Explanation:
It is given that,
Length of solenoid, l = 28 cm = 0.28 m
Area of cross section, A = 0.475 cm² = 4.75 × 10⁻⁵ m²
Current, I = 85 A
(a) The inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]
[tex]L=\dfrac{4\pi\times 10^{-7}\times (645)^2\times 4.75\times 10^{-5}}{0.28}[/tex]
L = 0.0000886 H
[tex]L=8.86\times 10^{-5}\ H[/tex]
(b) Energy contained in the coil's magnetic field is given by :
[tex]U=\dfrac{1}{2}LI^2[/tex]
[tex]U=\dfrac{1}{2}\times 8.86\times 10^{-5}\ H\times (85\ A)^2[/tex]
U = 0.3200675 Joules
or
U = 0.321 Joules
Hence, this is the required solution.
A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle?
Answer:
6.6 x 10^5 Nm^2/C
Explanation:
E = 125000 N/C
Area, A = length x width = 2.5 x 5 = 12.5 m^2
θ = 65 degree
electric flux, φ = E A Cosθ
φ = 125000 x 12.5 x Cos 65
φ = 6.6 x 10^5 Nm^2/C
Here, we are required to evaluate the electric flux through the rectangle.
The electric flux through the rectangle is;
φ = 6.6 × 10^5 Nm²/C
The electric flux through a body is given as;
electric flux, φ = E A Cosθ
where;.
φ = electric fluxE = Electric field magnitudeA = Areaθ = angle between the electric field vector and the vector normalThe Area of the rectangle, A = 2.5 × 5
Therefore, A = 12.5m².
Therefore, the electric flux, φ = E A Cosθ
φ = 125000 × 12.5 × Cos65
φ = 660,341 Nm²/C.
φ = 6.6 × 10^5 Nm²/C
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A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, vector B = 1.50 k T. If the current is in the positive x direction, what is the magnetic force on the section of wire?
Answer:
The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].
Explanation:
Given that,
Current [tex]I = 2.60\hat{i}\ A[/tex]
Length = 0.750 m
Magnetic field [tex]B = 1.50\hat{k}\ T[/tex]
We need to calculate the magnetic force on the section of wire
Using formula of magnetic force
[tex]\vec{F}=l\vec{I}\times\vec{B}[/tex]
[tex]\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}[/tex]
Since, [tex]\hat{i}\times\hat{k}=-\hat{j}[/tex]
[tex]\vec{F}=-2.925\hat{j}\ N[/tex]
Hence, The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].
If a material has an index of refraction of 1.61, Determine the speed of light through this medium
Answer:
1.86 x 10^8 m/s
Explanation:
n = 1.61
The formula for the refractive index is given by
n = speed of light in vacuum / speed of light in material
n = c / v
v = c / n
v = (3 x 10^8) / 1.61
v = 1.86 x 10^8 m/s
The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.
Explanation:The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.
Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.
By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.
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The amplitude of the voltage across an inductor can be greater than the amplitude of the generator EMF in an AC RLC circuit. Consider an RLC circuit with Epeak = 90 V, R = 21.1 capital omega, L = 1.05 H, and C = 2.6 µ F. Find the amplitude of the voltage across the inductor at resonance.
Answer:
2710.66 V
Explanation:
Eo = 90 V, R = 21.1 ohm, L = 1.05 H, C = 2.6 x 10^-6 F
At resonance, the angular frequency is given by
[tex]\omega _{0}=\frac{1}{\sqrt{LC}}[/tex]
[tex]\omega _{0}=\frac{1}{\sqrt{1.05\times 2.6\times 10^{-6}}}=605.23 rad/s[/tex]
XL = ω0 x L = 605.23 x 1.05 = 635.5 ohm
Xc = 1 / ω0 C = 1 / (605.23 x 2.6 x 10^-6) = 635.5 ohm
In case of resonance, the impedance is equal to teh resistance of circuit.
Z = R = 21.1 ohm
I0 = E0 / Z = 90 / 21.1 = 4.265 A
Voltage across inductor
VL = I0 x XL
VL = 4.265 x 635.5 = 2710.66 V
To find the amplitude of the voltage across the inductor at resonance in an AC RLC circuit, we can calculate the resonant frequency using the formula fo = 1/(2*π√(LC)). Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.
Explanation:To find the amplitude of the voltage across the inductor at resonance, we need to calculate the resonant frequency using the formula fo = 1/(2*π√(LC)).
Substituting the given values, fo = 1/(2 * 3.1416 * √(1.05 * 2.6 * 10^-6)) Hz.
Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.
Substituting the values and the calculated resonant frequency, we can find the amplitude of the voltage across the inductor.
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At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 4.1 m/s, and 65-kg person feels a 455-N force pressing against his back. What is the radius of a chamber?
Answer:
[tex]r=2.4m[/tex]
Explanation:
We have to use the centripetal force equation
[tex]Fc=\frac{mv^{2} }{r}[/tex]
we need the radious so we have to isolate "r" and we get
[tex]r=\frac{mv^{2} }{Fc}[/tex]
replacing m=65 kg, v= 4.1 m/s and Fc=455N we get
[tex]r=\frac{65*4.1^{2} }{455}[/tex]
[tex]r=2.4m[/tex]
The radius of the amusement park chamber is 2.4m
An empty beaker weighs 40.25 g. When completely filled with water, the beaker and its contents have a total mass of 317.45 g. What volume does the beaker hold? Use ????=1.00 g/mL as the density of water.
Answer:
The beaker holds 277.2 mL
Explanation:
Empty weight of beaker = 40.25 g
Weight of beaker with water = 317.45 g
Weight of water = 317.45 - 40.25 = 277.2 g
Density of water = 1 g/mL
We have
Mass = Volume x density
277.2 = Volume x 1
Volume = 277.2 mL
The beaker holds 277.2 mL
The volume of the beaker is 277.2 mL, obtained by subtracting the mass of the empty beaker from the total mass with water and using the density of water as 1.00 g/mL.
To find the volume of the beaker, we need to calculate the mass of the water it holds. First, subtract the mass of the empty beaker from the total mass with the water: 317.45 g - 40.25 g = 277.2 g. Since the density of water is 1.00 g/mL, the mass of water in grams is numerically equal to its volume in mL. Therefore, the beaker holds a volume of 277.2 mL of water.
Two objects, with masses m1 and m2, are originally a distance r apart, and the magnitude of the gravitational force on each one is F. The objects are moved a distance 2r apart. The new gravitational force: A) decreases 2 times
B) increases 2 times
C) decreases 4 times D) remains unchanged
E) increases 4 times
Explanation:
It is given that, two objects, with masses m₁ and m₂, are originally a distance r apart, and the magnitude of the gravitational force on each one is F. If the objects are moved a distance 2 r apart, we need to find the new gravitational force.
The gravitational force is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]...........(1)
Let the force is F' when the objects are moved a distance 2 r apart. So,
[tex]F=G\dfrac{m_1m_2}{(2r)^2}[/tex]
[tex]F'=\dfrac{1}{4}G\dfrac{m_1m_2}{r^2}[/tex]
[tex]F'=\dfrac{1}{4}F[/tex] (from equation (1))
So, it is clear that the magnitude of new force is F/4 i.e. decreases by a factor is 1/4. Hence, this is the required solution.
A hydrogen atom in the ground state absorbs a photon that increases its principle quantum number by 3. How many different wavelengths could the atom emit on the decay path back to the ground state? (Ignore spin-orbit coupling for this question).
Answer:
[tex]N = 3[/tex]
so these are
either 3 to 2 or 2 to 1 or 3 to 1
Explanation:
As we know that total number photons emitted from nth state to ground state is given by
[tex]N = ^nC_2[/tex]
[tex]N = \frac{n(n-1)}{2}[/tex]
here it shows that if electron has transition from nth excited state to ground state then total number of photons is given by above equation
here we know that higher state is n = 3
so total number of photons is given as
[tex]N = \frac{3(3-1)}{2}[/tex]
[tex]N = 3[/tex]
so these are
either 3 to 2 or 2 to 1 or 3 to 1
A positively charged bead having a mass of 1.00 g falls from rest in a vacuum from a height of 5.00 m in a uniform vertical electric field with a magnitude of 1.00 ✕ 104 N/C. The bead hits the ground at a speed of 21.9 m/s.
a) Determine the direction of the electric field (up or down)
b)Determine the Charge on the bead ___ µC
Answer:
a)
down direction.
b)
3.82 µC
Explanation:
a)
Consider the motion of the positively charged bead in vertical direction
y = vertical displacement of charged bead = 5 m
a = acceleration of charged bead = ?
v₀ = initial velocity of bead = 0 m/s
v = final velocity of bead = 21.9 m/s
using the equation
v² = v₀² + 2 a y
inserting the values
21.9² = 0² + 2 a (5)
a = 47.96 m/s²
m = mass of the bead = 1 g = 0.001 kg
F = force by the electric field
Force equation for the motion of the bead in electric field is given as
mg + F = ma
(0.001) (9.8) + F = (0.001) (47.96)
F = 0.0382 N
Since the electric force due to electric field comes out to be positive, the electric force acts in down direction. we also know that a positive charge experience electric force in the same direction as electric field. hence the electric field is in down direction.
b)
q = magnitude of charge on the bead
E = electric field = 1 x 10⁴ N/C
Electric force is given as
F = q E
0.0382 = q (1 x 10⁴)
q = 3.82 x 10⁻⁶ C
q = 3.82 µC
(a) The electric field direction is down as it contributes to the increased speed of the falling bead. (b) The charge on the bead is calculated to be 3.8 µC.
Let's address the given problem step-by-step:
(a) Determine the direction of the electric field
We know that the bead is positively charged and falls from rest in a vacuum. Gravity pulls the bead downward by itself, but the bead hits the ground at a speed greater than it would under gravity alone (21.9 m/s compared to the ~9.9 m/s due to gravitational acceleration over 5.00 meters). Therefore, the electric field must be contributing additional force downward to achieve this extra speed. Thus, the electric field must be pointing down.
(b) Determine the Charge on the bead in µC
First, calculate the work done by the electric field on the bead:
Gravitational Potential Energy:
Initial PE = mgh = 1.00*10⁻³kg * 9.8 m/s² * 5.00 m = 0.049 J
Final Kinetic Energy (KE): = 1/2 * m * v² = 0.5 * 1.00*10⁻³kg * (21.9 m/s)² = 0.239 J
Total work done by the electric field:
WE = KE - PE = 0.239 J - 0.049 J = 0.19 JUsing WE = qEd, we can solve for the charge q:
q = WE / (Ed)q = 0.19 J / (1.00*10⁴ N/C * 5.00 m)q = 0.19 J / 5.00 * 10⁴ N/CConvert the charge to µC:
q = 3.8 * 10⁻⁶ Cq = 3.8 µCSo, the charge on the bead is 3.8 µC.What will happen to a negatively charged particle placed at rest between points that have electric potentials 0V and 9V? Explain.b)Discuss the meanings of the words “potential,” “potential difference,” and “voltage.” Give examples to help illustrate what you’re talking aboutc)How canthe knowledge of electric potential help us understand the motion of electrically charged particles in electric circuits? Answer in your notebookand on the whiteboard.
Hey there!
a) The electric field is in direction of decreasing value of potential so the electric field here will points towards 0v from 9 v. But the charge on particle is negative so the force will be towards the 9v point. and so the charge will move towards the 9 V point.
b) Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location while.the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Electric potential is always a relative quantity because no one can find out absolute potential at a point, so in reality only potential difference exists. We can assume potential at certain point (say at infinity it is zero) only then we are able to define potential at every point.
c) The knowlendge of electric potential helps us in finding the work done on the particle which is used in work energy theorem to find out the chanrge in kinetic energy and other stuff.
Hope this helps!
A square wire, 20 cm on the side, moves at constant speed parallel to the xy-plane into a region where there is a uniform magnetic field = (0.1 T) .The induced electric current in the coil is 10 mA and its resistance is 10Ω. What is the speed of the wire?
Answer:
The speed of the wire is 5 m/s.
Explanation:
Given that,
Length = 20 cm
Magnetic field = 0.1 T
Current = 10 mA
Resistance [tex]R= 10\Omega[/tex]
We need to calculate the speed of the wire
Using formula of emf
[tex]\epsilon=Bvl[/tex]
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value of [tex]\epsilon[/tex] into the formula of current
[tex]I=\dfrac{Bvl}{R}[/tex]
[tex]v=\dfrac{IR}{Bl}[/tex]
[tex]v=\dfrac{10\times10^{-3}\times10}{0.1\times20\times10^{-2}}[/tex]
[tex]v= 5\ m/s[/tex]
Hence, The speed of the wire is 5 m/s.
A newly discovered element has two isotopes. One has an atomic weight of 120.9038 amu with 57.25% abundance. The other has an atomic weight of 122.8831 amu. What is the atomic weight of the element? 1. 123.45 amu 2. 121.54 amu 3. 121.17 amu 4. 121.75 amu 5. 122.15 amu 6. 122.38 amu
Answer : The atomic weight of the element is, 121.75 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
As we are given that,
Mass of isotope 1 = 120.9038 amu
Percentage abundance of isotope 1 = 57.25 %
Fractional abundance of isotope 1 = 0.5725
Mass of isotope 2 = 122.8831 amu
Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %
Fractional abundance of isotope 2 = 0.4275
Now put all the given values in above formula, we get:
[tex]\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)][/tex]
[tex]\text{Average atomic mass of element}=121.75amu[/tex]
Therefore, the atomic weight of the element is, 121.75 amu
The atomic weight of an element with two isotopes of 120.9038 amu at 57.25% abundance and 122.8831 amu is calculated as 121.75 amu after rounding to two decimal places.
The atomic weight of an element with two isotopes can be calculated using the weighted average formula. This takes into account the atomic masses of the isotopes and their relative abundances.
Since one isotope has an atomic weight of 120.9038 amu with 57.25% abundance and we know the atomic weight of the other isotope is 122.8831 amu, we can calculate the average atomic mass as follows:
The percent abundance of the second isotope must be 100% - 57.25% = 42.75%, since the two abundances must add up to 100%.
Atomic weight = (isotope1_mass * isotope1_abundance) + (isotope2_mass * isotope2_abundance)
Substituting our values in gives us:
Atomic weight = (120.9038 amu * 0.5725) + (122.8831 amu * 0.4275)
Atomic weight = (69.1866935 amu) + (52.53009825 amu)
Atomic weight = 121.71679175 amu
After rounding to two decimal places, the average atomic weight of the element is 121.72 amu which is closest to option 4, 121.75 amu.
A flat disk 1.0 m in diameter is oriented so that the plane of the disk makes an angle of π/6 radians with a uniform electric field. If the field strength is 487.0 N/C, find the electric flux through the surface. A) 1 10m N·m2/C B) 420 N·m2/C C) 61 π N·m2/C D) 1700/T N·m2/C
Final answer:
The electric flux through a surface can be calculated using the formula Φ = EA cos θ, where Φ is the electric flux, E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface. In this question, the electric flux through the given disk can be calculated using the given values and the formula. The answer is 61π N·m²/C, which corresponds to option C.
Explanation:
The electric flux through a surface can be calculated using the formula:
Φ = EA cos θ
where Φ is the electric flux, E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.
In this question, the electric field strength is given as 487.0 N/C, the diameter of the disk is 1.0 m (radius = 0.5 m), and the angle between the plane of the disk and the electric field is π/6 radians.
Using the formula, we can calculate the electric flux as follows:
Φ = (487.0 N/C)(π * (0.5 m)^2) * cos(π/6)
Φ = 61π Nm²/C
Therefore, the electric flux through the surface is 61π N·m²/C, which corresponds to option C.
Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form: a. bromine-87, lanthanum-146, and more neutrons. b. a nucleus with 56 protons, a second with a total of 94 neutrons and protons, and 2 additional neutrons.
Answer: a) [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]
b) [tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]
Explanation:
A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.
a) The given reaction is [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n[/tex]
Now, as the mass on both reactant and product side must be equal:
[tex]235+1=87+146+x[/tex]
[tex]x=3[/tex]
Thus three neutrons are produced and nuclear equation will be: [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]
b) For the another fission reaction:
[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}[/tex]
To calculate A:
Total mass on reactant side = total mass on product side
235 + 1 = A + 94 + 2
A = 140
To calculate Z:
Total atomic number on reactant side = total atomic number on product side
92 + 0 = 56 + Z + 0
Z = 36
As Krypton has atomic number of 36,Thus the nuclear equation will be :
[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]
Final answer:
The fission of uranium-235 can yield multiple products including bromine-87, lanthanum-146, and additional neutrons. Part a explains the specific fission process forming these products, while part b outlines a scenario producing a nucleus with 56 protons and another with 94 nucleons, alongside more neutrons.
Explanation:
The question relates to the process of nuclear fission, specifically the fission of uranium-235 (U-235) when it captures a neutron. Nuclear fission involves a large nucleus splitting into smaller nuclei along with the release of energy and additional neutrons. These neutrons can initiate further fission reactions, creating a chain reaction.
For part a (forming bromine-87 and lanthanum-146):
U-235 + n → Br-87 + La-146 + 3n
This equation shows that upon absorbing a neutron, U-235 splits into bromine-87, lanthanum-146, and releases 3 more neutrons.
For part b (forming a nucleus with 56 protons and another with a total of 94 nucleons):
A nucleus with 56 protons corresponds to iron (Fe), due to the atomic number. A second nucleus having a total of 94 nucleons with an unspecified proton-to-neutron ratio could refer to an element like plutonium-94, considering the total number of protons and neutrons. However, the provided information primarily focuses on the fission products rather than specifics on this second nucleus. Additionally, 2 more neutrons are produced. A precise equation for this scenario cannot be formulated without more specific information on the second fragment.
A hoop of mass and radius and a solid cylinder of mass and radius are rolling along level ground with the same translational speed. Which object has the greater total kinetic energy?
Answer:
the kinetic energy of hoop will be more than kinetic energy of solid cylinder
Explanation:
kinetic energy of rolling is given as
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]
here we know that for pure rolling we have
[tex]v = R \omega[/tex]
now we also know that
[tex]I = mk^2[/tex]
here k = radius of gyration
now we have
[tex]KE = \frac{1}{2}mv^2( 1 + \frac{k^2}{R^2})[/tex]
now we know that for
hoop
[tex]\frac{k^2}{R^2} = 1[/tex]
for Solid cylinder
[tex]\frac{k^2}{R^2} = \frac{1}{2}[/tex]
now the kinetic energy of hoop will be more than kinetic energy of solid cylinder
What are the two design conditions necessary before an ion chamber can be considered a Bragg-Grey chamber? What could such a chamber be used for?
Answer and Explanation:
The two necessary design conditions before an ion chamber can be considered a Bragg-Grey chamber are:
Cavity dimensions or thickness of the cavity material 'g' should be as small as possible in comparision to charge particles that strike the cavity in order to prevent any disturbance in the charged particle field.The total energy or the absorbed dose of the cavity material 'g' inside the cavity is completely deposited by the charged particle crossing their paths.Such a chamber can be used for the purpose of calculation and measurement of the absorbed ionizing radiation dose inside the chamber. Basically, ionization chambers are used as detectors for X-ray absorption spectroscopy.
Charge 1 of +4 micro-coulombs is placed at the origin, charge 2 of +18 micro-coulombs is placed at x = +0.32 m, y = -0.59 m, charge 3 of -2 micro-coulombs is placed at x = -0.79 m, y = 0 m. Measured counter-clockwise in degrees, what is the angle of the total electric force vector on charge 1?
CORRECT ANSWER: 122.36
Answer:
122.36
Explanation:
The distance (d) between Charge 1 and 2 can be calculated as:
[tex]d=\sqrt{0.32^2+0.59^2}=0.67m[/tex]
The force between them is given as
[tex]F_1=\frac{1}{4\pi \epsilon_0}\frac{4*18*10^-12}{0.67^2}= 1.44N[/tex]
The angle of this force with positive x-axis is given as
[tex]\theta_1=90^{\circ}+\tan^{-1}\frac{0.32}{0.59}=118.47^{\circ}[/tex]
Now,
The force between 1 and 3 is
[tex]F_2=\frac{1}{4\pi \epsilon_0}\frac{4*2*10^{-12}}{0.79^2}= 0.115N[/tex]
As the force is attractive it is along negative x direction so the angle will be given as = [tex]\theta_2 = 180^{\circ}[/tex]
So the negative x component of the resultant force will be calculated as
= [tex]1.44\cos(180-118.47)^{\circ}+0.115=0.801[/tex]
And the positive y component = [tex]1.44\sin(180-118.47)^{\circ}=1.26[/tex]
So the angle of the resultant with positive x axis will be
[tex]90^{\circ}+\tan^{-1}\frac{0.801}{1.26}=122.36^{\circ}[/tex]
You push your physics book 1.2 m along a horizontal tabletop with a horizontal push of 10 N while the opposing force of friction is 5.9 N. Find the work done by the force of 10 N push.
the work done on the book by the force of friction.
the work done on the book by the force of gravity.
Final answer:
The work done by the force of push is 12 J. The work done by the force of friction is 7.08 J. The work done on the book by the force of gravity is zero.
Explanation:
To find the work done by the force of push, we can use the formula:
Work = Force x Distance
In this case, the force of push is 10 N and the distance is 1.2 m. So the work done by the force of push is:
Work = 10 N x 1.2 m = 12 J
To find the work done by the force of friction, we can use the same formula. In this case, the force of friction is 5.9 N and the distance is also 1.2 m. So the work done by the force of friction is:
Work = 5.9 N x 1.2 m = 7.08 J
The work done on the book by the force of gravity is zero since the book is being moved horizontally and gravity acts vertically.
A 2170 kg space station orbits Earth at an altitude of 5.27 x 10^5 m. Find the magnitude of the force with which the space station attracts Earth. The mass and mean radius of Earth are 5.98 x 10^24 kg and 6.37 x 10^6 m, respectively.
Answer:
F = 18195.59 N or F = 18196 rounded up
Explanation:
force = GMm/d^2
G = 6.67x10^-11
M = 5.98x10^24 kg
m = 2170kg
d = 6370000 + 527000 = 6897000m
putting all values (6.67x10^-11)(5.98x10^24)(2170)/(6897000^2) = 18195.59....
The magnitude of the force with which the space station attracts Earth is about 1.82 × 10⁴ Newton
[tex]\texttt{ }[/tex]
Further explanationNewton's gravitational law states that the force of attraction between two objects can be formulated as follows:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
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Given:
mass of space station = m = 2170 kg
radius of the orbit = R = 5.27 × 10⁵ + 6.37 × 10⁶ = 6.897 × 10⁶ m/s
mass of Earth = M = 5.98 × 10²⁴ kg
Asked:
Gravitational Force = F = ?
Solution:
[tex]F = G \frac{M.m}{R^2}[/tex]
[tex]F = 6.67 \times 10^{-11} \times \frac{5.98 \times 10^{24} \times 2170}{(6.897 \times 10^6)^2}[/tex]
[tex]F \approx 1.82 \times 10^4 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Gravitational Fields
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
The absolute potential at a distance of 2.0 m from a negative point charge is-100. V. What is the absolute potential 4.0 m away from the same point charge? A)-50. V B) -25. V C)-0.40 kV D)-020 kV E)-0.50 kV
Answer:
The absolute potential 4.0 m away from the same point charge is -50 V.
(A) is correct option.
Explanation:
Given that,
Distance = 2.0 m
Potential = -100 V
Absolute potential = 4.0 m
We need to calculate the charge
Using formula of potential
[tex]V=\dfrac{kq}{r}[/tex]
Where, V = potential
q = charge
r = distance
Put the value into the formula
[tex]-100=\dfrac{9\times10^{9}\times q}{2.0}[/tex]
[tex]q=\dfrac{200}{9\times10^{9}}[/tex]
[tex]q=-22.2\times10^{-9}\ C[/tex]
We need to calculate the potential
Using formula of potential
[tex]V=\dfrac{9\times10^{9}\times(-22.2\times10^{-9})}{4.0}[/tex]
[tex]V=-49.95\ V[/tex]
[tex]V=-50\ V[/tex]
Hence, The absolute potential 4.0 m away from the same point charge is -50 V.
A driver in a car speeding to the right at 24m/s suddenly hits the brakes and goes into a skid, finally coming to rest. The coefficient of static and kinetic friction between the tires and the road are s = .8 and k = 0.7. How far does the car skid?
Answer:
42 m
Explanation:
[tex]v_{o}[/tex] = initial velocity of the car = 24 m/s
[tex]v_{f}[/tex] = final velocity of the car = 0 m/s
μ = coefficient of kinetic friction = 0.7
g = acceleration due to gravity = 9.8 m/s²
a = acceleration due to kinetic frictional force = - μg = - (0.7)(9.8) = - 6.86 m/s²
d = distance through which the car skids
Using the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d[/tex]
Inserting the values
[tex]0^{2} = 24^{2} + 2 (- 6.86) d[/tex]
d = 42 m
Calculate the centripetal force on a 6.0 kg sensor that is attached at the end of a 100.0-m a long wind turbine blade that is rotating at 1.25 rev/s.
Answer:
Centripetal force, F = 18486.75 N
Explanation:
It is given that,
Mass of the sensor, m = 6 kg
It is attached at the end of 100 m long wind turbine, d = 100 m
So, the radius of the wind turbine, r = 50 m
Angular velocity, [tex]\omega=1.25\ rev/s=7.85\ rad/s[/tex]
The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
Since, [tex]v=r\omega[/tex]
[tex]F=mr\omega^2[/tex]
[tex]F=6\ kg\times 50\ m\times (7.85\ rad/s)^2[/tex]
F = 18486.75 N
So, the centripetal force is 18486.75 N. Hence, this is the required solution.
A1.0µF capacitor has a potential difference of 6.0 V applied across its plates. If the potential difference acro its plates is increased to 8.0 V, how much additional energy does the capacitor store? If the dielectric constant is changed from 1 to 76.5 how does this change the amount of charge stored on the capacitor plates?
Answer:
14 x 10⁻⁶ J
1377 x 10⁻⁶ J
Explanation:
C = Capacitance of the capacitor = 1 x 10⁻⁶ F
ΔV = Original potential difference across the plates = 6.0 Volts
U₀ = Original energy stored in the capacitor
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² eq-1
ΔV' = Potential difference across the plates after increase = 8.0 Volts
U'₀ = New energy stored in the capacitor
New energy stored in the capacitor is given as
U'₀ = (0.5) C ΔV'² eq-2
U = Additional energy stored
Additional energy stored by the capacitor is given as
U = U'₀ - U₀
U = (0.5) C ΔV'² - (0.5) C ΔV²
U = (0.5) (1 x 10⁻⁶) (8)² - (0.5) (1 x 10⁻⁶) (6)²
U = 14 x 10⁻⁶ J
[tex]k_{final}[/tex] = final dielectric constant = 76.5
[tex]k_{initial}[/tex] = initial dielectric constant = 1
Energy stored in the capacitor is directly proportional to the dielectric constant, hence increase in the energy is given as
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² = (0.5) (1 x 10⁻⁶) (6)² = 18 x 10⁻⁶ J
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
[tex]U_{inc} = (76.5 - 1)(18\times 10^{-6})[/tex]
[tex]U_{inc} = 1377\times 10^{-6})[/tex]
The amount of charge stored on the capacitor plates increases by a factor of [tex]$76.5}$[/tex].
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where [tex]$E$[/tex] is the energy, [tex]$C$[/tex] is the capacitance, and [tex]$V$[/tex] is the potential difference.
Initially, the energy stored in the capacitor with a potential difference of 6.0 V is:
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (6.0)^2 \][/tex]
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 36 \][/tex]
[tex]\[ E_1 = 18 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_1 = 18 \text{ µJ} \][/tex] µJ
After increasing the potential difference to 8.0 V, the energy stored in the capacitor is:
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (8.0)^2 \][/tex]
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 64 \][/tex]
[tex]\[ E_2 = 32 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_2 = 32 \text{ µJ} \][/tex] µJ
The additional energy stored is the difference between [tex]$E_2$[/tex] and [tex]$E_1$[/tex]:
[tex]\[ \Delta E = E_2 - E_1 \][/tex]
[tex]\[ \Delta E = 32 - 18 \][/tex]
[tex]\[ \Delta E = 14 \text{ µJ} \][/tex] µJ
Now, let's consider the effect of changing the dielectric constant from 1 to 76.5. The capacitance of a capacitor with a dielectric material is given by:
[tex]\[ C = \frac{k \epsilon_0 A}{d} \][/tex]
Since the dielectric constant [tex]$k$[/tex] is the only quantity changing, the new capacitance [tex]$C'$[/tex] is:
[tex]\[ C' = kC \][/tex]
where [tex]$C$[/tex] is the original capacitance with [tex]$k = 1$[/tex].
Therefore, [tex]$C' = 76.5 \times 1.0 \times 10^{-6} \text{ F}$[/tex].
The charge stored on a capacitor is given by:
[tex]\[ Q = CV \][/tex]
With the new capacitance and the same potential difference of 8.0 V, the new charge [tex]$Q'$[/tex] is:
[tex]\[ Q' = C'V \][/tex]
[tex]\[ Q' = 76.5 \times 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q' = 612 \times 10^{-6} \text{ C} \][/tex]
The original charge [tex]$Q$[/tex] with [tex]$k = 1$[/tex] and [tex]$V = 8.0 \text{ V}$[/tex] was:
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q = 8 \times 10^{-6} \text{ C} \][/tex]
The ratio of the new charge to the original charge is:
[tex]\[ \frac{Q'}{Q} = \frac{612 \times 10^{-6}}{8 \times 10^{-6}} \][/tex]
[tex]\[ \frac{Q'}{Q} = 76.5 \][/tex]
16 A 20-pewton force daurected west and a 5 newton force directed north act concurently on a 5 kg object. Draw the resultant vector of the two forces, calculate its magnitude, and determine the acceleration of the object
Answer:
20.62 N
4.123 m/s^2
Explanation:
A = 20 N west
B = 5 N North
m = 5 kg
Both the forces acting at right angle
Use the formula of resultant of two vectors.
Let r be the magnitude of resultant of two vectors.
[tex]R = \sqrt{A^{2} + B^{2} + 2 A B Cos\theta}[/tex]
[tex]R = \sqrt{20^{2} + 5^{2} + 2 \times 20 \times 5 \times Cos90}[/tex]
R = 20.62 N
Let a be the acceleeration.
a = Net force / mass = R / m = 20.62 / 5
a = 4.123 m/s^2
The acceleration of high-speed aircraft is sometimes expressed in g's (in multiples of the standard acceleration of gravity). Determine the upward force, in N, that a 78.03-kg person would experience in an aircraft whose acceleration is 5.05 g's. Assume the standard acceleration of gravity is 8.23 m/s^2.
Answer:
3888.23 N
Explanation:
m = mass of the person = 78.03 kg
g = standard acceleration due to gravity = 8.23 m/s²
a = acceleration of the aircraft = 5.05 g = 5.05 x 8.23 = 41.6 m/s²
F = upward force the person experience
Force equation for the motion of person is given as
F - mg = ma
Inserting the values
F - (78.03)(8.23) = (78.03)(41.6)
F = 3888.23 N
A recent study found that electrons that have energies between 3.45 eV and 20.9 eV can cause breaks in a DNA molecule even though they do not ionize the molecule. If a single photon were to transfer its energy to a single electron, what range of light wavelengths could cause DNA breaks?
Answer:
5.95 nm to 33.6 nm
Explanation:
Energy of a single photon can be written as:
[tex]E = \frac{hc}{\lambda}[/tex]
where, h is the Planck's constant, c is the speed of light and λ is the wavelength of light.
Consider the lowest energy of an electron that can break a DNA = 3.45 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒3.45 eV = 5.52×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{5.52 \times 10^{-19} J} = 3.60\times 10^{-7} m = 360 nm[/tex]
Consider the highest energy of an electron that can break a DNA = 20.9 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒20.9 eV = 33.4×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{33.4 \times 10^{-19} J} = 0.595\times 10^{-7} m = 59.5 nm[/tex]
The wavelength range of light that can cause DNA breaks is approximately 59.4 nm to 360 nm. This corresponds to ultraviolet light and part of the visible spectrum.
To find the range of light wavelengths that can cause DNA breaks, we need to convert the given energy range of 3.45 eV to 20.9 eV into wavelengths.
The energy of a photon (E) is related to its wavelength (λ) by the equation:
E = hc/λ
where h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light .(3.00 × 10⁸m/s), and λ is the wavelength in meters.
First, convert the energy from electron volts (eV) to Joules (J):
1 eV = 1.602 × 10⁻¹⁹ JSo, the energy range in Joules is:
3.45 eV = 3.45 × 1.602 × 10⁻¹⁹J = 5.53 × 10⁻¹⁹ J20.9 eV = 20.9 × 1.602 × 10⁻¹⁹J = 3.35 × 10⁻¹⁸ JNext, use the energy-wavelength relation to find the wavelengths:
For 3.45 eV: λ = hc/E = (6.626 × 10⁻³⁴J·s × 3.00 × 10⁸ m/s) / 5.53 × 10⁻¹⁹J ≈ 3.60 × 10⁻⁷ m ≈ 360 nmFor 20.9 eV: λ = hc/E = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / 3.35 × 10⁻¹⁸ J ≈ 5.94 × 10⁻⁸ m ≈ 59.4 nmThus, the range of wavelengths that can cause DNA breaks is approximately 59.4 nm to 360 nm, corresponding to the ultraviolet (UV) and part of the visible spectrum.
A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the person and the kayak is 75 kg, what is the magnitude of the net force acting on the kayak?
Answer:
Magnitude of the net force acting on the kayak = 39.61 N
Explanation:
Considering motion of kayak:-
Initial velocity, u = 0 m/s
Distance , s = 0.40 m
Final velocity, v = 0.65 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0.65² = 0² + 2 x a x 0.4
a = 0.53 m/s²
We have force, F = ma
Mass, m = 75 kg
F = ma = 75 x 0.53 = 39.61 N
Magnitude of the net force acting on the kayak = 39.61 N
What is an Exothermic Reaction? What is an Endothermic Reaction?
Answer:
Answer to the question:
Explanation:
Exothermic Reaction
It is called an exothermic reaction to any chemical reaction that releases energy, either as light or heat, or what is the same: with a negative variation of enthalpy; that is to say: ΔH < 0. Therefore it is understood that exothermic reactions release energy.
Endothermic Reaction
It is called an endothermic reaction to any chemical reaction that absorbs energy, usually in the form of heat.
If we talk about enthalpy (H), an endothermic reaction is one that has a variation of enthalpy ΔH> 0. That is, the energy possessed by the products is greater than that of the reagents.
Does resistance in a circuit remain constant if the temperature changes?
Answer:
No the resistance of a given circuit does not remain constant if the temperature of the circuit changes.
Explanation:
The resistance of any resistor used in a circuit depends upon the temperature of that resistor. This can be mathematically represented as follows
[tex]R(t)=R_{0}(1+\alpha \Delta t)[/tex]
Where,
R(t) is resistance of any resistor at temperature t
[tex]R_{o}[/tex] is the resistance of the resistor at time of fabrication
α is temperature coefficient of resistivity it's value is different for different materials
This change in the resistance is the cumulative effect of:
1) Variation of resistivity with temperature
2) Change in dimensions of the resistor with change in temperature
Final answer:
Resistance in a circuit changes with temperature due to increased atomic vibrations affecting electron movement.
Explanation:
Resistance in a circuit does not remain constant if the temperature changes. As temperature increases, the resistance of a conductor typically increases due to the atoms vibrating more rapidly, causing more collisions for the electrons passing through.
This change in resistance with temperature is a common phenomenon seen in various materials. It generally increases with increasing temperature due to more frequent electron collisions within the conductor.
Ohm's Law, which states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, assumes constant temperature.
In practical terms, if you were to graph resistance against temperature, for some materials, you would notice a linear increase in resistance for small temperature changes, while for large changes, the relationship can be nonlinear.
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the axis of the loop. Somebody carefully grabs the edges of the loop and begins pulling them apart such that the area of the loop increases at a rate of 20 cm2/s. What is the magnitude of the induced EMF in the loop?
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
[tex]\frac{dA}{dt}[/tex] = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B [tex]\frac{dA}{dt}[/tex]
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV