formula B = μ₀nI and assuming the length of the solenoid can be estimated from the diameter of the wire and the number of turns.
To calculate the magnetic field inside a solenoid, we use Ampere's law, which is in the form B =
μ₀nI when the core is air. The term μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Just by looking at the problem, we don't need to use the diameter of the wire, as the number of turns per unit length (n) is the critical value needed in the formula and they are given directly.
The number of turns per unit length can be found by dividing the number of turns by the length of the solenoid. However, the length of the solenoid is not provided, so we would typically divide the number of turns (200) by the solenoid's length. Assuming that the coils are tightly wound with adjacent coils touching and the diameter of the wire is 2.0 mm, the length can be estimated as the diameter times the number of turns (0.002 m * 200 = 0.4 m).
So, substituting the values into the equation B = μ₀nI, where μ₀ = 4π x 10^-7 T·m/A, n = 200 turns / 0.4 m = 500 turns/m, and I = 0.10 A, we get:
B = (4π x 10^-7 T·m/A) * (500 turns/m) * (0.10 A) = 2π x 10^-5 (T)
Thus, the magnitude of the magnetic field at the center of the solenoid is 2π x 10^-5 Tesla.
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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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]
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.
Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of the following statements accurately describes the motion of this particle?
a. The particle travels 5.0 m during each second.
b. The particle travels 5.0 m only during the first second.
c. The speed of the particle increases by 5.0 m s during each second.
d. The speed of the particle increases by 5.0 m s2 during each second.
Answer:
Correct option is (C)
Explanation:
u = 0 m/s
a = 5 m/s^2
(a) Distance traveled by the particle in first second
s = ut + 1/2 at^2
s = 0 + 0.5 x 5 x 1 x 1 = 2.5 m
So, this option is wrong.
(b) As we observe by the part (a) that the particle travels a distance of 2.5 m in first second so, this option is wrong.
(c) The acceleration of the particle is 5m/s^2, it means the speed of teh particle increases by 5 m/s every second. So, this is true.
(d) As we observe by part (c), the speed of the particle increases by 5 m/s every second so this option is wrong.
The correct statement that describes the motion of a particle accelerating at a rate of 5.0 m/s^2 from rest is that the speed of the particle increases by 5.0 m/s during each second, based on the basic kinematics equation.
Explanation:The subject of this question falls under Physics, specifically kinematics. The statement that accurately describes the motion of the particle is option c. "The speed of the particle increases by 5.0 m/s during each second." This is because acceleration is defined as the change in velocity or speed per unit time. So if the particle is accelerating at a rate of 5.0 m/s2, this means its speed increases by 5.0 m/s for every second that passes.
This can be understood by the basic kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity (which is zero in this case, as the particle is starting from rest), a is the acceleration and t is the time. So, each second, the speed increases by the acceleration, which in this case is 5 m/s.
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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.
An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain? (Note: The density of iron is 7800 kg/m3 and the density of seawater is 1025 km/m3)
Answer:
723.54 N
Explanation:
Weight of anchor = 833 N
density of iron = 7800 mg/m^3
density of sea water = 1025 kg/m^3
According to the Archimedes principle, if a body is immersed wholly or partly in a liquid it experiences a loss in weight of the body which is equal to the weight of liquid displaced by the body.
The loss in the weight of body is equal to the buoyant force acting on the body.
The formula for the buoyant force acting on the body
= volume of body x density of liquid x acceleration due to gravity
Weight of anchor = mass x acceleration due to gravity
833 = m x 9.8
m = 85 kg
mass of anchor = Volume of anchor x density of iron
85 = V x 7800
V = 0.01089 m^3
Buoyant force on the anchor = Volume of anchor x density of sea water x g
= 0.01089 x 1025 x 9.8 = 109.46 N
So, the tension in the chain = Apparent weight of the anchor
= Weight - Buoyant force
= 833 - 109.46 = 723.54 N
Thus, the tension in the chain is 723.54 N.
Final answer:
Around 13.14% of an iron anchor's weight will be supported by buoyant force when it is submerged in saltwater, calculated based on the densities of iron and seawater.
Explanation:
The question revolves around determining what fraction of an iron anchor's weight will be supported by buoyant force when it is submerged in saltwater. The buoyant force acting on any object submerged in a fluid is equal to the weight of the fluid displaced by the object. This principle is articulated by Archimedes' principle. For the iron anchor, the fraction supported by the buoyant force is directly linked to the densities of iron (7800 kg/m3) and the seawater (1025 kg/m3) in which it is submerged.
To find this fraction, we use the formula: Fraction Supported = Density of Seawater / Density of Iron.
Therefore, Fraction Supported = 1025 / 7800 ≈ 0.1314.
So, when an iron anchor is submerged in seawater, approximately 13.14% of its weight is supported by the buoyant force.
Two private airplanes are taxiing at a small airport. Jim is in plane A rolling due south with respect to the ground. Samantha is in plane B rolling due west with respect to the ground. Samantha is in front of Jim and to his left. In what direction(s), relative to himself, does Jim see Samantha’s plane moving?
Answer:
Answered
Explanation:
[tex]V_A= -v_1j[/tex] ( in the south direction)
[tex]V_B= -v_2i[/tex] (in the west direction)
[tex]V_{AB}= -V_B-V_A[/tex]
[tex]= -v_2i-(-v_1j)[/tex]
so direction of v_1 and v_2 are north and west respectively.
we used the formula for relative velocity
[tex]V_{AB}= V_B- V_A[/tex]
is the velocity of B with respect to A.
Final answer:
From Jim's perspective, Samantha's plane appears to move in a primarily westward direction but slightly towards him, resembling a southwest direction, due to the orthogonal paths of their movements.
Explanation:
Two private airplanes are taxiing at a small airport. Jim is in plane A rolling due south with respect to the ground. Samantha is in plane B rolling due west with respect to the ground. Samantha is in front of Jim and to his left. In what direction(s), relative to himself, does Jim see Samantha’s plane moving?
Since Jim is moving south and Samantha is moving west, from Jim’s perspective, Samantha’s plane would be seen moving to his front-left but more towards the left due to the orthogonal (right angle) relationship between their movement paths. Therefore, Samantha’s plane would appear to be moving in a direction that is primarily westward but slightly towards Jim, making it seem to be moving in a southwest direction relative to Jim's position.
Consider a slab of face area A and thickness L. Suppose that L = 33 cm, A = 55 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 129°C and TC = 21°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.
Answer:
721.8 Joule per second
Explanation:
L = 33 cm = 0.33 m
A = 55 cm^2 = 0.0055 m^2
Th = 129°C
Tc = 21°C
k = 401 W/mK
The rate of flow of heat is given by the formula
[tex]H =\frac{K A \left ( T_{h}-T_{c} \right )}{L}[/tex]
[tex]H =\frac {401 \times 0.0055 \times \left (129-21\right )}{0.33}[/tex]
H = 721.8 Joule per second
Thus, the rate of flow of heat is given by 721.8 Joule per second.
Final answer:
To find the conduction rate through the slab, use the formula Q/t = kA(TH - TC)/L. With the given values for copper's thermal conductivity, area, thickness, and temperature difference, the conduction rate is calculated to be 74,454 Watts or 74.454 kW.
Explanation:
To find the conduction rate through a copper slab, we can use the formula for heat transfer through conduction: Q/t = kA(TH - TC)/L. In this formula, Q/t is the rate of heat transfer (conduction rate), k represents the thermal conductivity of copper, A is the surface area of the slab, TH and TC are the temperatures of the hot and cold faces respectively, and L is the thickness of the slab.
Using the given values, L = 0.33 m (converted from 33 cm), A = 0.0055 m2 (converted from 55 cm2), k = 401 W/m·K, TH = 129°C, and TC = 21°C, the conduction rate can be calculated as follows:
Q/t = 401 W/m·K × 0.0055 m2 × (129°C - 21°C) / 0.33 m
Q/t = 401 W/m·K × 0.0055 m2 × 108 K / 0.33 m
Q/t = 74,454 W/m·K × m2 / m
Q/t = 74,454 W
Therefore, the conduction rate through the copper slab is 74,454 Watts or 74.454 kW.
Define an astronomical unit. Choose all that apply. a. 8.3 minutes Average distance from Earth to the Sun. b. Distance that light travels in one year. c. Average distance from any planet to the Sun. d. 93 miles 150 million km
Answer:
a. 8.3 minutes average distance from earth to the sun
d. 93 miles or 150 million km
Explanation:
The distance between the earth and the sun is defined as an astronomical unit (AU). It takes 8.3 minutes to go from earth to the sun at the speed of light. That distance has a length of 150 million Kilometers or 93 miles.
It is common to see in planet charts that distance to the sun are compared in astronomical units. In the case of Mars is 1.524 AU away from the sun.
A wind is blowing a small ice fishing shed along the ice to the east with a 800N force. The shed weighs 2,200N. What would the largest coefficient of friction need to be to keep the small house moving?
Answer:
0.36
Explanation:
The maximum force of friction exerted by the surface is given by:
[tex]F_f = \mu N[/tex] (1)
where
[tex]\mu[/tex] is the coefficient of friction
N is the normal reaction
The shed's weight is 2200 N. Since there is no motion along the vertical direction, the normal reaction is equal and opposite to the weight, so
N = 2200 N
The horizontal force that is pushing the shed is
F = 800 N
In order for it to keep moving, the force of friction (which acts horizontally in the opposite direction) must be not greater than this value. So the maximum force of friction must be
[tex]F_f = 800 N[/tex]
And substituting the values into eq.(1), we can find the maximum value of the coefficient of friction:
[tex]\mu = \frac{F_f}{N}=\frac{800}{2200}=0.36[/tex]
The largest coefficient of friction required to keep the ice fishing shed moving east with a force of 800N, when it weighs 2,200N, is roughly 0.3636.
To determine the largest coefficient of friction required to keep the ice fishing shed moving, you need to understand the balance between the applied force (the wind) and the resistive force due to friction. The force of friction is calculated using the formula:
Ffriction = μ * N
where μ is the coefficient of friction and N is the normal force. The normal force on the shed is equal to its weight when the surface is flat, which is given as 2,200N. The applied force is to the east with 800N.
To just keep moving indicates that the forces are balanced, so the frictional force equals the applied wind force. Therefore, to find the largest coefficient of friction (μ), we set the frictional force equal to the applied force (800N):
μ * 2,200N = 800N
μ = 800N / 2,200N
μ = 0.3636
Thus, the largest coefficient of friction required to keep the shed moving is approximately 0.3636.
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
(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]
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.
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]
A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current.
Answer:
speed of current is 5 mile/hr
Explanation:
GIVEN DATA:
speed of motorboat = 15 miles/hr relative with water
let c is speed of current
15-c is speed of boat at upstream
15+c is speed of boat at downstream
we know that
travel time=distance/speed
[tex]\frac{10}{15-c} +\frac{10}{15+c} = 1.5[/tex]
150+10c+150-10c=1.5(15-c)(15+c)
300=1.5(225-c^2)
300=337.5-1.5c^2
200=225-c^2
c^2=25
c = 5
so speed of current is 5 mile/hr
Final answer:
To find the speed of the current for a boat trip upstream and downstream taking a total of 1.5 hours at a constant speed of 15 mph relative to the water, we derive and solve an equation based on time taken for each part of the journey. The solution reveals that the current's speed is 3 mph.
Explanation:
The student asked how to find the speed of the current when a motorboat, traveling at a constant speed of 15 miles per hour relative to the water, went 10 miles upstream and then returned downstream, with the total trip taking 1.5 hours.
Step-by-Step Solution
Let c represent the speed of the current. Hence, the speed of the motorboat upstream is (15 - c) mph, and downstream is (15 + c) mph.The time to travel upstream is the distance/speed = 10 / (15 - c) hours, and the time to travel downstream is 10 / (15 + c) hours.Since the total time for both parts of the trip is 1.5 hours, we can write the equation: 10 / (15 - c) + 10 / (15 + c) = 1.5.To solve for c, multiply each term by the common denominator, which is (15 - c)(15 + c), to clear the fractions and simplify the equation to find the value of c.By solving the equation, we find that the speed of the current is 3 miles per hour.
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]
A wire with mass 43.0 g is stretched so that its ends are tied down at points a distance 81.0 cm apart. The wire vibrates in its fundamental mode with frequency 59.0 Hz and with an amplitude at the antinodes of 0.350 cm.
a) What is the speed of propagation of transverse waves in the wire?
v = ______ m/s
b) Compute the tension in the wire.
F = _____ N
c) Find the magnitude of the maximum transverse velocity of particles in the wire.
= _____ m/s
d) Find the magnitude of the maximum acceleration of particles in the wire.
= m/s2
Answer:
Explanation:
mass, m = 43 g = 0.043 kg
L = 81 cm = 0.81 m
frequency, f = 59 Hz
Amplitude, A = 0.35 cm
(a)
Frequency
[tex]f=\frac{v}{2L}[/tex]
v = 59 x 2 x 0.81 = 95.58 m/s
(b)
Let F be the tension in the string
[tex]v=\sqrt{\frac{F}{\mu }}[/tex]
where, μ is mass per unit length
[tex]\mu =\frac{0.043}{0.81}=0.053 kg/m[/tex]
[tex]95.58=\sqrt{\frac{F}{0.053 }}[/tex]
[tex]F = 95.58\times 95.58 \times 0.053[/tex]
F = 484.18 N
(c)
Maximum velocity
v = ω A = 2 π f A
v = 2 x 3.14 x 59 x 0.0035 = 1.3 m/s
(d)
Maximum acceleration
a = ω² A
a = (2 π f )² x 0.0035
a = ( 2 x 3.14 x 59)² x 0.0035
a = 480.5 m/s^2
A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police car. The driver of the SUV suspects he may be exceeding the speed limit, so just as he passes the police car he lets the SUV slow down at a constant 1.80 m/s2. How much time elapses from when the SUV passes the police car to when the police car passes the SUV?
Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. [tex]x=x_{0}+vt[/tex]
2. [tex]x=x_{0}+v_{0}t+\frac{at^{2}}{2}[/tex]
Since both cars will travel the same distance x, we can equal both formulas and solve for t:
[tex]vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}[/tex]
We simplify the fraction present and rearrange for our formula so that it equals 0:
[tex]0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0[/tex]
In the very last step we factored a common factor t. There is two possible solutions to the equation at [tex]t=0[/tex] and:
[tex]0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s[/tex]
What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at [tex]t=0 s[/tex] (when the SUV passed the police car) and [tex]t=15.56s[/tex](when the police car catches up to the SUV)
28. Magnesium bromide is a binary ionic compound. From its formula, MgBr2, how do you know that Mg is the metal? (Pages 567-568)
29. Element X has five valence electrons, element Y has one valence electron, and element Z has one valence electron. Which two of these elements are most likely to have similar properties? Explain your answer. (Page 503)
Answer:
See Below
Explanation:
28. In ionic compounds the metal is always at the left.
MgBr2
Cu(SO4)2
CaCO3
FeO
Li2O
etc
29.
Elements from same group are more likely to have similar properties.
Since Y and Z have one valence electron they are from the same group, therefore are more likely to have similar properties
Rectangle A has side lengths of 6\text{ cm}6 cm6, space, c, m and 3.5\text{ cm}3.5 cm3, point, 5, space, c, m. The side lengths of rectangle B are proportional to the side lengths of rectangle A. What could be the side lengths of rectangle B? Choose 2 answers: Choose 2 answers: (Choice A) A 3\text{ cm}3 cm3, space, c, m and 1.75\text{ cm}1.75 cm1, point, 75, space, c, m (Choice B) B 5\text{ cm}5 cm5, space, c, m and 2.5\text{ cm}2.5 cm2, point, 5, space, c, m (Choice C) C 7\text{ cm}7 cm7, space, c, m and 7\text{ cm}7 cm7, space, c, m (Choice D) D 12\text{ cm}12 cm12, space, c, m and 5\text{ cm}5 cm5, space, c, m (Choice E) E 5.25\text{ cm}5.25 cm5, point, 25, space, c, m and 9\text{ cm}9 cm9, space, c, m
Answer:
Explanation:
Gotta work on writing the problem, kida hard to read with the coding text. Anyway, a rectangle with one side of 6 cm and another at 3.5 cm, then rectangle B is proportional to that. Which basically means rectangle B will be multiplied or divided by some number.
To make the concept a little simpler imagine it was a square with sides of length 2. a proportional square twice as big would have lengths of 4 and half as big would have lengths of 1.
In this case we need measurements where one is 6 multiplied or divided by a number x, and then 3.5 has the same action done on it with the same number. so again, twice as big would be 12 and 7 while half as big would be 3 and 1.75 anyway, let's look at the options.
A) 3 and 1.75
B) 5 and 2.5
C) 7 and 7
D) 12 and 5
E) 5.25 and 9
Just to have ti written, the original rectangle is 6 by 3.5
So A is obvious since I used it as an example, this is a proportional rectangle half the size of A.
The rest are harder to check except C. 7 and 7 make a square, but we know the side lengths have to be different, so C is easy to tell it's not right.
B we have to check. Since it's not as obvious as A, we have to check with math more carefully. What do we have to multiply 6 by to get 5? Also, you always want to make the largest side of each rectangle proportional to each other. Anyway, 6 times 5/6 is 5, so for B to be right, 3.5 times 5/6 needs to be equal to 2.5. Does it? Nope, its 2.9166 repeating. So B is not right.
I'm not going to go through so much explanation for the others, if you don't get something let me know.
D we could also know quickly isn't right since I told you the measurements if the rectangle is changed to one with a side length of 12. We would need the other to be 7, so 12 and 5 isn't right.
E is our last possibility, since we need to pick two, but let's check anyway. we need to multiply 6 by 7/8 to get 5.25. 3.5 times 7/8 is 3.0625, so not 9. Good thing we checked here, because only one answer looks to be right. Are you sure you put the right numbers? Or did I misinterpret some?
Answer:
A and E
Explanation:
If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
a. towing line
b. bow line
c. starboard (right) line
d. port (left) line
Answer:
Bow Line
Explanation:
If the wind or current is pushing your boat away from the dock, bow line should be secured first.
1- We should cast off the bow and stern lines.
2-With the help of an oar or boat hook, keep the boat clear of the dock.
3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.
4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.
. Light travels at a speed of about 300 000 km/s. a. Express this value in scientific notation. b. Convert this value to meters per hour. c. What distance in centimeters does light travel in 1 s? 6
Answer:
The answer to your question is below
Explanation:
Data
light speed = 300 000 km/s
a) Express it in scientific notation
to do it, we just move the decimal point 5 places to the left
300 000 = 3.0 x 10 ⁵ km/s
b) Convert this value to meters per hour
(300 000 km/s)(1000 m/1 km)(3600 s/1 h) = 300000x1000x3600 / 1x1x1
= 1.08 x 10¹² m/h
c) What distance in centimeters does light travel in 1 s?
data
v = 300 000 km/s
d = ?
t = 1 s
formula v = d/t we clear distance d = vxt
d = 300000 x 1 = 300000 km
d = 300000000 m = 30000000000 cm
A. Expression of 300000 km/s in scientific notation.
Scientific notation is a most reliable way of writing a very large or very small number in a convenient way.
To express 300000 km/s in scientific notation, move the decimal point from the last zero number to the non zero number. Include a multiple of 10 raised to power of the number of move as shown below:
Number to express = 300000 Km/s
The Non zero number is => 3
Number of moves = 5
Scientific notation = 3.0 × 10⁵ Km/sTherefore, the scientific notation of 300000 Km/s is 3.0 × 10⁵ Km/s
B. Converting 300000 Km/s to m/h
We'll begin by converting 300000 Km/s to m/s. This can be obtained as follow :
1 Km/s = 1000 m/s
Therefore,
[tex]300000 Km/s =\frac{300000 Km/s * 1000 m/s}{1 Km/s } \\[/tex]
300000 Km/s = 300000000 m/sFinally, we shall convert 300000000 m/s to m/h
1 m/s = 3600 m/h
Thus,
[tex]300000000 m/s = \frac{300000000 m/s * 3600 m/h}{1 m/s}[/tex]
300000000 m/s = 1.08×10¹² m/hTherefore,
300000 Km/s = 1.08×10¹² m/hC. Determination of the distance (in cm) travelled in 1 second.
We'll begin by converting 300000 km/s to cm/s.
This can be obtained as follow:
1 km/s = 100000 cm/s
Therefore,
[tex]300000 km/s = \frac{300000 km/s * 100000 cm/s }{1 km/s} \\[/tex]
300000 km/s = 3×10¹⁰ cm/sFinally, we shall determine the distance (in cm). This can be obtained as follow:
Speed = 3×10¹⁰ cm/s
Time = 1 s
Distance =?[tex]Speed = \frac{Distance}{Time}\\\\3*10^{10} = \frac{Distance}{1}[/tex]
Cross multiply
Distance = 3×10¹⁰ × 1
Distance = 3×10¹⁰ cmTherefore, light travels a distance of 3×10¹⁰ cm in 1 second.
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The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m3. Convert this volume to liters and express the result in standard exponential notation.
Answer:
[tex]3,666,500\ m^3 = 3.666\times 10^9\ Liters[/tex]
Explanation:
Given that
Volume
[tex]V=3,666,500\ m^3[/tex]
As we know that
[tex]1\ m^3 = 1000\ Liters[/tex]
[tex]1\ m^3 = 10^3\ Liters[/tex]
So
[tex]3,666,500\ m^3 = 3,666,500\times 10^3\ Liters[/tex]
[tex]3,666,500\ m^3 = 3,666,5\times 10^5\ Liters[/tex]
[tex]3,666,500\ m^3 = 3.666\times 10^9\ Liters[/tex]
So we cab say that [tex]3,666,500\ m^3[/tex] is equal to [tex]3.666\times 10^9\ Liters[/tex].
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
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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In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector →C has a magnitude of 15.0 m and is angled 20.0° counterclockwise from the −x direction. What are (a) the magnitude and (b) the angle (relative to +x) of →B?
Answer:
Explanation: Ok, first caracterize the two vectors that we know.
A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m
now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x
C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m
where i and j refers to the versors associated to te x axis and the y axis respectively.
in a sum of vectors, we must decompose in components, so: ax + bx = cx and ay + by = cy. From this two equations we can obtain B.
bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m
by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m
Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.
Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°
So the total angle will be 180° + 28.8° = 208.8°.
For the magnitude of B, lets call it B', we can use the angle that we just obtained.
bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) = 26.58 m.
So the magnitude of B is 26.58 m.
Final answer:
The magnitude of vector B is 3.0 m and its angle relative to +x direction is 60.0°.
Explanation:
To find the magnitude and angle of vector →B, we can use vector addition. The magnitude of →B can be found using the equation:
|→B| = |→C| - |→A|
Substituting the given magnitudes, we have:
|→B| = 15.0 m - 12.0 m = 3.0 m
Next, we can find the angle of →B using trigonometry. Since →A is angled 40.0° counterclockwise from the +x direction, and →C is angled 20.0° counterclockwise from the −x direction, the angle of →B can be found by subtracting these angles:
θ = (20.0° - (-40.0°)) = 60.0°
Therefore, the magnitude of →B is 3.0 m and its angle relative to the +x direction is 60.0°.
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.