Answer:
(a). The critical angle for the liquid when surrounded by air is 44.37°
(b). The angle of refraction is 26.17°.
Explanation:
Given that,
Incidence angle = 26.7°
Refraction angle = 18.3°
(a). We need to calculate the refraction of liquid
Using Snell's law
[tex]n=\dfrac{\sin i}{\sin r}[/tex]
Put the value into the formula
[tex]n=\dfrac{\sin 26.7}{\sin 18.3}[/tex]
[tex]n=1.43[/tex]
We need to critical angle for the liquid when surrounded by air
Using formula of critical angle
[tex]C=\sin^{-1}(\dfrac{1}{n})[/tex]
Put the value into the formula
[tex]C=\sin^{-1}(\dfrac{1}{1.43})[/tex]
[tex]C=44.37^{\circ}[/tex]
(b). Given that,
Incidence angle = 37.5°
Speed of light in mineral [tex]v=2.17\times10^{8}\ m/s[/tex]
We need to calculate the index of refraction
Using formula of index of refraction
[tex]n=\dfrac{c}{v}[/tex]
Put the value into the formula
[tex]n=\dfrac{3\times10^{8}}{2.17\times10^{8}}[/tex]
[tex]n=1.38[/tex]
We need to calculate the angle of refraction
Using Snell's law
[tex]n=\dfrac{\sin i}{\sin r}[/tex]
[tex]\sin r=\dfrac{\sin i}{n}[/tex]
Put the value into the formula
[tex]\sin r=\dfrac{\sin 37.5}{1.38}[/tex]
[tex]r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})[/tex]
[tex]r=26.17^{\circ}[/tex]
Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°
(b). The angle of refraction is 26.17°.
Answer
a) Angle of incidence i = 26.7°
Angle of refraction r = 18.3°
From Snell’s law index of refraction of the liquid
[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]
[tex]n = \dfrac{sin\ 26.7^0}{sin\ 18.3^0}[/tex]
n = 1.43
So, critical angle
[tex]C= sin^{-1}(\dfrac{1}{n})[/tex]
[tex]C= sin^{-1}(\dfrac{1}{1.43})[/tex]
C = 44.33°
b) Angle of incidence, i = 37.5°
speed of light in mineral oil , v = 2.17 x 10⁸ m/s
speed of light in air, c = 3 x 10⁸ m/s
refractive index of the oil
[tex]n = \dfrac{c}{v}[/tex]
[tex]n = \dfrac{3\times 10^8}{2.17\times 10^8}[/tex]
n = 1.38
again using Snell's law
[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]
[tex]sin\ r = \dfrac{sin\ i}{n}[/tex]
[tex]sin\ r = \dfrac{sin\ 37.5^0}{1.38}[/tex]
[tex] r = sin^{-1}(0.441)[/tex]
r = 26.18°
hence, the angle of refraction is equal to r = 26.18°
Measure the distance from the drop point in Brazil to the drop point in Angola. Use that number in your calculation. Given that this portion of Pangaea broke apart 200,000,000 years ago, calculate how fast South America and Africa are separating in cm/year? (Hint: Speed= Distance/Time)
To develop this point we will start by finding the approximate coordinates of the points that were connected at the time of the Pangaea between Brazil and Angola. These coordinates are presented below.
1 . Brazil - Latitude: 18 0 07’ 55.56” S Longitude: 39 0 35’ 14.50” W 2.
Angola - Latitude: 9 0 08’ 50.02” S Longitude: 13 0 02’ 32.11” E
Using a tool for calculating distances between these two points we will notice that its distance is 576,155,570.12 cm
Applying the equation given in the statement we will have to,
[tex]v = \frac{x}{t} \rightarrow v = Velocity, x = Distance, t = Time[/tex]
200,000,000 years have passed and the movement was previously found, so the speed of travel is,
[tex]v = \frac{576,155,570.12cm}{200.000.000 years}[/tex]
[tex]v = 2.88 cm/year[/tex]
Therefore the velocity is 2.88 cm per year.
Tim and Rick both can run at speed v_r and walk at speed v_w, with v_r > v_w They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.How long does it take Rick to cover the distance D?Express the time taken by Rick in terms of v_r, v_w, and D.
Answer:
The time taken by Rick is Δt = (D/2 / v_w ) + (D/2 / v_r)
Explanation:
Hi there!
The equation of average velocity (v) is the following:
v = Δx / Δt
Where:
Δx = traveled distance.
Δt = elapsed time.
During the first half of Rick´s journey, the average velocity can be written as follows:
v_w = D/2 / Δt1
Solving for Δt1:
Δt1 = D/2 / v_w
For the second half of the trip:
v_r = D/2 / Δt2
Δt2 = D/2 / v_r
The time it takes Rick to cover the distance D will be equal to Δt1 + Δt2
Δt = Δt1 + Δt12
Δt = (D/2 / v_w ) + (D/2 / v_r)
The time taken by Rick is Δt = (D/2 / v_w ) + (D/2 / v_r)
A spacecraft is in a 400-km by 600-km LEO. How long (in minutes) does it take to coast from the perigee to the apogee?
To solve this problem we will use Kepler's third law for which the period is defined as
[tex]T= 2\pi \sqrt{\frac{a^3}{GM}}[/tex]
The perigee altitude is the shortest distance between Earth's surface and Satellite.
The average distance of the perigee and apogee of a satellite can be defined as
[tex]a = R+\frac{r_1+r_2}{2}[/tex]
Here,
R = Radius of Earth
[tex]r_1[/tex]= Lower orbit
[tex]r_2[/tex]= Higher orbit
Replacing we have,
[tex]a = 6378.1+\frac{400+600}{2}= 6878.1km[/tex]
Time Taken to fly from perigee to apogee equals to half of orbital speed is
[tex]t = \frac{T}{2}[/tex]
[tex]t = \pi \sqrt{\frac{a^3}{GM}}[/tex]
Replacing,
[tex]t =\pi \sqrt{\frac{(6878.1*10^3)^3}{(6.67*10^{-11})(6*10^{24})}}[/tex]
[tex]t = 2832.79s[/tex]
[tex]t = \text{47min 12.8s}[/tex]
Therefore will take around to 47 min and 12.8s to coast from perigee to the apogee.
Final answer:
The task requires understanding of orbital mechanics and Kepler's laws to calculate the coasting time from perigee to apogee in a specific orbit, but the information provided is not sufficient for a direct calculation without additional details on velocity or using specific orbital mechanics equations.
Explanation:
The question asks about the time it takes for a spacecraft to coast from the perigee (the point in the orbit closest to Earth) to the apogee (the point in the orbit farthest from Earth) in a low Earth orbit (LEO) with altitudes of 400 km and 600 km respectively. To solve this, we would typically use Kepler's laws and orbital mechanics equations. However, the information provided does not directly enable the calculation of the coasting time without additional steps or assumptions about the spacecraft's orbit, such as its velocity, the mass of the Earth, and orbital mechanics principles that allow for the calculation of orbital period. An approximate method to find the orbital period (the total time for one complete orbit) involves using the semi-major axis of the orbit, which can be derived from the given perigee and apogee. However, without the specific velocities or applying advanced physics formulas, we can't provide an accurate answer.
Two small plastic spheres are given positive electrical charges. When they are 16.0 cm apart, the repulsive force between them has magnitude 0.200 N.
a)What is the charge on each sphere if the two charges are equal?
b)What is the charge on each sphere if one sphere has four times the charge of the other?
The smaller charge is approximately **3.37 x 10⁻⁷ C** and the larger charge is approximately **1.35 x 10⁻⁶ C**.
These are approximate values due to rounding during calculations.
Solving for the Charges on the Spheres:
Case (a): Equal Charges
1. **Apply Coulomb's Law:** The force between two charged objects is given by Coulomb's Law:
[tex]$$F = k \cdot \frac{q_1 \cdot q_2}{r^2}$$[/tex]
where:
* F is the force (0.200 N)
* k is Coulomb's constant (8.99 x 10^9 N m²/C²)
* q₁ and q₂ are the charges on the spheres (which are equal in this case)
* r is the distance between the spheres (0.16 m)
2. **Plug in the values and solve for q₁:**
[tex]$$0.200 = 8.99 \times 10^9 \cdot \frac{q_1^2}{(0.16)^2}$$$$q_1^2 = \frac{0.200 \cdot (0.16)^2}{8.99 \times 10^9}$$$$q_1 = \sqrt{ \frac{0.200 \cdot (0.16)^2}{8.99 \times 10^9}} \approx 7.54 \times 10^{-7} \text{ C}$$[/tex]
Therefore, the charge on each sphere is approximately **7.54 x 10⁻⁷ C**.
Case (b): One Sphere has Four Times the Charge
1. **Let q₁ be the smaller charge and q₂ be the larger charge:** We know q₂ = 4q₁.
2. **Apply Coulomb's Law again:
[tex]$$0.200 = 8.99 \times 10^9 \cdot \frac{q_1 \cdot q_2}{(0.16)^2}$$$$0.200 = 8.99 \times 10^9 \cdot \frac{q_1 \cdot (4q_1)}{(0.16)^2}$$3. **Substitute and solve for q₁:**$$0.200 = 8.99 \times 10^9 \cdot \frac{16q_1^2}{(0.16)^2}$$$$q_1^2 = \frac{0.200 \cdot (0.16)^2}{8.99 \times 10^9 \cdot 16}$$$$q_1 = \sqrt{ \frac{0.200 \cdot (0.16)^2}{8.99 \times 10^9 \cdot 16}} \approx 3.37 \times 10^{-7} \text{ C}$$4. **Find the larger charge (q₂):**$$q_2 = 4q_1 = 4 \cdot (3.37 \times 10^{-7} \text{ C}[/tex]
If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 50 N·m, what minimum force must be exerted by the mechanic at the end of a 22 cm-long wrench to loosen the nut? Answer in units of N.
Answer:
227.27 N
Explanation:
Torque: This can be defined as the force that tend to cause an object to twist or rotate. the torque of a body is also called moment of a body. The S.I unit of torque is N.m.
Mathematically
T = F×d ....................... Equation 1
Making F the subject of the equation,
F = T/d ..................... Equation 2
T = torque required to loosen the nut, F = force applied , d = distance between the force and the center of the nut.
Given: T = 50 N.m, d = 22 cm = (22/100) = 0.22 m.
F = 50/0.22
F = 227.27 N.
Thus the minimum force required to loosen the nut = 227.27 N
I’m not accelerating, so the net (vertical) force on me, while I’m sitting here doing this lab is _________
Answer:
Net force is zero
Explanation:
According to the Newton's second law, the net force on the body is equal to the product of mas of body and the acceleration.
here acceleration is equal to zero so net force is also zero because mass of an object can never be zero.
The displacement of a wave traveling in the positive x-direction is:
y(x,t) = (3.5cm) cos (2.7x−92t), where x is in m and t is in s.
Part A) What is the frequency of this wave?
Part B) What is the wavelength of this wave?
Part C) What is the speed of this wave?
The frequency, wavelength, and speed of the wave are equal to 14.65 Hz, 0.4299m, and 6.298 Hz.
What are frequency and wavelength?The frequency can be described as the number of oscillations or cycles in 1 second. The SI unit of frequency has per second or hertz.
The wavelength can be defined as the distance between the two adjacent crests or troughs on a wave that is separated by a distance.
The standard equation of wave can be expressed as:
[tex]{\displaystyle y(x,t) = Acos (\frac{2\pi x}{\lambda} \pm 2\pi ft )[/tex]
Given the equation of the wave: y(x,t) = (3.5cm) cos (2.7x−92t)
Therefore, the wavelength can be calculated as:
2π/ λ = 2.7
λ = 2.7/2π
λ = 0.4299 m
The frequency of the given wave can be calculated as:
2πf = 92
f = 92/2π
f = 14.65 Hz
The speed of the wave can determine from the above-mentioned relationship:
V = νλ
V = 14.65 × 0.4299
V = 6.29 m/s
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Which is the most appropriate units for expressing angular velocity in kinematics problems?
Answer:
[tex]\frac{rad}{s}[/tex]
Explanation:
The most appropriate units for expressing the angular velocity are the [tex]\frac{rad}{s}[/tex], since they are the units assigned for this magnitude in the International System of Units. Which is derived from the fact that radians are dimensionless and the second is the unit assigned for time in this unit system.
Final answer:
The most appropriate units for expressing angular velocity in kinematics problems are radians per second (rad/s). This unit arises from the definition of angular velocity as the change in angular position (ΔΘ) over time (Δt).
Explanation:
When discussing angular velocity in kinematics problems, the most appropriate unit to express it is radians per second (rad/s). Angular velocity is the rate of change of the angular position over time, defined as ΔΘ/Δt. For example, if a disc makes one-fourth of a revolution in 0.0150 seconds, its angular velocity would be π/2 rad / 0.0150 s, equaling 104.7 rad/s.
In situations where the rotation rate is given in revolutions per second or cycles per second, we can convert it to angular velocity by multiplying by 2π rad, as one complete revolution is equivalent to 2π radians. It is also important to note that the centripetal acceleration in circular motion is related to the angular velocity, and the units are critical in deriving related kinematics equations.
In the deep space between galaxies, the density of atoms is as low as 106 atoms/m3, and the temperature is a frigid 2.7 K. What is the pressure (in Pa)?
Answer: 3.73 × 10^-17 Pa
Explanation:
N/V= 10^6 atom/m^3
T=2.7k
Kb=1.38 ×10^-23 J/K
NA= 6.02 × 10^23 mol^-1
R= 8.31J/mol.K
PV= NaKbT
PV= N/V × KbT
P= 10^6 × 1.38 × 10^-23× 2.7
Pressure= 3.73×10^-17 Pa
The pressure in pascal of the deep space whose density of atoms is as low as 10⁶ atoms/m3 is 3.73 × 10-¹⁷ Pa.
How to calculate pressure?The pressure of a space can be calculated using the following expression:
PV= N/V × KbT
Where;
P = pressureV = volumeKb = temperature constantT = temperatureR = gas law constantN/V = 10⁶ atom/m³T = 2.7KKb = 1.38 ×10^-23 J/KNA= 6.02 × 10²³ mol-¹R= 8.31J/mol.KP = N/V × KbT
P= 10⁶ × 1.38 × 10-²³ × 2.7
Pressure = 3.73 × 10-¹⁷ Pa
Therefore, the pressure in pascal of the deep space whose density of atoms is as low as 10⁶ atoms/m3 is 3.73 × 10-¹⁷ Pa.
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A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 4.7-m-deep valley, then climbs to the top of a hill that is 5.4 m above the first section of track. Assume any effects of friction or of air resistance are negligible.
(a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill?
(b) Can we affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom?
Explanation:
From the curve given in the attachment,
a) minimum speed v0 required if the car is to travel beyond the top of the hill
climb to the top = 5.4 m
Therefore, [tex]V_0=\sqrt{2gh}[/tex]
[tex]V_0=\sqrt{2(9.81)(5.4)}[/tex] = 10.29 m/s
b) No, we cannot affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom as this speed does not depend upon depth of the valley.
A gadget of mass 21.85 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up into 3 fragments flying away from each other. The first fragment has mass m1 = 6.42 kg and speed v1 = 6.8 m/s while the second fragment has mass m2 = 8.26 kg and speed v2 = 3.54 m/s. The angle between the velocity vectors ~v1 and ~v2 is θ12 = 64 ◦ . What is the speed v3 of the third fragment? Answer in units of m/s.
To solve this problem we will apply the concepts related to the conservation of momentum. For this purpose we will determine the velocities in the three body in the vertical and horizontal components. Once the system of equations is obtained, we will proceed to find the angle and the speed at which the third fragment is directed.
Mass of third part is
[tex]m_3 = m-(m_1+m_2)[/tex]
[tex]m_3= 21.85-6.42-8.26[/tex]
[tex]m_3 =7.17 kg[/tex]
Assume that [tex]m_1[/tex] is along X-axis we have that [tex]m_2[/tex] makes an angle is 64 degrees with x-axis and [tex]m_3[/tex] makes an angle [tex]\theta[/tex] with x-axis.
Using law of conservation of momentum along X-axis
[tex]0 = (6.42*6.8)+(8.26*3.54*cos(64))+(7.71v_3 cos\theta)[/tex]
[tex](7.71v_3 cos\theta) = 56.4741[/tex]
[tex]v_3 cos\theta = 7.3247[/tex] [tex]\rightarrow \text{Equation 1}[/tex]
Now applying the same through the Y-axis.
[tex]0=0+8.26*3.54*sin(64\°) + 7.71*v_3*sin\theta[/tex]
[tex]-8.26*3.54*sin(64\°)=7.71*v_3*sin\theta[/tex]
[tex]v_3*sin\theta = -3.409[/tex] [tex]\rightarrow \text{Equation 2}[/tex]
If we divide the equation 1 with the equation 2 we have that
[tex]\frac{v_3cos\theta}{v_3 sin\theta } = \frac{7.3247}{-3.409}[/tex]
[tex]tan\theta = \frac{7.3247}{-3.409}[/tex]
[tex]\theta = tan^{-1} (\frac{7.3247}{-3.409})[/tex]
[tex]\theta = -65.04\°[/tex]
Using this angle in the second equation we have that velocity 3 is,
[tex]v_3 = \frac{-3.409}{sin(-65.04)}[/tex]
[tex]v_3 = 3.7601m/s[/tex]
Therefore the speed of the third fragment is [tex]3.7601\frac{m}{s} \angle -65.04\°[/tex]
Replacing an object attached to a spring with an object having 1 4 the original mass will change the frequency of oscillation of the system by a factor of?
Answer:
f = 2 f₀
Explanation:
A mass-spring system has angular velocity
w = √ k / m
The angular velocity is related to the frequency
w = 2π f
f = 1 /2π √k / m
If we change the mass for another that is ¼ of the initial mass
m = ¼ mo
We replace
f = 1 / 2π √(k 4 / mo)
f = (1 / 2π √k/m₀) √ 4
f = 2 f₀
In summary the frequency doubles from the initial frequency
The frequency of oscillation of the system by a factor of should be considered as the f = 2 f₀.
Frequency of oscillation:Since
A mass-spring system has angular velocity that applied the following equation
w = √ k / m
Also,
The angular velocity should be related to the frequency
w = 2π f
f = 1 /2π √k / m
Now
In the case when we change the mass for another that is ¼ of the initial mass
So,
m = ¼ mo
Now
f = 1 / 2π √(k 4 / mo)
f = (1 / 2π √k/m₀) √ 4
f = 2 f₀
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A chair is at rest on the floor. The chair absorbs thermal energy from the floor, and begins moving spontaneously with kinetic energy equal to the thermal energy absorbed. This process violates ___________.
A- Both the 1st and 2nd laws of thermodynamicsB- Only the 1st law of thermodynamicsC- Neither the 1st nor the 2nd law of thermodynamicsD- Only the 2nd law of thermodynamics
Answer:
D- Only the 2nd law of thermodynamics
Explanation:
It violates 2nd law because according to 2nd law of thermodynamics, it is impossible that the sole result of a process is is to absorb energy and do equivalent amount of work. so some heat must lose to surrounding which is not specified here. so it violates 2nd law.
so option D is correct
The process violates the 2nd law of thermodynamics. The chair absorbing thermal energy from the floor and beginning to move spontaneously with kinetic energy equal to the thermal energy absorbed implies a perfect conversion of heat into work, which is against the principle of entropy increase stated by the 2nd law.
Explanation:This process violates D- Only the 2nd law of thermodynamics.
The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system.
In the scenario described where a chair absorbs thermal energy from the floor and begins moving spontaneously with kinetic energy equal to the thermal energy absorbed, it would mean that 100% of the thermal energy has been converted into kinetic energy. This is a violation of the second law of thermodynamics because it implies a perfect conversion of heat energy into work with no increase in entropy, i.e., the thermal energy is fully converted into kinetic energy without any of it being 'wasted' or dispersed.
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A mosquito of mass 0.15 mg is found to be flying at a speed of 50 cm/s with an uncertainty of 0.5 mm/s. (a) How precisely may its position be known? (b) Does this inherent uncertainty present any hindrance to the application of classical mechanics?
(a) The uncertainty principle reveals that the position of a flying mosquito can be known with an extremely high precision that doesn't affect the application of classical mechanics. (b) The inherent uncertainty calculated is extremely small.
(a) The Heisenberg principle states that the more precisely the position (Δx) of a particle is known, the less precisely its momentum (Δp) can be known, and vice versa. This is quantitatively expressed as ΔxΔp ≥ ħ/2, where ħ is the reduced Planck's constant (approximately 1.055 × 10⁻³⁴ J⋅s).
Given the speed (v) of the mosquito is 50 cm/s with an uncertainty in velocity (Δv) of 0.5 mm/s, and the mass (m) of the mosquito is 0.15 mg, we first convert these to SI units: v = 0.5 m/s, Δv = 5 × 10⁻⁴ m/s, and m = 0.15 × 10⁻⁶ kg. The uncertainty in momentum, Δp, is mΔv = (0.15 × 10⁻⁶ kg)(5 × 10⁻⁴ m/s) = 7.5 × 10⁻¹¹ kg⋅m/s.
Using the uncertainty principle, Δx ≥ ħ / (2Δp), where Δp is the momentum uncertainty calculated above. Plugging in values, Δx ≥ (1.055 × 10⁻³⁴ J⋅s) / (2 × 7.5 × 10⁻¹¹ kg⋅m/s) ≈ 7.033 × 10⁻²⁵ meters. This calculation shows how precisely the mosquito's position can be known.
(b) The inherent uncertainty calculated is extremely small, particularly when dealing with macroscopic objects like a mosquito. Therefore, this uncertainty does not present any hindrance to the application of classical mechanics, which comfortably applies at the scale of everyday objects.
We experience fictitious forces due to: a. Rotation of a reference frame b. Inertial reference frames c. Translational motion d. Universal gravitation.
Answer:
A.
Explanation:
A fictional force (also called force of inertia, pseudo-force, or force of d'Alembert, 5), is a force that appears when describing a movement with respect to a non-inertial reference system, and that therefore it does not correspond to a genuine force in the context of the description of the movement that Newton's laws are enunciated for inertial reference systems.
The forces of inertia are, therefore, corrective terms to the real forces, which ensure that the formalism of Newton's laws can be applied unchanged to phenomena described with respect to a non-inertial reference system. The correct answer is A.
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s.
To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer [tex]f_{obs}[/tex] is more than the frequency emitted by the source. The expression to find the frequency received by the person is,
[tex]f_{obs} = f_s (\frac{v_w}{v_w-v_s})[/tex]
[tex]f_s[/tex]= Frequency of the source
[tex]v_w[/tex]= Speed of sound
[tex]v_s[/tex]= Speed of source
The velocity of the ambulance is
[tex]v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]v_s = 30.55m/s[/tex]
Replacing at the expression to frequency of observer we have,
[tex]f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})[/tex]
[tex]f_{obs} = 878Hz[/tex]
Therefore the frequency receive by observer is 878Hz
The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are used, the second-order Bragg reflection from a set of parallel planes in a(n) gold crystal is observed at an angle θ = 20.62°. If the spacing between these planes corresponds to the unit cell length (d = a), calculate the radius of a(n) gold atom.
Answer:
r = 1.45 Å
Explanation:
given,
λ = 1.436 Å
θ = 20.62°
d = a
n = 2
metal gold crystallizes in a face centered cubic unit cell
Radius of the gold atom = ?
using Bragg's Law
n λ = 2 d sin θ
2 x 1.436 Å = 2 a sin 20.62°
a = 4.077 Å
We know relation of radius for face centered cubic unit cell
[tex]a = \dfrac{4r}{\sqrt{2}}[/tex]
[tex]4.077= \dfrac{4\times r}{\sqrt{2}}[/tex]
r = 1.45 Å
the radius of a(n) gold atom. is equal to 1.45 Å
A hypodermic syringe contains a medicine with the density of water (see figure below). The barrel of the syringe has a cross-sectional area A = 2.20 10-5 m2, and the needle has a cross-sectional area a = 1.00 10-8 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force F with arrow of magnitude 2.05 N acts on the plunger, making medicine squirt horizontally from the needle. Determine the speed of the medicine as it leaves the needle's tip. m/s
The speed of the medicine as it leaves the needle's tip is the same as the speed of the medicine inside the syringe, which is 4.0 mm/s.
Explanation:To determine the speed of the medicine as it leaves the needle's tip, we can use Bernoulli's equation.
Bernoulli's equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline in a fluid flow.
Applying Bernoulli's equation, we have:
P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2
Where P1 and P2 are the pressures, v1 and v2 are the velocities, ρ is the density of the fluid, g is the acceleration due to gravity, and h1 and h2 are the heights.
In this case, the medicine is moving horizontally, so the heights (h1 and h2) are the same. Also, the pressure everywhere is 1.00 atm, so P1 and P2 are equal. Additionally, the density of the medicine is the same as water. Therefore, the equation simplifies to:
(1/2)ρv1² = (1/2)ρv2²
Cancelling out the common terms, we have:
v1² = v2²
Taking the square root of both sides, we find:
v1 = v2
Therefore, the speed of the medicine as it leaves the needle's tip is the same as the speed of the medicine inside the syringe, which is 4.0 mm/s.
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. The charge entering the positive terminal of an element is ???? = 10 sin 4???????? m????, while the voltage across the element is ???? = 2 cos 4???????? ????. (i) Find the power delivered to the element at t = 0.3s. (ii) Calculate the energy delivered to the element between 0 and 0.6s.
Answer:
P (t = 0.3) = 164.5 mW
W ( 0 < t < 0.6) = 78.34 mJ
Explanation:
Given:
q (t) = 10*sin(4*pi*t) mC
V (t) = 2 *cos(4*pi*t) V
part a
The current i (t) flowing through the element is obtained as follows:
i (t) = dq / dt
= d (10*sin(4*pi*t)) / dt
= 40 * pi * cos (4*pi*t) mA
Next P(t) delivered to the element is obtained as follows:
P (t) = i (t)*V(t)
= 40 * pi * cos (4*pi*t) * 2 *cos(4*pi*t)
= 80*pi*(cos(4*pi*t))^2 mW
Finally the power delivered to element @ t = 0.3 s
P (t=0.3) = 80*pi*(cos(4*pi*0.3))^2 = 164.50 mW
Answer: P (t = 0.3) = 164.5 mW
part b
Energy delivered to the element time 0 to 0.6 s is obtained as follows:
[tex]W (0 <t<0.6) = \int\limits {P(t)} \, dt\\\\ =\int {80*pi*(cos(4*pi*t))^2} \, dt\\\\= (5 sin (8*pi*t) + 40*pi*t )\limits^0.6_0 \\\\= 78.33715mJ[/tex]
Answer: W ( 0 < t < 0.6) = 78.34 mJ
(a) A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant net force), what was the average speed during this 3 s interval?
Answer:
During the 3 s interval, the average velocity was 4 m/s.
Explanation:
Hi there!
The average velocity (AV) is calculated as follows:
AV = Δx / Δt
Where:
Δx = traveled distance.
Δt = elapsed time.
The traveled distance (x) is calculated as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since x0 and v0 are equal to zero, the equation gets reduced to:
x = 1/2 · a · t²
Since the acceleration is constant, it can be calculated with this equation:
a = v/t
a = 8 m/s / 3 s
a = 8/3 m/s²
Then, the traveled distance will be:
x = 1/2 · a · t²
x = 1/2 · 8/3 m/s² · (3 s)²
x = 12 m
And the average velocity will be:
AV = Δx / Δt
AV = 12 m / 3 s = 4 m/s
During the 3 s interval, the average velocity was 4 m/s.
The small piston of a hydraulic lift has a crosssectional area of 2.23 cm2 and the large piston 297 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN? (In service stations, this force is usually exerted by compressed air.) Answer in units of N.
Answer:
The force that must be applied to the small piston = 18.02 N.
Explanation:
Hydraulic Press: This is a device that produce a very large force to compress something e.g printing press.
From pascal's principle,
In an hydraulic press,
f/a = F/A......................... Equation 1
Where f = force applied to the small piston, a = area of the small piston, F = force required to lift the load/ force produced at the large piston. A = Area of the large piston.
Making f the subject of the equation above,
f = F×a/A.............................. Equation 2
Given: F = 2.4 kN = 2400 N, a = 2.23 cm², A = 297 cm².
Substituting into equation 2
f = 2400(2.23)/297
f = 18.02 N.
Thus the force that must be applied to the small piston = 18.02 N.
A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
Final answer:
The question from the student involves the concept of thermal expansion in physics, where the goal is to determine at what temperature a copper cylinder's volume becomes 0.163% larger than its original volume at 21.1°C.
Explanation:
The student's question is about thermal expansion, which is a concept in physics specifically relating to how the volume of a solid changes with temperature. This falls under the broader subject of thermodynamics. The student is given the initial volume and temperature of the copper cylinder and is asked to find the temperature at which its volume is 0.163% larger. To solve this, we need to use the linear expansion formula for solids.
The formula for the volume expansion of solids is
V = V₀(1 + βΔT), where V is the final volume, V₀ is the initial volume, β is the coefficient of volume expansion for copper, and ΔT is the change in temperature. To find the new temperature, we first need to express the 0.163% increase in volume as a decimal, which gives us 0.00163. We can then rearrange the formula to solve for ΔT. After finding ΔT, we add it to the initial temperature of 21.1°C to find the final temperature.
Plugging in the numbers:
0.00163 = 3(16.5 x 10⁻⁶)ΔT
ΔT = 21.3 °C
Therefore, T2 = T1 + ΔT = 21.1 °C + 21.3 °C = 42.4 °C
The temperature at which the copper cylinder's volume will be 0.163% larger than at 21.1 °C is 42.4 °C.
An astronaut in an orbiting space craft attaches a mass m to a string and whirls it around in uniform circular motion. The radius of the circle is R, the speed of the mass is v, and the tension in the string is F. If the mass, radius, and speed were all to double the tension required to maintain uniform circular motion would be
Answer:
[tex]F'=4F[/tex]
Explanation:
According to Newton's second law, the tension in the string is equal to the centripetal force, since the mass is under an uniform circular motion:
[tex]F=F_c\\F=ma_c[/tex]
Here [tex]a_c[/tex] is the centripetal acceleration, which is defined as:
[tex]a_c=\frac{v^2}{r}[/tex]
So, replacing:
[tex]F=m\frac{v^2}{r}[/tex]
In this case we have [tex]m'=2m[/tex], [tex]v'=2v[/tex] and [tex]r'=2r[/tex]. Thus, the tension required to mantain uniform circular motion is:
[tex]F'=m'\frac{v'^2}{r'}\\F'=2m\frac{(2v)^2}{2r}\\F'=4m\frac{v^2}{r}\\F'=4F[/tex]
Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C directed downward. At 600 m above the ground, the electric field is 110 N/C downward. What is the average volume charge density in the layer of air between these two elevations?
Answer:
[tex]1.475\times 10^{-13}\ C/m^3[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
A = Area
h = Altitude = 600 m
Electric flux through the top would be
[tex]-110A[/tex] (negative as the electric field is going into the volume)
At the bottom
[tex]120A[/tex]
Total flux through the volume
[tex]\phi=120-110\\\Rightarrow \phi=10A[/tex]
Electric flux is given by
[tex]\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0[/tex]
Charge per volume is given by
[tex]\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3[/tex]
The volume charge density is [tex]1.475\times 10^{-13}\ C/m^3[/tex]
Why isn't Coulomb's law valid for large conducting objects, even if they are spherically symmetrical?
Answer:
The Coulomb’s Law is as follows
[tex] \vec{F} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]
According to this law, the force between two charged objects can be calculated using the distance between the objects. If the objects are large, then it is not possible to determine the distance, r, between that object and the other object. Because, the edge of the object contain charges as well as the center of the object.
In that case, you need to separate the object into infinitesimal points, apply the formula to those points, then integrate over the large object to find the force between objects.
What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?
Answer:
[tex]\lambda=1090nm[/tex]
Explanation:
Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:
[tex]\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
Where [tex]R_H[/tex] is the Rydberg constant for hydrogen and [tex]n_1[/tex], [tex]n_2[/tex] are the lower energy state and the higher energy state, respectively.
[tex]\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm[/tex]
We are usually not aware of the electric force acting between two everyday objects because
a.) most everyday objects have as many plus charges as minus charges
b.) the electric force is invisible
c.) the electric force is one of the weakest forces in nature
d.) the electric force is due to microscopic-sized particles such as electrons and protons
Answer:We are usually not aware of the electric force acting between two everyday objects because most everyday objects have as many plus charges as minus charges. Option A
Explanation:An electric force is exerted between any two charged objects( either positive or negative). Objects with the same charge will repel each other, and objects with opposite charge will attract each other. The strength of the electric force between any two charged objects depends on the amount of charge that each object contains and on the distance between the two charges. Electric charges are generated all around us due to different surfaces bearing different types of charges. We are usually not aware of it because the quantity of positive charges equals the number of negative charges.
The electric force between everyday objects is typically not noticeable because these objects are generally electrically neutral, having an equal number of positive and negative charges cancelling out any major electrostatic force.
Explanation:We are usually not aware of the electric force acting between two everyday objects because most everyday objects have as many plus charges as minus charges, which results in them being electrically neutral. This neutrality means that there are no noticeable electrostatic forces being exerted between such objects. In contrast, if everyday objects were not electrically neutral, we would experience significant electrostatic interactions on a daily basis. Furthermore, the electric force is actually one of the four fundamental forces in nature and is much stronger than gravity (considering two charged particles like protons). This force can be felt in the form of both attractive and repulsive forces depending on the nature of the charges involved. It is this force that governs several interactions at the atomic level including the bond between atoms in a molecule and the forces that we perceive as contact forces, like friction.
Moreover, Coulomb's law describes how the electric force varies as the product of the charges and inversely with the square of the distance between them. The fact that electric force can be both attractive and repulsive, whereas gravity is always attractive, can lead to forces cancelling out when objects have both positive and negative charges evenly distributed, as is the case with most macroscopic objects we encounter.
(1) Develop an equation that relates the rms voltage of a sine wave to its peak-to-peak voltage. a. If a sine wave has a peak-to-peak value of 1.5V, a frequency of 3kHz, and a phase of 0 radians, what is the rms voltage
Answer:
(A) Equation will be [tex]v=v_msin\omega t=0.75sin(18840t)[/tex]
(B) RMS value of voltage will be 0.530 volt
Explanation:
We have given peak to peak voltage of ac wave = 1.5 volt
Peak to peak voltage of ac wave is equal to 2 times of peak voltage
So [tex]2v_{peak}=1.5volt[/tex]
[tex]v_{peak}=\frac{1.5}{2}=0.75volt[/tex]
Frequency of ac wave is given f = 3 kHz
So angular frequency [tex]\omega =2\pi f[/tex] = 2×3.14×3000 = 18840 rad/sec
So expression of equation will be [tex]v=v_msin\omega t=0.75sin(18840t)[/tex] ( As phase difference is 0 )
Now we have to find the rms value of voltage
So rms voltage will be equal to [tex]v_{rms}=\frac{v_{peak}}{\sqrt{2}}=\frac{0.75}{1.414}=0.530volt[/tex]
The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.
Answer:
v = 3×10^8 m/s
s= 384,400 km= 3.84×10^8 m/s
t = ?
v = s/t = 2s/t
t = 2s/v
t = (2×3.84×10^8) ÷ 3×10^8
t = 2.56 seconds
Explanation:
Earth's moon is the brightest object in our
night sky and the closest celestial body. Its
presence and proximity play a huge role in
making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.
The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696
km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."
The speed of light is used to determine the time it takes for light to travel from Earth to the Moon and back. By applying the formula Time = Distance / Speed, the total round trip time can be calculated.
The speed of light is 3.00×108 m/s. To calculate how long it takes for light to travel from Earth to the Moon and back, we need to consider the distance. The average distance from Earth to the Moon is about 384,400 km. Using the formula Time = Distance / Speed, the total round trip time would be approximately 2.56 seconds.
Two identical cylinders, A and B, contain the same type of gas at the same pressure. Cylinder A has twice as much gas as cylinder B. Which is true?a. TA = TBb. Not enough information to make a comparison.c. TA > TBd. TA < TB
Answer:
D. TA < TB
Explanation:
From general gas equation, we know that:
PV = nRT
PV/R = nT
where,
P = pressure of gas
V = volume of gas
R = General gas constant
T = temperature of gas
n = no. of moles of gas
FOR CYLINDER A:
PV/R = (nA)(TA) _____ eqn (1)
FOR CYLINDER B:
PV/R = (nB)(TB) _____ eqn (2)
Because, Pressure, Volume are constant for both cylinders.
Comparing eqn (1) and (2)
(nA)(TA) = (nB)(TB)
It is given that the amount of gas in cylinder A is twice as much as the gas in cylinder B. This means the number moles in cylinder A are twice as much as no. of moles in cylinder B.
nA = 2(nB)
using this in eqn:
2(nB)(TA) = (nB)(TB)
TA = (1/2)(TB)
TA = 0.5 TB
Therefore it is clear that the correct option is:
D. TA<TB
OPTION A.
According to the ideal gas law, two cylinders containing the same type of gas at the same pressure, where one has twice the amount of gas, will have the same temperature because the gas law considers the number of moles of gas, not the volume. Therefore, TA = TB.
Explanation:The question is asking about the relationship between the amount of gas and the temperature in two containers. Using the formula PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant and T is the temperature, the question provided that the pressures and the gases in Cylinder A and B are identical but the volume of gas in Cylinder A is twice that of Cylinder B. This indicates that Cylinder A contains twice the moles of gas(n) as Cylinder B, since 'n' is proportional to 'V' when 'P','R' and 'T' are constant.
So, if Cylinder A contains twice as much gas as Cylinder B, it implicitly means it contains twice the moles of gas. But since their pressures are identical, by the ideal gas law (PV=nRT), therefore, their temperatures must also be the same. So, TA = TB.
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