The shear deformation in a copper block due to applied forces can be calculated using the formula Δx = (F * L₀) / (S * A), where F is the force applied, L₀ is the initial length of the block, S is the shear modulus of copper, and A is the area of the surface on which the force is applied. The angle can then be calculated by taking the inverse tangent of the ratio of deformation (Δx) to the original length (L₀) of the block.
Explanation:The subject of the question is related to the physics concept of shear strain and shear stress. The shear strain caused on a material, in this case copper, is defined by the sideways deformation of the material due to the applied force, and it is given by the formula Δx = (F * L₀) / (S * A). Here, F represents the force applied which is 3.07 X 10 6 N, L₀ is the height of the block which is 0.242 m. The shear modulus S of copper is given as 4.2 X 10¹⁰ N/m², and A is the area of the surface on which the force is applied, which can be calculated as 0.242 m² since a cube has all its sides equal. By substituting these values in the equation, we can calculate the value of the deformation Δx, which will give us an idea of how the shape of the block has been altered. The angle of shear can be obtained by taking the inverse tangent of the ratio of deformation to the original height of the cube.
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Calculate the wavelength of 100-MHz microwaves in muscle and in fat.
Answer:
Wavelength of microwaves is 3 m.
Explanation:
In this question, we need to find the wavelength of 100 MHz microwaves in muscle and in fat.
Frequency of the microwaves, [tex]\nu=100\ MHz=100\times 10^6\ Hz=10^8\ Hz[/tex]
The relation between the frequency and the wavelength is given by :
[tex]c=\nu\times \lambda[/tex]
[tex]\lambda=\dfrac{c}{\nu}[/tex]
Where
c is the sped of light
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{10^8\ Hz}[/tex]
[tex]\lambda=3\ m[/tex]
So, the wavelength of microwaves is 3 m. Hence, this is the required solution.
A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat makes. If bats emit a chirp at a frequency of 7.84 104 Hz, and if the speed of sound in air is 343 m/s, what is the smallest insect a bat can detect?
Answer:
0.4375 cm
Explanation:
f = frequency of the chirp emitted by the bats = 7.84 x 10⁴ Hz
v = speed of sound in air = 343 m/s
λ = smallest wavelength = size of the smallest insect a bat can detect
Using the equation
v = f λ
inserting the values
343 = (7.84 x 10⁴) λ
λ = [tex]\frac{343}{7.84\times 10^{4}}[/tex]
λ = 43.75 x 10⁻⁴ m
λ = 0.4375 cm
lectric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?
Answer:
The resistance of the heater wire is of R= 0.68 Ω.
Explanation:
1 cal/s = 4.184 W
P= 50 cal/s = 209.2 W
V= 12V
P= V* I
I= P/V
I= 17.43 A
P= I² * R
R= P / I²
R= 0.68 Ω
An object at the surface of Earth (at a distance R from the center of Earth) weighs 166 N. What is its weight (in N) at a distance 4R from the center of Earth? Round your answer to the nearest tenth.
Answer:
The weight at a distance 4R from the center of earth is 10.37 N.
Explanation:
Given that,
Weight = 166 N
Distance = 4R
Let m be the mass of the object.
We know that,
Mass of earth [tex]M_{e}=5.98\times10^{24}\ kg[/tex]
Gravitational constant[tex]G = 6.67\times10^{-11}\ N-m^2/kg^2[/tex]
Radius of earth [tex]R = 6.38\times10^{6}\ m[/tex]
We need to calculate the weight at a distance 4 R from the center of earth
Using formula of gravitational force
[tex]W = \dfrac{GmM_{e}}{R^2}[/tex]
Put the value in to the formula
[tex]166=\dfrac{6.67\times10^{-11}\times m\times5.98\times10^{24}}{(6.38\times10^{6})^2}[/tex]
[tex]m=\dfrac{166\times(6.38\times10^{6})^2}{6.67\times10^{-11}\times5.98\times10^{24}}[/tex]
[tex]m=16.94 kg[/tex]
Now, Again using formula of gravitational
[tex]W=\dfrac{6.67\times10^{-11}\times 16.94\times5.98\times10^{24}}{(4\times6.38\times10^{6})^2}[/tex]
[tex]W=10.37 N[/tex]
Hence, The weight at a distance 4R from the center of earth is 10.37 N.
The position of an electron is measured within an uncertainty of 0.100 nm. What will be its minimum position uncertainty 2.00 s later? {3.32 x 106 m}
Answer:
Minimum uncertainty in position is [tex]\Delta x= 1157808.48\ m[/tex]
Explanation:
It is given that,
Uncertainty in the position of an electron, [tex]\Delta x=0.1\ nm=0.1\times 10^{-9}\ m[/tex]
According to uncertainty principle,
[tex]\Delta x.\Delta p\geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta x.m\Delta v\geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta v\geq \dfrac{h}{4\pi \times \Delta x\times m}[/tex]
[tex]\Delta v\geq \dfrac{6.62\times 10^{-34}\ J-s}{4\pi \times 0.1\times 10^{-9}\ m\times 9.1\times 10^{-31}\ kg}[/tex]
[tex]\Delta v\geq 578904.24\ m/s[/tex]
Let [tex]\Delta x[/tex] is the uncertainty in position after 2 seconds such that,
[tex]\Delta x=\Delta v\times t[/tex]
[tex]\Delta x=578904.24\ m/s\times 2\ s[/tex]
[tex]\Delta x= 1157808.48\ m[/tex]
or
[tex]\Delta x= 1.15\times 10^6\ m[/tex]
Hence, this is the required solution.
How far would you go if you traveled at a speed of 30 miles per hour for 3 hours?
Answer:
90 miles
Explanation:
speed = 30 miles per hour, time = 3 hour
The formula for the speed is given by
speed = distance / time
Distance = speed x time
distance = 30 x 3 = 90 miles
Answer:
90 Miles
Explanation:
This is more simple than you thought...
The equation is 30 miles per hour,in 3 hours you would travel (30*3)=90 miles.
Formula used Distance = Speed x Time
Speed = 30 mph
Time = 3 hours
So, Distance = Speed * Time = 30 * 3 = 90 miles.
So all in all, the answer is 90 miles
A particle has charge -1.95 nC. (a) Find the magnitude and direction of the electric field due to this particle at a point 0.225 m directly above it magnitude direction | Select ' N/C (b) At what distance from this particle does its electric field have a magnitude of 10.5 N/C?
Answer:
a)
346.67 N/C, downward
b)
1.3 m
Explanation:
(a)
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
r = distance of location from the charged particle = 0.225 m
E = magnitude of electric field at the location
Magnitude of electric field at the location is given as
[tex]E = \frac{kq}{r^{2}}[/tex]
Inserting the values
[tex]E = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{(0.225)^{2}}[/tex]
E = 346.67 N/C
a negative charge produce electric field towards itself.
Direction : downward
(b)
E = magnitude of electric field at the location = 10.5 N/C
r = distance of location from the charged particle = ?
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
Magnitude of electric field at the location is given as
[tex]E = \frac{kq}{r^{2}}[/tex]
Inserting the values
[tex]10.5 = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{r^{2}}[/tex]
r = 1.3 m
The magnitude and direction of the electric field due to a particle of charge -1.95 nC at a point 0.225 m directly above it is approximately -3.57 x 10^5 N/C towards the particle. The distance from this particle at which its electric field has a magnitude of 10.5 N/C is approximately 0.036 m or 36 mm.
The magnitude of the electric field (E) due to a charged particle can be calculated using Coulomb's Law, which states E = k*Q/r^2, where k is Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2), Q is the charge of the particle, and r is the distance from the particle.
(a) Incorporating the given values: Q = -1.95 nC = -1.95 × 10^-9 C, and r = 0.225 m, we find E = k*Q/r^2 = (8.99 × 10^9 N m^2/C^2) * (-1.95 × 10^-9 C) / (0.225 m)^2 ≈ -3.57 x 10^5 N/C. The negative sign indicates the direction of the electric field is towards the particle.
(b) To find the distance at which the electric field has a magnitude of 10.5 N/C, we rearrange the equation to solve for r. That is r = sqrt(k*Q/E). By plugging in the given values, we find r = sqrt((8.99 × 10^9 N m^2/C^2 * -1.95 × 10^-9 C) / 10.5 N/C) ≈ 0.036 m or 36 mm.
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A 3.9 kg block is pushed along a horizontal floor by a force ModifyingAbove Upper F With right-arrow of magnitude 27 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.22. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.
Answer:
a) 12.23 N
b) 2.2 m/s²
Explanation:
m = mass of the block = 3.9 kg
F = applied force = 27 N
θ = angle of the applied force with the horizontal = 40°
μ = Coefficient of kinetic friction = 0.22
[tex]F_{n}[/tex] = normal force
[tex]F_{g}[/tex] = weight of the block = mg
Along the vertical direction, force equation is given as
[tex]F_{n}[/tex] = F Sinθ + [tex]F_{g}[/tex]
[tex]F_{n}[/tex] = F Sinθ + mg
Kinetic frictional force is given as
f = μ [tex]F_{n}[/tex]
f = μ (F Sinθ + mg)
f = (0.22) (27 Sin40 + (3.9)(9.8))
f = 12.23 N
b)
Force equation along the horizontal direction is given as
F Cosθ - f = ma
27 Cos40 - 12.23 = 3.9 a
a = 2.2 m/s²
Rhodium has an atomic radius of 0.1345 nm and a density 12.41g/cm^3. De-termine if it comes in FCC or BCC structure.
Answer:
Rhodium has FCC structure.
Explanation:
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
1) If it FCC cubic lattice
Number of atom in unit cell of FCC (Z) = 4
Atomic radius of Rh= 0.1345 nm = [tex]1.345\times 10^{-8} cm[/tex]
Edge length = a
For FCC, a = 2.828 × r :
a = [tex]2.828\times 1.345\times 10^{-8} cm=3.80366\times 10^{-8}cm[/tex]
Density of Rh= [tex]12.41 g/cm^3[/tex]
Atomic mass of Rh(M) = 102.91 g/mol
On substituting all the given values , we will get the value of 'a'.
[tex]12.41 g/cm^3=\frac{4\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.80366\times 10^{-8}cm)^{3}}[/tex]
[tex]12.41 g/cm^3\approx 12.4214 g/cm^3[/tex]
2) If it BCC cubic lattice
Number of atom in unit cell of BCC (Z) = 2
Atomic radius of Rh= 0.1345 nm = [tex]1.345\times 10^{-8} cm[/tex]
Edge length = a
For BCC, a = 2.309 × r :
a = [tex]2.828\times 1.345\times 10^{-8} cm=3.105605\times 10^{-8}cm[/tex]
Density of Rh= [tex]12.41 g/cm^3[/tex]
Atomic mass of Rh(M) = 102.91 g/mol
On substituting all the given values , we will get the value of 'a'.
[tex]12.41 g/cm^3=\frac{2\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.105605\times 10^{-8}cm)^{3}}[/tex]
[tex]12.41 g/cm^3[/tex] ≠ [tex]11.41 g/cm^3[/tex]
Rhodium has FCC structure.
Rhodium has a FCC structure because the edge length of the unit cell is smaller than 4 times the atomic radius divided by the square root of 3.
Explanation:BCC stands for Body-Centered Cubic structure and FCC stands for Face-Centered Cubic structure. To determine which structure rhodium has, we need to compare the atomic radius of rhodium with the edge length of the unit cell for both structures.
For BCC, the diagonal distance of the body-centered cube is equal to 4 times the atomic radius. Therefore, the edge length of the unit cell for BCC is given by 4 times the atomic radius divided by √3.
For FCC, the face diagonal distance of the face-centered cube is equal to 2 times the atomic radius. Therefore, the edge length of the unit cell for FCC is given by 2 times the atomic radius.
By comparing the given atomic radius of 0.1345 nm to the calculated edge lengths, we can determine that rhodium has a FCC structure because 2 times the atomic radius (2 x 0.1345 nm) is smaller than 4 times the atomic radius divided by √3 (4 x 0.1345 nm / √3).
Which of the following is NOT true for a spherical concave mirror? a. it cannot produce virtual images
b. its focal length is half its radius
c. its focal length is positive
d. it can produce upright images
Answer:
Its focal length is positive
Explanation:
A concave mirror is shown in attached figure. The distance from the pole to the focus of the mirror is called its focal length. Spherical mirrors are a part of a sphere.
As per conventions, we know that the axis opposite to x axis is taken as negative.
So, it is clear that the focal length of spherical concave mirror is negative.
Hence, the incorrect option is (c) " its focal length is positive".
Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the ratio of the power drawn from the battery for the two cases (i.e. P series: P parallel)? Note: The battery supplies voltage V in both cases. A) 1:N B) 1:N2 C) N:1 D) N2:1
Answer:
option (b)
Explanation:
Let the resistance of each resistor is R.
In series combination,
The effective resistance is Rs.
rs = r + R + R + .... + n times = NR
Let V be the source of potential difference.
Power in series
Ps = v^2 / Rs = V^2 / NR ..... (1)
In parallel combination
the effective resistance is Rp
1 / Rp = 1 / R + 1 / R + .... + N times
1 / Rp = N / R
Rp = R / N
Power is parallel
Rp = v^2 / Rp = N V^2 / R ..... (2)
Divide equation (1) by equation (2) we get
Ps / Pp = 1 / N^2
When tuning a guitar, by comparing the frequency of a string that is struck against a standard sound source (of known frequency), what does the one adjusting the tension in the string listen to?
A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 m/s, and the crests of the waves ahead of the duck are spaced 0.12 m apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?
Answer:
a)
0.245 m/s
b)
0.904 m
Explanation:
a)
[tex]v_{d}[/tex] = speed of duck ahead of wave
[tex]v_{s}[/tex] = speed of surface wave = 0.32 m/s
T = time for paddling = 1.6 s
d = spacing between the waves = 0.12 m
speed of duck ahead of wave is given as
[tex]v_{d}[/tex] = [tex]v_{s}[/tex] - [tex]\frac{d}{T}[/tex]
[tex]v_{d}[/tex] = 0.32 - [tex]\frac{0.12}{1.6}[/tex]
[tex]v_{d}[/tex] = 0.245 m/s
b)
[tex]v_{w}[/tex] = speed of wave behind the duck
speed of wave behind the duck is given as
[tex]v_{w}[/tex] = [tex]v_{s}[/tex] + [tex]v_{d}[/tex]
[tex]v_{w}[/tex] = 0.32 + 0.245
[tex]v_{w}[/tex] = 0.565 m/s
D = spacing between the crests
spacing between the crests is given as
D = [tex]v_{w}[/tex] T
D = (0.565) (1.6)
D = 0.904 m
A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If the impedance of this circuit is 45.0 Ω at resonance, what is its impedance at a frequency twice the resonance frequency?
Answer:
61.85 ohm
Explanation:
L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm
Let ω0 be the resonant frequency.
[tex]\omega _{0}=\frac{1}{\sqrt{LC}}[/tex]
[tex]\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}[/tex]
ω0 = 2357 rad/s
ω = 2 x 2357 = 4714 rad/s
XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm
Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm
Impedance, Z = [tex]\sqrt{R^{2}+\left ( XL - Xc \right )^{2}}[/tex]
Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm
Thus, the impedance at double the resonant frequency is 61.85 ohm.
In an LRC circuit, impedance at resonance is equal to resistance. When frequency is doubled, the impedance will increase as the inductive reactance becomes greater than the capacitive reactance. Calculation of the new impedance would involve re-computing inductive and capacitive reactances and substituting these in the impedance formula.
Explanation:In an LRC series circuit, the impedance (Z) at resonance is equal to the resistance (R), because the reactance of the inductor (L) and the capacitor (C) cancel each other. The resonant frequency is determined by the values of L and C. If we increase the frequency to twice the resonance frequency, the reactance of the inductor becomes higher than the reactance of the capacitor, leading to an increase in total impedance.
Impedance (Z) is given by the square root formula Z = sqrt( R^2 + (X_L - X_C)^2 ). X_L and X_C are the inductive reactance (2πfL) and capacitive reactance (1/2πfC) respectively. If the frequency (f) is doubled, then X_L will be doubled while X_C will be halved.
We can calculate X_L and X_C under these conditions and substitute these values into the impedance formula to find the new impedance. Without the values of R, L, and C, it's not possible to give a numeric answer to this question, but the overall concept of how frequency affects impedance in an LRC circuit can be understood from the explanation.
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A.
Calculate the specific weight, density and specific gravity of one litre of a liquid, which weighs 7N.
Select one:
1. 7000 N/m3, 713.5 kg/m3, 0.7135
2. 700 N/m3, 71.35 kg/m3, 0.07135
3. 70 N/m3, 7.135 kg/m3, 0.007135
4. None of the above.
B.
The multiplying factor for converting one stoke into m2/s is
Select one:
1. 102
2. 104
3. 10-2
4. 10-4
Answer:
A) Option 1 is the correct answer.
B) Option 4 is the correct answer.
Explanation:
A) Weight of liquid = 7 N
Volume of liquid = 1 L = 0.001 m³
Specific weight = [tex]\frac{7}{0.001}=7000N/m^3[/tex]
Density = [tex]\frac{7000}{9.81}=713.5kg/m^3[/tex]
Specific gravity = [tex]\frac{713.5}{1000}=0.7135[/tex]
Option 1 is the correct answer.
B) The Stokes(St) is the cgs physical unit for kinematic viscosity, named after George Gabriel Stokes.
We have
1 St = 10⁻⁴ m²/s
Option 4 is the correct answer.
After evaluating all the options we have:
A. From all of the options of specific weight, density, and specific gravity of 1 liter of liquid, the correct is option 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. The correct option of the multiplying factor for converting one stoke into m²/s is 4: 10⁻⁴.
A. Let's calculate the specific weight, density, and specific gravity of 1 L of the liquid that weighs 7 N.
The specific weight is given by:[tex] \gamma = dg [/tex] (1)
Where:
γ: is the specific weight
d: is the density
g: is the gravity = 9.81 m/s²
We need to find the density which is:
[tex] d = \frac{m}{V} [/tex] (2)
Where:
m: is the mass
V: is the volume = 1 L = 0.001 m³
The mass can be found knowing that the liquid weighs (W) 7 N, so:
[tex] W = mg [/tex]
[tex] m = \frac{W}{g} [/tex] (3)
By entering equations (3) and (2) into (1) we have:
[tex] \gamma = dg = \frac{mg}{V} = \frac{W}{V} = \frac{7 N}{0.001 m^{3}} = 7000 N/m^{3} [/tex]
Hence, the specific weight is 7000 N/m³.
The density can be found as follows:[tex] d = \frac{m}{V} = \frac{W}{gV} = \frac{7 N}{9.81 m/s^{2}*0.001 m^{3}} = 713.5 kg/m^{3} [/tex]
Then, the density is 713.5 kg/m³.
The specific gravity (SG) of a liquid can be calculated with the following equation:[tex] SG = \frac{d}{d_{H_{2}O}} = \frac{713.5 kg/m^{3}}{1000 kg/m^{3}} = 0.7135 [/tex]
Hence, the specific gravity is 0.7135.
Therefore, the correct option is 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. A Stokes is a measurement unit of kinematic viscosity.
One m²/s is equal to 10⁴ stokes, so to convert 1 stokes to m²/s we need to multiply for 10⁻⁴.
Hence, the correct option is 4: 10⁻⁴.
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At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values of L and C. larger than R equal to R Which of the following does the quality factor of the circuit depend on? (Select all that apply.)
Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
[tex]Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2[/tex]
but we know that at resonance [tex]X_L=X_C[/tex]
putting [tex]X_L=X_C[/tex] in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
[tex]Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}[/tex] so from given expression it is clear that quality factor depends on R L and C
What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
Answer:
Weight, w = mg
Mass is an intrinsic property.
Explanation:
Mass is the measure of amount of matter in an object. It is an intrinsic that is unchanging property of a body. Mass cannot be destroyed nor be created.
Weight is the product of mass and acceleration due to gravity.
It changes with value of acceleration due to gravity. That is weight in Earth not equal to weight in moon since the value of acceleration due to gravity is different. So, weight is not an intrinsic property. It is a changing property for a body.
Weight, w = mg, where m is the mass and g is the acceleration due to gravity value.
Mass is an intrinsic and constant property of a body representing its matter content, while weight, the force of gravity on a body, varies with the gravitational environment. Despite common misuse in everyday language, they are distinct concepts in physics.
Explanation:Mass and weight are two different but closely related concepts. Mass is an intrinsic property of a body, representing the amount of matter it contains, and it remains constant regardless of the body's location, be it on Earth, the moon, or in orbit.
On the other hand, weight is the force exerted on a body due to gravity. It's a product of the body's mass and the acceleration due to gravity, represented by the formula 'Weight = Mass x Gravity'. Unlike mass, weight changes with the gravitational environment. For example, a person's weight on the moon is only one-sixth of their weight on Earth due to the moon's lower gravity, while their mass remains the same.
In everyday language, mass and weight are often used interchangeably, but in the field of physics, it is important to distinguish between them. For example, when we refer to our 'weight' in kilograms, we're technically referring to our mass. The proper unit of weight, consistent with its definition as a force, is the newton in the International System of Units (SI).
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Taking the speed of light in vacuum to be 3.000 x 10^8 m/s, find the speed of light in: a. air b. diamond c. crown glass d. water Data: nair =1.0003; ndiamond = 2.420; nwater = 1.340 ncrown glass = 1.500
Explanation:
The speed of light in vacuum is, c = 3 × 10⁸ m/s
We have to find the speed of light :
(a) In air :
[tex]n_1_{air}=1.0003[/tex]
The equation of refractive index is given as :
[tex]n_1=\dfrac{c}{v_1}[/tex]
[tex]v_1=\dfrac{c}{n_1}[/tex]
[tex]v_1=\dfrac{3\times 10^8\ m/s}{1.0003}[/tex]
[tex]v_1=299910026.9\ m/s[/tex]
[tex]v_1=2.99\times 10^8\ m/s[/tex]
(b) In diamond :
[tex]n_2_{diamond}=2.42[/tex]
The equation of refractive index is given as :
[tex]n_2=\dfrac{c}{v_2}[/tex]
[tex]v_2=\dfrac{c}{n_2}[/tex]
[tex]v_2=\dfrac{3\times 10^8\ m/s}{2.42}[/tex]
[tex]v_2=123966942.1\ m/s[/tex]
[tex]v_2=1.23\times 10^8\ m/s[/tex]
(c) In crown glass :
[tex]n_3_{glass}=1.5[/tex]
The equation of refractive index is given as :
[tex]n_3=\dfrac{c}{v_3}[/tex]
[tex]v_3=\dfrac{c}{n_3}[/tex]
[tex]v_3=\dfrac{3\times 10^8\ m/s}{1.5}[/tex]
[tex]v_3=200000000\ m/s[/tex]
[tex]v_3=2\times 10^8\ m/s[/tex]
(4) In water :
[tex]n_4_{glass}=1.34[/tex]
The equation of refractive index is given as :
[tex]n_4=\dfrac{c}{v_4}[/tex]
[tex]v_4=\dfrac{c}{n_4}[/tex]
[tex]v_4=\dfrac{3\times 10^8\ m/s}{1.34}[/tex]
[tex]v_4=223880597.01\ m/s[/tex]
[tex]v_4=2.23\times 10^8\ m/s[/tex]
Hence, this is the required solution.
Calculate the speed of an electron that has fallen through a potential difference of
(a) 125 volts and
(b) 125 megavolts.
Explanation:
We need to find the speed of an electron that has fallen through a potential difference of 125 volts. It can be calculated using the De-broglie hypothesis as :
(a) V = 125 volts
[tex]\dfrac{1}{2}mv^2=qV[/tex]
Where
v = speed of electron
V is potential difference
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 125\ V}{9.1\times 10^{-31}}}[/tex]
v = 6629935.44 m/s
[tex]v=6.62\times 10^6\ m/s[/tex]
(b) V = 125 megavolts
[tex]V=1.25\times 10^8\ V[/tex]
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.25\times 10^8\ V}{9.1\times 10^{-31}}}[/tex]
[tex]v=6.62\times 10^9\ m/s[/tex]
Hence, this is the required solution.
1. An object on Earth and the same object on the Moon would have a difference in
a. weight
b. mass
c. weight and mass
d. none of the above
2. How does doubling the mass of one object and tripling the distance between another object change the gravitational force between them?
FG = G M1 M2 / r2
a. Force changes by 2/3
b. Force changes by 2/9
c. Force increases by 9
d. Force decreases by 3
e. No change in force
3. According to the scientific definition of work, pushing on a rock accomplishes no work unless there is
a. a net force.
b. movement.
c. an opposing force.
d. movement in the same direction as the direction of the force.
4. A car going 30 mph has a kinetic energy of 10,000 Joules. How much kinetic energy does it have if it goes 60 mph?
a. 40,000 Joules
b. 10,000 Joules
c. 5,000 Joules
d. 2,500 Joules
e. 20,000 Joules
5. The specific heat of soil is 0.20 kcal/kgC° and the specific heat of water is 1.00 kcal/kgC°. This means that if 1 kg of soil and 1 kg of water each receive 1 kcal of energy, ideally,
a. the water will be 5°C.
b. the water will be warmer than the soil by 0.8°C.
c. the soil will be 5°C.
d. the water will warm by 1°C, and the soil will warm by 0.2°C.
Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.
Answer 1: a. weight
Mass and weight are very different concepts.
Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.
On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass [tex]m[/tex] of the body by the acceleration of gravity [tex]g[/tex]:
[tex]W=m.g[/tex]
Then, since the Earth and the Moon have different values of gravity, the weight of an object in each place will vary, but its mass will not.
Answer 2: b. Force changes by 2/9
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex] (1)
Where:
[tex]F[/tex] is the module of the force exerted between both bodies
[tex]G[/tex] is the universal gravitation constant
[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.
[tex]r[/tex] is the distance between both bodies
If we double the mass of one object (for example [tex]2m_{1}[/tex]) and triple the distance between both (for example [tex]3r[/tex]). The equation (1) will be rewritten as:
[tex]F=G\frac{2m_{1}m_{2}}{(3r)^2}[/tex] (2)
[tex]F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}[/tex] (3)
If we compare (1) and (2) we will be able to see the force changes by 2/9.
Answer 3: b. movement
The Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
[tex]W=(F)(d)[/tex]
Now, when they are not parallel, both directions form an angle, let's call it [tex]\alpha[/tex]. In that case the expression to calculate the Work is:
[tex]W=Fdcos{\alpha}[/tex]
Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).
Answer 4: a. 40,000 JoulesThe Kinetic Energy is given by:
[tex]K=\frac{1}{2}mV^{2}[/tex] (4)
Where [tex]m[/tex] is the mass of the body and [tex]V[/tex] its velocity
For the first case (kinetic energy [tex]K_{1}=10000J[/tex] for a car at [tex]V_{1}=30 mph=13.4112m/s[/tex]):
[tex]K_{1}=\frac{1}{2}mV_{1}^{2}[/tex] (5)
Finding [tex]m[/tex]:
[tex]m=\frac{2K_{1}}{V_{1}^{2}}[/tex] (6)
[tex]m=\frac{2(10000J)}{(13.4112m/s)^{2}}[/tex] (7)
[tex]m=111.197kg[/tex] (8)
For the second case (unknown kinetic energy [tex]K_{2}[/tex] for a car with the same mass at [tex]V_{2}=60 mph=26.8224m/s[/tex]):
[tex]K_{2}=\frac{1}{2}mV_{2}^{2}[/tex] (9)
[tex]K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}[/tex] (10)
[tex]K_{2}=40000J[/tex] (11)
Answer 5: c. the soil will be 5°C
The formula to calculate the amount of calories [tex]Q[/tex] is:
[tex]Q=m. c. \Delta T[/tex] (12)
Where:
[tex]m[/tex] is the mass
[tex]c[/tex] is the specific heat of the element. For water is [tex]c_{w}=1 kcal/g\°C[/tex] and for soil is [tex]c_{s}=0.20 kcal/g\°C[/tex]
[tex]\Delta T[/tex] is the variation in temperature (the amount we want to find for both elements)
This means we have to clear [tex]\Delta T[/tex] from (12) :
[tex]\Delta T=\frac{Q}{m.c} [/tex] (13)
For Water:
[tex]\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}} [/tex] (14)
[tex]\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}[/tex] (15)
[tex]\Delta T_{w}=1\°C)}[/tex] (16)
For Soil:
[tex]\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}} [/tex] (17)
[tex]\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}[/tex] (18)
[tex]\Delta T_{s}=5\°C)}[/tex] (19)
Hence the correct option is c.
Answer:
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Two cars are travelling with the same speed and the drivers hit the brakes at the same time. The deceleration of one car is a quarter that of the other. By what factor do the distances required for two cars to come to a stop differ?
Answer:
The ratio of stopping distances is 4 i.e by a factor 4 the stopping distances differ
Explanation:
Using 3rd equation of motion we have
For car 1
[tex]v_{1}^{^{2}}=u_{1}^{2}+2a_{1}s_{1}[/tex]
For car 2 [tex]v_{2}^{^{2}}=u_{2}^{2}+2a_{2}s_{2}[/tex]
Since the initial speed of both the cars are equal and when the cars stop the final velocities of both the cars become zero thus the above equations reduce to
[tex]u^{2}=-2a_{1}s_{1}\\\\s_{1}=\frac{-u^{2}}{2a_{1}}[/tex].............(i)
Similarly for car 2 we have
[tex]s_{2}=\frac{-u^{2}}{2a_{2}}[/tex]..................(ii)\
Taking ratio of i and ii we get
[tex]\frac{s_{1}}{s_{2}}=\frac{a_{2}}{a_{1}}[/tex]
Let
[tex]\frac{a_{2}}{a_{1}}=4[/tex]
Thus
[tex]\frac{s_{1}}{s_{2}}=4[/tex]
The ratio of stopping distances is 4
Final answer:
The stopping distances of two cars with different deceleration rates will differ by the inverse ratio of their decelerations. If one car's deceleration is a quarter of the other's, the car with the lower deceleration will require four times the stopping distance of the other car.
Explanation:
The question at hand is about how the stopping distances of two cars differ when their deceleration rates are different. If one car's deceleration is a quarter of the other's, then to find the factor by which the stopping distances differ, we can use the equation of motion [tex]v^2 = u^2 + 2as[/tex] v is the final velocity, u is the initial velocity, a is the deceleration, and s is the stopping distance.
Since the final velocity (v) for both cars will be zero (they come to a stop), and assuming the initial velocities (u) are the same for both cars, we can set the equation for both as follows: 0 = u^2 + 2a1s1 and 0 = u^2 + 2a2s2, with a1 being the higher deceleration of the first car and a2 being a quarter of that, s1 and s2 being the stopping distances respectively.
When we solve the equations for s1 and s2, we find that s1 is proportional to 1/a1 and s2 is proportional to 1/a2, so if a2 = 1/4 a1, then the ratio of the stopping distances s2/s1 is equal to the inverse ratio of the decelerations, which is 4. Hence, the car with the lower deceleration requires four times the distance to stop compared to the car with the higher deceleration.
Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreased to d/4. How does the gravitational force between these two objects change as a result of the decrease?
Answer:
Increased by 16 times
Explanation:
F = Gravitational force between two bodies
G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²
m₁ = Mass of one body
m₂ = Mass of other body
d = distance between the two bodies
[tex]F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}[/tex]
[tex]F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F[/tex]
∴Force will increase 16 times
The weight of a product is measured in pounds. A sample of 50 units is taken from a recent production. The sample yielded X⎯⎯⎯ = 75 lb, and we know that σ2 = 100 lb. Calculate a 99 percent confidence interval for μ.
Answer:
The 99% confidence interval for the weights = [71.36lb, 78.64lb]
Explanation:
Mean weight
[tex]\bar{x} =75 lb[/tex]
Variance of weights
[tex]\sigma^2 =100lb[/tex]
Standard deviation,
[tex]\sigma =\sqrt{100}=10lb[/tex]
Confidence interval is given by
[tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]
For 99% confidence interval Z = 2.576,
Number of weights, n = 50
Substituting
[tex]75-2.576\times \frac{10}{\sqrt{50}}\leq \mu\leq 75+2.576\times \frac{10}{\sqrt{50}}\\\\71.36\leq \mu\leq 78.64[/tex]
The 99% confidence interval for the weights = [71.36lb, 78.64lb]
The 99 percent confidence interval for the population mean (μ) can be calculated using the formula X ± Z*(σ/√n) where X is the sample mean, Z is the Z-score for a 99% confidence interval (approximately 2.57), σ is the standard deviation of the population and n is the sample size. Substituting the given values results in 75 ± 2.57*(10/√50).
Explanation:To calculate a 99 percent confidence interval for the population mean (μ), you can use the formula:
X± Z*(σ/√n)
Where:
X is the sample mean, which is given as 75 pounds.Z is the Z-score associated with the desired confidence level. For a 99% confidence interval, the Z-score is approximately 2.57. This can be found using the standard normal distribution table or online Z-score calculator.σ is the standard deviation of the population, the square root of given variance 100 lb, which is 10 lb.n is the sample size, which is given as 50.Substituting these values into the formula gives the following 99 percent confidence interval:
75 ± 2.57*(10/√50).
Calculate the range and you will find your answer for the 99 percent confidence interval for μ.
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wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right has a mass of 60 kg and is sitting 1.8 m from the pivot point. What is the net torque on the system?
Answer:
4.5 Nm (Anticlockwise)
Explanation:
Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.
Anticlockwise Torque = 75 x 1.5 = 112.5 Nm
clockwise Torque = 60 x 1.8 = 108 Nm
Net torque = Anticlockwise torque - clockwise torque
Net Torque = 112.5 - 108 = 4.5 Nm (Anticlockwise)
It takes 0.16 g of helium (He) to fill a balloon. How many grams of nitrogen (N2) would be required to fill the balloon to the same pressure, volume, and temperature?
Answer: The mass of nitrogen gas required to fill the balloon is 1.12 grams.
Explanation:
We are given:
Mass of helium gas = 0.16 g
We need to calculate the mass of nitrogen gas that can fill the balloon at same pressure, volume and temperature. This means that the moles of both the gases filling up balloon will be same.
So, to calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of Helium = 0.16 g
Molar mass of helium = 4.00 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of helium}=\frac{0.16g}{4g/mol}=0.04mol[/tex]
Now, calculating the mass of nitrogen gas using same above equation, we get:
Moles of nitrogen gas = 0.04 moles
Molar mass of nitrogen gas = 28.01 g/mol
Putting values in above equation, we get:
[tex]0.04mol=\frac{\text{Mass of }N_2}{28.01g/mol}\\\\\text{Mass of }N_2=1.12g[/tex]
Hence, the mass of nitrogen gas required to fill the balloon is 1.12 grams.
To fill the balloon with nitrogen (N2) under the same conditions as helium (He), you would need 1.12 grams, based on the molar masses of each gas and Avogadro's law.
Explanation:The subject matter of your question lies in the field of chemistry, specifically the concepts of Avogadro's law and the ideal gas law. Avogadro's law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. Therefore, given that N2 and He are both gases, the same volume of each gas will contain the same number of molecules.
The ideal gas law can also be applied here. The molar mass of nitrogen (N₂) is 28.01 g/mol, and the molar mass of helium (He) is 4 g/mol. Given that we know the mass of helium required to fill the balloon (0.16 g), we can use the relationship between molar mass and the actual mass to determine the equivalent mass for nitrogen.
This relationship is represented by the ratio of the molar mass of nitrogen to the molar mass of helium, which is 28.01 / 4 = 7. Thus, to fill the balloon with nitrogen under the same conditions of pressure, volume, and temperature, the required mass would be 7 times that of the mass of the helium, or 0.16 * 7 = 1.12 g.
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The diameter of a 12-gauge copper wire is about 0.790 mm. If the drift velocity of the electrons is 3.25 mm/s what is the electron current in the wire? The number of electron carriers in 1.0 cm3 of copper is 8.5 × 1022.
Answer:
21.6 A
Explanation:
n = number density of free electrons in copper = 8.5 x 10²² cm⁻³ = 8.5 x 10²⁸ m⁻³
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
d = diameter of copper wire = 0.790 mm = 0.790 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.790 x 10⁻³)²
A = 4.89 x 10⁻⁷ m²
v = drift speed = 3.25 mm/s = 3.25 x 10⁻³ m /s
the electric current is given as
i = n e A v
i = (8.5 x 10²⁸) (1.6 x 10⁻¹⁹) (4.89 x 10⁻⁷ ) (3.25 x 10⁻³)
i = 21.6 A
After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 5.50 kg of ice and the temperature of the ground was 43.0 °C, calculate the energy that is required to melt all the ice at 0 °C. The heat of fusion for water is 80.0 cal/g.
Answer:
[tex]Q = 4.40 \times 10^5 Cal[/tex]
Explanation:
Here we know that initial temperature of ice is given as
[tex]T = 0^o C[/tex]
now the latent heat of ice is given as
[tex]L = 80 Cal/g[/tex]
now we also know that the mass of ice is
[tex]m = 5.50 kg[/tex]
so here we know that heat required to change the phase of the ice is given as
[tex]Q = mL[/tex]
[tex]Q = (5.50 \times 10^3)(80)[/tex]
[tex]Q = 4.40 \times 10^5 Cal[/tex]
The total energy required to melt all the 5.50 kg (or 5500g) of ice is 440,000 calories. This is calculated using the given heat of fusion (Lf) value of 80 cal/g for water and the heat exchange formula, Q=mLf.
Explanation:To calculate the amount of energy required to melt all the ice, we need to use the formula for heat exchange: Q = mLf. Where Q is the heat required, m is the mass, and Lf is the heat of fusion. Given that Lf is 80 cal/g for water, the mass m is 5.50 kg (or 5500g), and you are trying to find Q, you can simply replace the known quantities into the equation:
Q = (5500g) * (80cal/g)
So, the total energy required to melt all the ice is 440,000 calories. This heat is absorbed by the ice, providing the energy required to break the intermolecular bonds in the ice and facilitate the phase transition from solid ice to liquid water. The energy required for a phase change like this is significant, explaining why ice can take a while to melt even on a hot summer day.
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In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second piston with a radius of 10.8 cm. a) How large a force must the compressed air exert to lift a 1.44 × 104 N car
Answer:
847.45 N
Explanation:
F₁=force of exerted by smaller piston
F₂=force of exerted by larger piston=1.44×10⁴ N
A₁=Area of smaller piston= 2.62 cm =0.0265 m
A₂=Area of larger piston= 10.8 cm =0.108 m
Pressure exerted by both the pistons will be equal
[tex]P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times 0.108^2}\pi\times 0.0262^2\\\Rightarrow F_1=847.45\ N[/tex]
Hence, force exerted to lift a 14400 N car is 847.45 N
Assume the amplitude of the electric field in a plane electromagnetic wave is E1 and the amplitude of the magnetic field is B1. The source of the wave is then adjusted so that the amplitude of the electric field doubles to become 2E1. (i) What happens to the amplitude of the magnetic field in this process? It becomes four times larger. It becomes two times larger. It can stay constant. It becomes one-half as large. It becomes one-fourth as large. (ii) What happens to the intensity of the wave? It becomes four times larger. It becomes two times larger. It can stay constant. It becomes one-half as large. It becomes one-fourth as large. Need Help?
Answer:
I) It becomes two times larger
II) It becomes four times larger
Explanation:
I) Electric field is directly proportional to Magnetic field and as such, if one is increased, the other is also increased by the same proportion.
The formula is given as
[tex]E_{1} = cB_{1}[/tex]
c = speed of light, therefore
[tex]2E_{1} = c2B_{1}[/tex]
II) Intensity of is proportional to the square of amplitude, as such
If amplitude is doubled ([tex]X2^{2}[/tex]), intensity is [tex]X4^{}[/tex]
This means the intensity becomes four times bigger.
An electron (mass m=9.11×10−31kg) is accelerated from the rest in the uniform field E⃗ (E=1.45×104N/C) between two thin parallel charged plates. The separation of the plates is 1.90 cm . The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, with what speed does it leave the hole?
Explanation:
It is given that,
Mass of an electron, [tex]m=9.11\times 10^{-31}\ kg[/tex]
Electric field, [tex]E=1.45\times 10^4\ N/C[/tex]
Separation between the plates, d = 1.9 cm = 0.019 m
The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. We need to find the speed of the electron as it leave the hole.
The force due to accelerating electron is balanced by the electrostatic force i.e
qE = ma
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\ C\times 1.45\times 10^4\ N/C}{9.11\times 10^{-31}\ kg}[/tex]
[tex]a=2.54\times 10^{15}\ m/s^2[/tex]
Let v is the speed as it leave the hole. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]v=\sqrt{2ad}[/tex]
[tex]v=\sqrt{2\times 2.54\times 10^{15}\times 0.019\ m}[/tex]
v = 9824459.27 m/s
or
[tex]v=9.82\times 10^6\ m/s[/tex]
So, the speed of the electron as it leave the hole is [tex]9.82\times 10^6\ m/s[/tex]. Hence, this is the required solution.