Answer:
a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b) The maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium.
Explanation:
Given that :
Frequency (f) = 4.80 Hz
Amplitude (A) = 3.40 cm
a)
The total distance traveled by the sphere during one cycle of simple harmonic motion is:
d = 4A (where A is the Amplitude)
d = 4(3.40 cm)
d = 13.60 cm
Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b)
As we all know that:
[tex]x = Asin \omega t[/tex]
Differentiating the above expression with respect to x ; we have :
[tex]\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)[/tex]
[tex]v = A \omega cos \omega t[/tex]
Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;
Then:
[tex]v_{max} = A \omega[/tex]
We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium
substituting [tex]2 \pi f[/tex] for [tex]\omega[/tex] in the above expression;
[tex]v_{max} = A(2 \pi f)[/tex]
[tex]v_{max} = 3.40 cm (2 \pi *4.80)[/tex]
[tex]v_{max} = 102.54 \ cm/s[/tex]
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
[tex]v = A \omega cos \omega t[/tex]
By differentiation with respect to t;
[tex]\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)[/tex]
[tex]a =- A \omega^2 sin \omega t[/tex]
The maximum acceleration of the sphere is;
[tex]a_{max} =A \omega^2[/tex]
where;
[tex]w = 2 \pi f[/tex]
[tex]a_{max} = A(2 \pi f)^2[/tex]
where A= 3.40 cm = 0.034 m
[tex]a_{max} = 0.034*(2 \pi *4.80)^2[/tex]
[tex]a_{max} = 30.93 \ m/s^2[/tex]
The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of [tex]x = \pm A[/tex]
The sphere moves a total distance of 6.80 cm in one cycle. Its maximum speed is 102.5 cm/s as it passes through equilibrium and the maximum acceleration is at maximum excursion from equilibrium, with a magnitude of 18.5 m/s².
Explanation:The sphere's total distance moved in one cycle is twice the amplitude, as it moves to the maximum amplitude and back again. Therefore, in the case of this sphere, it moves through a total distance of 6.80 cm during one cycle. The maximum speed of an object in simple harmonic motion occurs as it passes through equilibrium. You can calculate this speed using the formula v = ωA, where ω is the angular frequency and A is the amplitude.
The angular frequency is ω = 2πf, where f is the frequency. Thus, the maximum speed of the sphere is approximately 102.5 cm/s. Lastly, the maximum magnitude of acceleration occurs at maximum excursion from equilibrium, and can be calculated with the formula derived from Newton's second law a = ω²A, thus the acceleration is 18.5 m/s².
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A common technique in analysis of scientific data is normalization. The purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. The goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. In this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. Suppose that in Part 2.2.2 of the experiment, students obtain a signal value of 185 mV at a distance of 4 cm and a value of 82 mV at a distance of 6.2 cm. Normalize the students' data to the value obtained at 4 cm. (Divide the signal value by 185.) Then calculate the theoretically expected (normalized) value at 6.2 cm.___________________ Normalized experimental value at 6.2 cm____________________Theoretically expected normalized value at 6.2 cm
Answer:
1. Normalized value at 6.2 cm = 0.443
2. Theoretical expected value = 0.416
Explanation:
1. Normalized experimental value is calculated as follows;
Normalized experimental value = 82/185
= 0.443
Therefore, normalized value at 6.2 cm = 0.443
2. Calculating the theoretical expected value using the relation;
V₁r₁² = V₂r₂²
V₂ = V₁(r₁/r₂)²
= 1* (4/6.2)²
= 1 *0.645²
= 0.416
Therefore, the theoretical expected value = 0.416
The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they act like a single solenoid?
Answer: 13.1 μH
Explanation:
Given
length of heating coil, l = 1 m
Diameter of heating coil, d = 0.8 cm = 8*10^-3 m
No of loops, N = 400
L = μN²A / l
where
μ = 4π*10^-7 = 1.26*10^-6 T
A = πd²/4 = (π * .008 * .008) / 4 = 6.4*10^-5 m²
L = μN²A / l
L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1
L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5
L = 1.31*10^-5
L = 13.1 μH
Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH
Answer:
Their total self-inductance assuming they act like a single solenoid is 10.11 μH
Explanation:
Given;
diameter of the heating coil, d = 0.800 cm
combined length of heating coil and hair dryer, [tex]l[/tex] = 1.0 m
number of turns, N = 400 turns
Formula for self-inductance is given as;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where
μ₀ is constant = 4π x 10⁻⁷ T.m/A
A is the area of the coil:
A = πd²/4
A = π (0.8 x 10⁻²)²/4
A = 5.027 x 10⁻⁵ m²
[tex]L = \frac{\mu_oN^2 A}{l } = \frac{4\pi *10^{-7}(400)^2 *5.027*10^{-5}}{1 } \\\\L =1.011 *10^{-5} \ H\\\\L = 10.11 \mu H[/tex]
Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH
Problem 24.3 The assembly is made from a steel hemisphere, rho st = 7. 80 Mg/m3 , and an aluminum cylinder, rho al = 2. 70 Mg/m3 . If the height of the cylinder is h = 180 mm, determine the location z of the mass center of the assembly.
Given:
[tex]\rho_s = 7.8 \ Mg/m^3[/tex][tex]\rho_a = 2.70 \ Mg/m^3[/tex]Height of cylinder, [tex]h = 180 \ mm[/tex]Now,
The mass of the steel hemisphere will be:
→ [tex]m_h = \rho_s\times \frac{2}{3} \pi r^2[/tex]
By putting the values, we get
[tex]= 7.8\times 10^3\times \frac{2}{3} \pi (0.16)^2[/tex]
[tex]= 66.9 \ kg[/tex]
and,
The mass of aluminum cylinder will be:
→ [tex]m_c = \rho_a \times \pi r^2 h[/tex]
[tex]= 2.7\times 10^3\times \pi (0.08)^2 (0.18)[/tex]
[tex]= 9.7 \ kg[/tex]
Now,
The mass center of steel hemisphere will be:
→ [tex]z_1 = r - \frac{3r}{8}[/tex]
[tex]= 0.16-(3\times \frac{0.16}{8} )[/tex]
[tex]= 0.1 \ m[/tex]
The mass center of aluminum hemisphere will be:
→ [tex]z_2 = r+\frac{h}{2}[/tex]
[tex]= 0.16+\frac{0.18}{2}[/tex]
[tex]= 0.25 \ m[/tex]
hence,
The center of mass will be:
→ [tex]Z = \frac{m_h z_1 +z_2 m_c}{m_s+m_c}[/tex]
By putting the values, we get
[tex]= \frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76}[/tex]
[tex]= 0.12 \ m[/tex]
Thus the above answer is right.
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To find the location (z) of the mass center of a steel hemisphere and an aluminum cylinder assembly, we first calculate the mass of each part using the given densities and geometric properties. We then use a formula to calculate 'z' based on the mass and z-coordinate of the center of mass of each part. The center of mass assumes the symmetric mass distributions.
Explanation:To determine the location z of the mass center of the assembly, we first need to calculate the mass of each part of the assembly using their given densities and geometrical properties.
The mass of the steel hemisphere (mSt) is calculated as density times the volume. The volume of the hemisphere can be calculated using the formula 2/3*Pi*r^3, where r is the radius of the hemisphere.
The mass of the aluminum cylinder (mAl) is calculated the same way, as the product of the cylinder's density and volume. The volume of a cylinder is given by the formula Pi*r^2*h, where r is the radius and h is the height of the cylinder, which is given as 180 mm or 0.12 m.
Once we have these masses, we then calculate the mass center z of the assembly, which is given by the formula z = (mSt * zSt + mAl * zAl) / (mSt + mAl), where zSt and zAl are the z-coordinates of the center of the mass of the steel hemisphere and the aluminum cylinder respectively.
Since the mass of each body acts as if it were at the center of the body, we can assume that the z-coordinate of the hemisphere (zSt) is at its geometric center, whereas for the cylinder (zAl), it would be at h/2 from the bottom end.
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Two slits separated by a distance of d = 0.12 mm are located at a distance of D = 0.63 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of λ = 540 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. What is the pathlength difference between the waves at the second maximum (m=2) on the screen?
Answer:
The path-length difference is [tex]dsin\theta=1.08*10^{-3}mm[/tex]
The angle is [tex]\theta = 0.5157^o[/tex]
Explanation:
From the question we are told that
The distance of separation is d = 0.12 mm = [tex]0.12*10^{-3} m[/tex]
The distance from the screen is D = 0.63 m
The wavelength is [tex]\lambda = 540nm = 540 *10^{-9}m[/tex]
From the question we can deduce that the the two maxima's are at the m=0 and m=2
Now the path difference for this second maxima is mathematically represented as
[tex]d sin \theta = m \lambda[/tex]
Where d[tex]dsin\theta[/tex] is the path length difference
Substituting values
[tex]dsin \theta = 2 * 540*10^{-9}[/tex]
[tex]dsin\theta = 1.08*10^{-6}m[/tex]
converting to mm
[tex]dsin\theta = 1.08*10^{-6} * 1000 mm[/tex]
[tex]dsin\theta=1.08*10^{-3}mm[/tex]
To obtain the angle we make [tex]\theta[/tex] the subject
[tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
Substituting values
[tex]\theta = sin ^{-1} [\frac{1.08*10^{-6}}{0.12*10^-3} ][/tex]
[tex]\theta = 0.5157^o[/tex]
The Pathlength difference between the waves at second maximum on the screen is; 1.08 × 10^(-6) m
We are given;
Distance between two slits; d = 0.12 mm = 0.12 × 10^(-3) m
Distance of slit from screen; D = 0.6 m
We want to find the path length at second maxima m = 2
λ = 540 nm = 540 × 10^(-9) m
Formula for Pathlength is;
dsin θ = mλ
Where mλ is the Pathlength difference.
Since at m = 0, the pathlength is zero,
Thus;
Pathlength difference = (2 × 540 × 10^(-9)) - 0
Pathlength difference = 1.08 × 10^(-6) m
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A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill? where X = 16.8.
Answer:h=19.4 m
Explanation:
Given
mass of automobile [tex]m=750\ kg[/tex]
Initial height of automobile [tex]h_o=5\ m[/tex]
Velocity at this instant [tex]v=16.8\ m/s[/tex]
If the car stops somewhere at a height [tex]h[/tex]
Thus conserving total energy we get
[tex]K_i+U_i=K_f+U_f[/tex]
[tex]\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh[/tex]
[tex]\frac{v^2}{2g}+h_o=h[/tex]
[tex]h=5+\frac{16.8^}{2\times 9.8}[/tex]
[tex]h=5+14.4[/tex]
[tex]h=19.4\ m[/tex]
A proton moves through a magnetic field at 26.1 % of the speed of light. At a location where the field has a magnitude of 0.00667 T and the proton's velocity makes an angle of 139 ∘ with the field, what is the magnitude F B of the magnetic force acting on the proton? Use c = 2.998 × 10 8 m/s for the speed of light and e = 1.602 × 10 − 19 C as the elementary charge.
Answer:
[tex]5.48\cdot 10^{-14} N[/tex]
Explanation:
When a charged particle is moving in a region with a magnetic field, the particle experiences a force perpendicular to its direction of motion. The magnitude of this force is given by
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is its velocity
B is the strength of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
In this problem we have:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]v=0.261 c[/tex] is the speed of the proton, where
[tex]c=2.998\cdot 10^8 m/s[/tex] is the speed of light
[tex]B=0.00667 T[/tex] is the strength of the magnetic field
[tex]\theta=139^{\circ}[/tex] is the angle between the direction of the proton and the magnetic field
Substituting, we find the magnitude of the force:
[tex]F=(1.6\cdot 10^{-19})(0.261\cdot 2.998\cdot 10^8)(0.00667)(sin 139^{\circ})=5.48\cdot 10^{-14} N[/tex]
The magnitude of the magnetic force acting on the proton is [tex]5.46*10^{-14} N[/tex]
Force on charge :
When a charge particle is moving in magnetic field.
Then force is given as, [tex]F=qvBsin(\theta)[/tex]
where
[tex]q[/tex] is the charge of the particle[tex]v[/tex] is velocity[tex]B[/tex] is the strength of the magnetic field[tex]\theta[/tex] is the angle between the direction of [tex]v[/tex] and [tex]B[/tex]Given that, [tex]q=1.6*10^{-19}C,v=0.261*3*10^{8}=7.8*10^{7} ,B=0.00667T,\theta=139[/tex]
Substitute all values in above relation.
[tex]F=1.6*10^{-19}*7.8*10^{7} *0.00667T*sin(139)\\\\F=5.46*10^{-14} N[/tex]
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For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.
Answer:
1.56 - 1.67
Explanation:
Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.
Mathematically, it is given as:
n = c/v
Where c is the speed of light in a vacuum and v is the speed of light in the medium.
Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.
When the speed is 1.8 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.8 * 10^8)
n = 1.67
When the speed is 1.92 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.92 * 10^8)
n = 1.56
Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.
Final answer:
The range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
Explanation:
The index of refraction, n, of a material can be calculated using the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.
For this specific optical medium, the speed of light varies from 1.80 × 10^8 m/s for violet light to 1.92 × 10^8 m/s for red light. To calculate the range of the index of refraction, we need to determine the ratio of the speed of light in a vacuum to the speed of light in the material for both violet and red light.
The range of the index of refraction, n, can be calculated as:
For violet light: n = c/v = (3.00 × 10^8 m/s) / (1.80 × 10^8 m/s) ≈ 1.67
For red light: n = c/v = (3.00 × 10^8 m/s) / (1.92 × 10^8 m/s) ≈ 1.56
Therefore, the range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
Calculate the magnitude of the magnetic force on a 270 m length of wire stretched between two towers and carrying a 150 A current. The Earth's magnetic field of 4.0×10−5 T makes an angle of 80 ∘ with the wire
Answer:
Force on the wire is equal to 15.95 N
Explanation:
We have given length of the wire l = 270 m
Current flowing in the wire i = 150 A
Magnetic field [tex]B=4\times 10^{-5}T[/tex]
Angle between magnetic field and wire [tex]\Theta =80^{\circ}[/tex]
We have to find the force on the wire
Force on current carrying conductor is equal to [tex]F=IBlsin\Theta[/tex]
So [tex]F=150\times 4\times 10^{-5}\times 270\times sin80^{\circ}=15.95N[/tex]
So force on the wire is equal to 15.95 N
Using the formula F = IlB sin(θ), with I = 150 A, l = 270 m, B = 4.0×10⁻⁵ T, and θ = 80°, the magnitude of the magnetic force on the wire is calculated to be approximately 158.76 Newtons.
Explanation:To calculate the magnitude of the magnetic force on the wire, we use the formula F = IlB sin(θ), where F represents the force, I is the current, l is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field. In this scenario, I = 150 A, l = 270 m, B = 4.0×10⁻⁵ T, and θ = 80°.
First, we calculate the sine of the angle:
sin(80°) ≈ 0.9848Plugging the values into the force equation gives:
F = (150 A) × (270 m) × (4.0×10⁻⁵ T) × 0.9848
F = 158.76 N (rounded to two decimal places)
Hence, the magnitude of the magnetic force on the wire is approximately 158.76 Newtons.
Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0.048 m. You wish the mass to have a kinetic energy of 14.0 Joules when released. How strong will the string need to be
Answer:
Tension in the string is equal to 58.33 N ( this will be the strength of the string )
Explanation:
We have given mass m = 1.7 kg
radius of the circle r = 0.48 m[tex]F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N[/tex]
Kinetic energy is given 14 J
Kinetic energy is equal to [tex]KE=\frac{1}{2}mv^2[/tex]
So [tex]\frac{1}{2}\times 1.7\times v^2=14[/tex]
[tex]v^2=16.47[/tex]
v = 4.05 m/sec
Centripetal force is equal to [tex]F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N[/tex]
So tension in the string will be equal to 58.33 N ( this will be the strength of the string )
Final answer:
The problem requires finding the tension in the string by using the given kinetic energy to calculate the velocity and then applying the centripetal force formula to find the required string strength.
Explanation:
The student question involves calculating the tension in a string when a mass is swung to obtain a certain kinetic energy. Considering the mass M has a kinetic energy of 14.0 Joules, we can use the relation for kinetic energy (KE) of a rotating body, which is KE = 1/2 m v^2, where 'm' is the mass and 'v' is the tangential velocity. From the given kinetic energy and mass, we can solve for v.
Once we have the velocity, we can use centripetal force formula, which is F = m v^2 / r, where 'm' is the mass, 'v' is the velocity, and 'r' is the radius. This formula will allow us to find the required strength of the string, which, in physics, is essentially the tension that the string must be able to withstand without breaking.
Note that the tension in the string will be equal to the centripetal force at the point of release when the mass is in horizontal circular motion.
It's a hot and sunny day and you decide to go for a drive. When you open your car door, it's a lot hotter than outside. Why is that
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 59.3 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy U that flows through area of 0.0225 m 2 over an interval of 12.3 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer: S = 4.67 W/m², U = 1.29 J
Explanation:
Given
Time of flow, t = 12.3 s
Area of flow, a = 0.0225 s
Amplitude, E = 59.3 V/m
Intensity, S = ?
I = E² / cμ, where
μ = permeability of free space
c = speed of light
E = E(max) / √2
E = 59.3 / √2
E = 41.93 V/m
I = 41.93² / (2.99*10^8 * 1.26*10^-6)
I = 1758.125 / 376.74
I = 4.67 W/m²
Energy that flows through
U = Iat
U = 4.67 * 0.0225 * 12.3
U = 1.29 J
Therefore, the intensity is 4.67 W/m² and the energy is 1.29J
Answer:
A) Intensity = 4.664 W/m²
B) U = 1.29J
Explanation:
A) The intensity of the wave is related to a time-averaged version of a quantity called the Poynting vector, and is given by the formula:
I = (E_rms/cμo)
Where;
c = speed of light which has a value of 3 x 10^(8) m/s
μo = permeability of free space which has a constant value of 4π x 10^(-7) N/A²
E_rms is root mean square value of electric field
In the question, we are given maximum amplitude of the electric field. In this case, we would have to calculate the "root-mean-square" or "rms" value through the relationship:
E_rms = E_max/√2
Thus, E_rms = 59.3/√2 = 41.93 V/m
Thus, Intensity, I = (E_rms/cμo)= [41.93²/(3 x 10^(8) x 4π x 10^(-7))]
I = 4.664 W/m²
B) The formula for the energy flowing is given by the formula ;
U = IAt
Where;
I is intensity
A is area
t is time in seconds
Thus, U = 4.664 x 0.0225 x 12.3 = 1.29J
People who do very detailed work close up, such as jewelers, often can see objects clearly at a much closer distance than the normal 25.0 cm . What is the power of the eyes of a woman who can see an object clearly at a distance of only 7.25 cm ? Assume a distance from the eye lens to the retina of 2.00 cm .
The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore
Explanation:
To find the power of the eye
we have the formula,
P=1/f= 1/d0 +1/di
Where,
do denotes the distance between eyes length and the object
di denotes the distance between eyes length and the image
Given data
do=7.25 cm di=2.00 cm
substitute in the formula
P=1/f= 1/d0 +1/di
P= 1/0.0725 +1/0.02=13.79
P=63.79 D
The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore
A 60.0-kg man jumps 1.70 m down onto a concrete walkway. His downward motion stops in 0.025 seconds. If he forgets to bend his knees, what force (N) is transmitted to his leg bones?
Answer:
The magnitude of the force transmitted to his leg bones is 1413.6 N
Explanation:
Recall that force is defined as the change in linear momentum per unit time:
[tex]F=\frac{P_f-P_i}{\Delta t}[/tex]
We can use this formula to find the force transmitted to his legs. We know that the final momentum ([tex]P_f[/tex]) is 0 since the person is not moving on the floor, but we need to find what the person's momentum was an instant before he touches the ground. Since we know he person's mass, all we need for the initial momentum is his velocity.
For such we use conservation of energy in free fall, knowing that he jumped from 1.7 meters:
[tex]Potential \,\,Energy \,\,at \,\,the \,\,top \,\,of \,\,the\,\, jump = U_i=m * g * h = 60*9.8*1.7 \,J\\Kinetic \,\,Energy \,\,when \,\,touching \,\,ground =KE_f= \frac{1}{2} m*v^2=\frac{60\,kg}{2} v^2\\\\KE_f=U_i\\\frac{60\,kg}{2} v^2=60*9.8*1.7 \,J\\v^2=2*9.8*1.7 \,\frac{m^2}{s^2} \\v=0.589\,\frac{m}{s}[/tex]
Now with this velocity, we know the [tex]P_i[/tex] (initial momentum) just before impact.
[tex]P_i=60 \,kg * 0.589 \frac{m}{s} =35.34 \,kg\,\frac{m}{s}[/tex]
And since the impact lasted 0.025 seconds, we can find the force using the first formula we recalled:
[tex]F=\frac{P_f-P_i}{\Delta t}=\frac{0-35.34_i}{\0.025}: N= -1413.6\,N[/tex]
so the magnitude of the force is 1413.6 N
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 5.20 m and, in starting, rotates according to θ = 0.170t2, where t is in seconds and θ is in radians. When t = 4.70 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?
Answer:
A) angular velocity; ω = 1.598 rad/s
B) linear velocity;V = 8.31 m/s
C) Tangential Acceleration;a_t = 1.768 m/s²
D) Radial Acceleration;a_r = 13.28 m/s²
Explanation:
We are given that;
Radius; r = 5.2m
Time;t = 4.7 sec
θ = 0.170t²
Thus, angular acceleration would be the second derivative of θ which is d²θ/dt²
Thus,α = d²θ/dt² = 0.34 rad/s²
A) Formula for angular velocity is;
ω = αt
Where α is angular acceleration and t is time.
Thus;ω = 0.34 x 4.7
ω = 1.598 rad/s
b) formula for linear velocity is given by; V = ωr
We have ω = 1.598 rad/s and r = 5.2m
Thus; V = 1.598 x 5.2
V = 8.31 m/s
c) formula for tangential acceleration is;
a_t = αr
a_t = 0.34 x 5.2
a_t = 1.768 m/s²
d) formula for radial acceleration is;
a_r = rω²
a_r = 5.2 x 1.598²
a_r = 13.28 m/s²
To find the magnitudes of the astronaut's angular velocity, linear velocity, tangential acceleration, and radial acceleration, differentiate the angular position equation, multiply the radius and the angular velocity to find the linear velocity, and use the rate of change of linear velocity to find the tangential acceleration. The radial acceleration is the product of the angular velocity and the linear velocity.
Explanation:The angular velocity can be found by differentiating the angular position equation with respect to time. The linear velocity is equal to the product of the radius and the angular velocity. The tangential acceleration is equal to the rate of change of linear velocity, while the radial acceleration is equal to the product of the angular velocity and the linear velocity.
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A car goes from 100 m/s to a full stop for in 9.5 seconds. What is the acceleration?
Answer:Calculate displacement of an object that is not acceleration, given initial position and velocity.
Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.
Four men racing up a river in their kayaks.
Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr).
We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
Notation: t, x, v, a
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is
Δ
t
=
t
f
−
t
0
, taking
t
0
=
0
means that
Δ
t
=
t
f
, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,
x
0
is the initial position and
v
0
is the initial velocity. We put no subscripts on the final values. That is,
t
is the final time,
x
is the final position, and
v
is the final velocity. This gives a simpler expression for elapsed time—now,
Δ
t
=
t
. It also simplifies the expression for displacement, which is now
Δ
x
=
x
−
x
0
. Also, it simplifies the expression for change in velocity, which is now
Δ
v
=
v
−
v
0
. To summarize, using the simplified notation, with the initial time taken to be zero,
Explanation:
What is the magnitude of the rate of change of the magnetic field inside the coil? You will need to calculate this algebraically before you can get a number. Write an expression for the magnetic field due to the wire at the location of the coil. Use the approximate formula, since the wire is very long. Remember the chain rule, and remember that ⅆ ⅆ .
Answer:
The answer to this questions are, (a) 0.685 * 10^-4 T/sec (b) 0.946 * 10^-6V
Explanation:
Solution
Recall
radius r = 0.02m
N = 11 turns
Instant I = 3 amperes
Velocity v =3.3m/s
x = 0.17 m
(a) What is the magnitude of the rate of change of the magnetic field inside the coil
db/dt =μ₀T/2π x²
Thus,
= 4π * 10^-7* 3* 3.3/2π * 0.17²
=685.12* 10^-7
which is now,
0.685 * 10^-4 T/sec
(b) What is the magnitude of the voltmeter reading.
μ = N (db/dt)πr²
Note: this includes all 11 turns of the coil
Thus,
= 11 * 0.6585* 10^-4* 3.14 * (0.02)²
= 946 * 10^-7
which is = 0.946 * 10^-6V
Note: Kindly find an attached copy or document of the complete question of this exercise
The rate of change of a magnetic field in a coil can be determined using Faraday's law and the chain rule. Starting with the formula for the magnetic field created by current in a straight wire, we differentiate with respect to time to get the rate of change.
Explanation:The magnitude of the rate of change of the magnetic field inside a coil can be represented algebraically using Faraday's law of induction. Starting with the equation B = μ * I / 2λr (where I is the current, r is the distance to the wire, and μ is the permeability of free space), we can find the rate of change of the magnetic field using the chain rule.
Faraday's law is represented as E = - dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is time. Since the magnetic flux Φ is the product of the magnetic field (B) and the area enclosed by the loop (A), we can express this as d/dt (B * A). By applying the chain rule, we can find the rate of change of the magnetic field in the coil.
Last, magnetic field B is determined by Ampère's law, often used in calculations involving magnetic fields around conductors and coils. For a long straight wire, field lines form concentric circles around the wire, following the right-hand rule.
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A spaceship is traveling toward the earth from the space colony on Asteroid 1040A. The ship is at the halfway point of the trip, passing Mars at a speed of 0.9c relative to the Mars frame of reference. At the same instant, a passenger on the spaceship receives a radio message from her boyfriend on 1040A and another from her sister on earth. According to the passenger on the ship, were these messages sent simultaneously or at different times? If at different times, which one was sent first? Explain your reason
Answer:
The message (signal) coming from the earth was sent first.
Explanation:
The radio messages reach Mars simultaneously, because the distance from Earth to Mars and that of from Mars to Asteroid is same. But the passenger in the spaceship is moving relative to Mars towards Earth and hence, the message from Earth reaches first.
According to passenger frame, the message (signal) coming from the Earth was sent first compared to message coming from Asteroid.
The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz.
a. What is the spring constant of the spring on which the head is mounted?
b. suppose the head is pushed 2.0 cm against the spring, then released. What is the head's maximum speed as it oscillates?
c. the amplitude of the head's oscillations decreases to 0.5cm in 4.0s. What is the head's damping constant?
Answer:
(a) 9.375 N/m
(b) 0.5024 m/s
(c) 0.01 kg/s
Explanation:
mass of head, m = 15 g = 0.015 kg
frequency, f = 4 Hz
Time period, T = 1 / f = 0.25 s
Let k is the spring constant.
(a)
The formula for the time period is
[tex]T=2\pi\sqrt{\frac{m}{K}}[/tex]
[tex]0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}[/tex]
[tex]0.04=\sqrt{\frac{0.015}{K}}[/tex]
K = 9.375 N/m
(b)
Amplitude, A = 2 cm
Let ω is the angular velocity.
Maximum velocity, v = A ω = A x 2πf
v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s
(c)
Let b is the damping constant.
A(t = 4s) = 0.5 cm
Ao = 2 cm
Using the formula of damping
[tex]\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}[/tex]
[tex]\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}[/tex]
[tex]0.25=e^{-133.3 b}[/tex]
Taking natural log on both the sides
ln (0.25) = - 133.3 b
- 133.3 b = - 1.386
b = 0.01 kg/s
This question involves the concepts of simple harmonic motion, spring constant, and amplitude.
a) The spring constant of the spring is "9.47 N/m".
b) The maximum speed of the head is "0.5 m/s".
c) The damping constant is "0.01 kg/s".
a)
We can find the spring constant of the spring by using the formula of frequency in the simple harmonic motion:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where,
f = frequency = 4 Hz
k = spring constant = ?
m = mass = 15 g = 0.015 kg
Therefore,
[tex]4\ Hz=\frac{1}{2\pi}\sqrt{\frac{k}{0.015\ kg}}\\\\(16\ Hz^2)(4\pi^2)(0.015\ kg)=k\\[/tex]
k = 9.47 N/m
b)
Maximum speed is simply given by the following formula:
[tex]v=A\omega[/tex]
where,
v = maximum speed = ?
A = Amplitude = 2 cm = 0.02 m
ω = angular freuency = 2πf
Therefore,
[tex]v=A(2\pi f)=(0.02\ m)(2\pi)(4\ Hz)[/tex]
v = 0.5 m/s
c)
using the following equation to find out the damping constant:
[tex]ln(\frac{A}{A_o})=-\frac{bt}{2m}[/tex]
where,
A = amplitude at t = 4 s = 0.5 cm
A₀ = initial amplitude = 2 cm
b = damping constant = ?
t = time = 4 s
m = mass = 15 g = 0.015 kg
Therefore,
[tex]ln(\frac{0.5\ cm}{2\ cm})=-\frac{b(4\ s)}{2*0.015\ kg}[/tex]
[tex]\frac{(-1.386)(2)(0.015\ kg)}{4\ s}=-b[/tex]
b = 0.01 kg/s
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A monochromatic laser is exciting hydrogen atoms from the n=2 state to the n=5 state.
(A) What is the wavelength λ of the laser? Express your answer to three significant digits in nanometers.
(B) Eventually, all of the excited hydrogen atoms will emit photons until they fall back to the ground state. How many different wavelengths can be observed in this process?
(C) What is the longest wavelength λ_max that is observed? Express your answer to three significant digits in nanometers.
(D) What is the shortest wavelength λ_min observed? Express your answer to three significant digits in nanometers.
Answer:
a) λ = 435 nm , c) c) λ = 4052 nm, d) λ= 95 nm
Explanation:
A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.
En = -13,606 / n² [eV]
therefore the energy of the transition is
ΔE = E₅ -E₂
ΔE = 13.606 (1 / n₂² - 1 / n₅²)
ΔE = 13.606 (1/2² - 1/5²)
ΔE = 2,85726 eV
now let's use Planck's equation
E = h f
the speed of light is related to wavelength and frequencies
c = λ f
f = c /λ
E = h c /λ
λ = h c / E
let's reduce the energy to the SI system
E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J
let's calculate
λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹
λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition
initial state n = 5
final state n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ = h c / E
λ = 4052 nm
n = 5
final ΔE (eV) λ (nm)
level
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6
c) λ = 4052 nm
d) λ= 95 nm
The monochromatic laser has a wavelength of 434 nm. When excited hydrogen atoms fall back to the ground state, 10 different wavelengths can be observed ranging from 97.3 nm to 121.6 nm.
Explanation:A monochromatic laser is used to excite hydrogen atoms from the n=2 state to the n=5 state. The wavelength of the laser can be calculated using the Rydberg formula for hydrogen, 1 / λ = RH (1/n1^2 - 1/n2^2), where RH is the Rydberg constant for hydrogen (1.097×10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level. After calculating, we get λ = 434 nm.
Eventually, as excited hydrogen atoms fall back to the ground state, there are 10 different transitions possible, corresponding to 10 different wavelengths of light.
The longest wavelength λ_max is observed when the electron falls from n=2 to n=1 state, λ_max = 121.6 nm.
The shortest wavelength λ_min is observed when an electron falls from n=5 to n=1, λ_min = 97.3 nm.
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A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume
Answer:
A) 100
B) 10
C) 0.01
Explanation:
Let
Fs = force on slave cylinder
Fm = force on master cylinder
Given that the designed to exert a force 100 times as large as the one put into it. That is
Fs = 100 Fm
Using Pascal's law
Fs/As = Fm/Am
100Fm/As = Fm/Am
100/As = 1/Am
Cross multiply
As= 100Am
As/Am = 100
b) What must be the ratio of their diameters
Using their areas ratio, area of a cylinder is πr^2
A = πr^2 = π(D/2)^2
π(Ds/2)^2 ÷ π(Dm/2)^2 = 100
(Ds/2)^2 ÷ (Dm/2)^2 = 100
(Ds/Dm)^2 = 100
Ds/Dm = 10
(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction
Let us consider the workdone at the input and at the output.
Work done = force × distance
Fs × Hs = Fm × Hm
100Fm × Hs = Fm × Hm
100Hs = Hm
100Hs/Hm = 1
Hs/Hm = 1/100
Hs/Hm = 0.01
A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, away. The sun emits a power of 3.8 * 1026 W. Using this information, at what distance would a candle that emits a power of 0.20 W just be visible?
To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The distance at which the candle would be visible is approximately 2.1 million kilometers.
Explanation:To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The luminosity of the Sun is 3.8 * 10^26 W. The candle's luminosity can be calculated using the ratio of its power to the power of the Sun. Luminosity is inversely proportional to the distance squared, so we can set up an equation with the ratio of the candle's luminosity to the Sun's luminosity equal to the ratio of the distance at which the candle is visible to the distance at which the Sun is barely visible:
(0.20 W) / (3.8 * 10^26 W) = (6.6 * 10^16 m)^2 / (x)^2
Cross-multiplying and solving for x, we find that the distance at which the candle would be visible is approximately 2.1 * 10^9 m, or 2.1 million kilometers.
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Approximately 4785 meters is the distance at which a candle would just be visible.
To solve this problem, we need to use the inverse square law for the intensity of light. The apparent brightness of a light source decreases with the square of the distance from the observer. The formula relating the brightness (intensity I) at a distance d is given by:
[tex]\[ I = \frac{P}{4 \pi d^2} \][/tex]
First, we find the intensity I at the distance where the sun-like star is barely visible:
[tex]\[ I = \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} \][/tex]
Plugging in the values:
[tex]\[ I = \frac{3.8 \times 10^{26}}{4 \pi (6.6 \times 10^{16})^2} \][/tex]
Now we need to find the distance [tex]\( d_{\text{candle}} \)[/tex] at which the candle with power [tex]\( P_{\text{candle}} \)[/tex] would have the same intensity I:
[tex]\[ I = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]
Setting the intensities equal:
[tex]\[ \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]
Solving for [tex]\( d_{\text{candle}} \)[/tex]:
[tex]\[ \frac{P_{\text{star}}}{d_{\text{star}}^2} = \frac{P_{\text{candle}}}{d_{\text{candle}}^2} \][/tex]
[tex]\[ d_{\text{candle}}^2 = d_{\text{star}}^2 \frac{P_{\text{candle}}}{P_{\text{star}}} \][/tex]
[tex]\[ d_{\text{candle}} = d_{\text{star}} \sqrt{\frac{P_{\text{candle}}}{P_{\text{star}}}} \][/tex]
Plugging in the values:
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{2 \times 10^{-1}}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{5.26 \times 10^{-28}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \times 7.25 \times 10^{-14} \][/tex]
[tex]\[ d_{\text{candle}} = 4.785 \times 10^3 \][/tex]
[tex]\[ d_{\text{candle}} \approx 4785 \, \text{meters} \][/tex]
Thus, a candle that emits a power of 0.20 W would just be visible at a distance of approximately 4785 meters.
Consider a comet with an elliptic orbit whose aphelion and perihelion distances are rA = 5.00109 km and rP = 8.00107 km. e. Find the speed of the comet at aphelion and at perihelion
Answer:
Explanation:
To find this speed find the attached document below;
The speed of a comet can be calculated at aphelion and perihelion by using the vis-viva equation, which involves the gravitational constant, the mass of the Sun, the distance of the comet from the Sun at specific points, and the semi-major axis of the comet's orbit.
Explanation:Calculating Comet Speed at Aphelion and PerihelionTo find the speed of a comet at aphelion (ra) and perihelion (rp), we will use the conservation of energy and angular momentum principles for orbital motion. The specific mechanical energy (the sum of kinetic and potential energy) in an elliptical orbit around a much larger body, like the sun, is constant at any point along the orbit. This allows us to create an equation relating the speed of the comet at its closest and furthest points from the Sun.
Using the vis-viva equation:
v = √GM(rac{2}{r} - rac{1}{a})
We can calculate the speed at perihelion (vp) and aphelion (va) using the given distances, where a is the semi-major axis (the average of aphelion and perihelion distances) and r is the comet's distance to the Sun at a specific point:
Calculate the semi-major axis a = (½)(ra + rp).Substitute a, ra, and rp into the vis-viva equation to find va and vp.For the given values of rA = 5.00 x 109 km and rP = 8.00 x 107 km, we calculate a, and then use these values to find the comet's speeds at its closest and furthest points from the Sun.
A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit specific enthalpy of 297.31 kj/kg. For negligible heat transfer with the surroundings, the exit velocity is
a) 223 m/s
b) 197 m/x
c) 90 m/s
d) 70 m/s
Answer: c) 90 m/s
Explanation:
Given
Invest velocity, v1 = 250 m/s
Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg
Outlet specific enthalpy, h2 = 297.31 kJ/kg = 297310 J/kg
Outlet velocity, v2 = ?
0 = Q(cv) - W(cv) + m[(h1 - h2) + 1/2(v1² - v2²) + g(z1 - z2)]
0 = Q(cv) + m[(h1 - h2) + 1/2(v1² - v2²)]
0 = [(h1 - h2) + 1/2(v1² - v2²)]
Substituting the values of the above, we get
0 = [(270110 - 297310) + 1/2 ( 250² - v²)
0 = [-27200 + 1/2 (62500 - v²)]
27200 = 1/2 (62500 - v²)
54400 = 62500 - v²
v² = 62500 - 54400
v² = 8100
v = √8100
v = 90 m/s
A dielectric cube of side a, centered at the origin, carries a "frozen-in" polarization P = kr, where k is a constant. Find all the bound charges and check that they add up to zero.
The total volume of bound charge is zero.
Explanation:
We have to the volume and surface bounded charge densities.
ρb = - Δ . p = - Δ .k ([tex]x^{X}[/tex] +[tex]y^{Y}[/tex] +[tex]x^{Y}[/tex])
= - 3k
On the top of the cube the surface charge density is
σb = p . z
= [tex]\frac{ka}{2}[/tex]
By symmetry this holds for all the other sides. The total bounded charge should be zero
Qtot = (-3k)a³ + 6 . [tex]\frac{ka}{2}[/tex] . a² = 0
σb = -3K σb = [tex]\frac{ka}{2}[/tex]
Qtot = 0
Hence, the total volume of bound charge is zero.
A piston is compressed from a volume of 8.47 L to 2.62 L against a constant pressure of 1.93 atm. In the process, there is a heat gain by the system of 360. J. 371.2 Incorrect: Your answer is incorrect. J
Final answer:
The question relates to thermodynamics in physics, focusing on gas expansion or compression in a piston and its related work, heat transfer, and temperature change.
Explanation:
The question involves the concept of thermodynamics, which is a branch of physics dealing with heat, work, and energy transfer. When gas in a piston expands or compresses, it can perform work on its surroundings, and there may also be a transfer of heat between the system and its surroundings. The specifics of the temperature change, work done, and energy transfer depend on factors such as initial and final volume, pressure, and the heat capacity of the system or the environment it's in contact with.
Two monatomic gases, helium and neon, are mixed in a sealed container and brought into thermal equilibrium at temperature T . If the molar mass of helium is 4.0 g/mol and the molar mass of neon is 20.2 g/mol , then _______.
A. all the atoms have the same average speed
B. the average speed of the neon atoms is greater than the average speed of the helium atoms
C. the average speed of the helium atoms is greater than the average speed of the neon atoms
D. the atoms diffuse from high temperature to low temperature
E. all the atoms have exactly the same velocity.
In a mixture of helium and neon gases at thermal equilibrium, the average speed of helium atoms will be greater than the average speed of neon atoms.
Explanation:In a mixture of two gases at thermal equilibrium, the average speed of the atoms will depend on their respective molar masses. The average speed of an atom can be calculated using the root mean square speed formula, v = sqrt(3kT/m), where v is the average speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the atom. In this case, helium has a smaller molar mass than neon, so according to the formula, the average speed of helium atoms will be greater than the average speed of neon atoms.
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The average speed of helium atoms is higher than that of neon atoms because the molar mass of helium is much smaller than that of neon. At the same temperature, lighter molecules move faster. Thus, the correct answer is C.
To determine the average speeds of helium and neon atoms in a mixture at thermal equilibrium at temperature T, we need to consider the relationship between temperature, mass, and molecular speed.
The average kinetic energy per atom at a given temperature T is the same for both helium and neon. This can be expressed as:
[tex]\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T[/tex]
where,
[tex]k_B[/tex]= Boltzmann constant
[tex]T[/tex]= temperature
[tex]m[/tex]= molar mass of the gas
Since both gases have the same average kinetic energy, the differences in their masses will influence their average speeds. Specifically, the equation for the root-mean-square speed (v_{rms}) of a gas is:
[tex]v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}[/tex]
Given:
Molar mass of helium (He) =[tex]4.0 g/mol[/tex]
Molar mass of neon (Ne) = [tex]20.2 g/mol[/tex]
We see that helium has a much smaller molar mass compared to neon. Therefore, for the same temperature T:
[tex]v_{rms, He} > v_{rms, Ne}[/tex]
This means the average speed of helium atoms is greater than that of neon atoms. Thus, the correct answer is:
C. the average speed of the helium atoms is greater than the average speed of the neon atoms.
Find the terminal speed of the rod if it has mass m = 2.8 grams , length l = 21 cm , and resistance R = 0.0011 Ω . It is falling in a uniform horizontal field B = 0.052 T . Neglect the resistance of the U-shaped conductor.
Answer:
0.253 m/s
Explanation:
As the conductor falls down, the magnetic flux throug the coil formed by the conductor and the the rest of the circuit changes, therefore an electromotive force is induced in the rod; its magnitude is given by
[tex]E=BvL[/tex]
where
B = 0.052 T is the strength of the magnetic field
v is the speed at which the rod is falling
L = 21 cm = 0.21 m is the length of the rod
Due to this electromotive force, a current is also induced in the rod and the circuit, and this current is given by
[tex]I=\frac{E}{R}[/tex]
where
[tex]R=0.0011 \Omega[/tex] is the resistance of the rod
So the current is
[tex]I=\frac{BvL}{R}[/tex] (1)
At the same time, we know that a current-carrying wire in a magnetic field experiences a force, which is given by
[tex]F_B = IBL[/tex] (2)
where in this case:
I is the induced current given by eq(1)
B is the strength of the magnetic field
L is the lenght of the rod
Inserting eq(1) into (2), we find that the magnetic force on the rod is:
[tex]F_B=\frac{BvL}{R}\cdot BL = \frac{B^2 L^2 v}{R}[/tex]
However, there is another force acting on the rod: the force of gravity, given by
[tex]F_g=mg[/tex]
where
[tex]m=2.8 g = 0.0028 kg[/tex] is the mass of the rod
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
The falling rod will reach its terminal speed when its net acceleration becomes zero; this occurs when the net force on it is zero, so when the magnetic force is balanced by the force of gravity, so when
[tex]F_B = F_g[/tex]
So
[tex]\frac{B^2 L^2v}{R}=mg[/tex]
And solving for v, we find the terminal speed:
[tex]v=\frac{mgR}{B^2L^2}=\frac{(0.0028)(9.8)(0.0011)}{(0.052)^2(0.21)^2}=0.253 m/s[/tex]
Which front formed widespread clouds rain or snow
Answer: A cold front occurs when cold, denser air replaces the rising, less dense air mass. The reason this front brings in the rain is that as the rising warm air cools (as it rises to the cooler upper atmosphere) the moisture in it condenses into clouds that precipitate down as rain or snow.
Answer:
cold front
Explanation:
Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual image of the mite. If the focal distance of the objective lens is shortened, then its magnification decreases. The converging objective lens forms a virtual image of the mite. If the focal distance of the objective lens is lengthened, then its magnification increases. The image formed by the objective lens should be located just inside the focal point of the eyepiece lens
Answer:
a) True. The image of the mite is virtual
e) True. The image must be within the focal length of the eyepiece len
Explanation:
Let's review the general characteristics of compound microscopes
Formed by two converging lenses
Magnification is
M = -L/fo 0.25/fe
Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length
Let's review the claims
a) True. The image of the mite is virtual
b) False. The effect is the opposite of the magnification equation
c) False. The objective lens forms a real image
d) False. As the seal distance increases the magnification decreases
e) True. The image must be within the focal length of the eyepiece len
A compound microscope uses an objective lens to create a real, inverted image larger than the object, and an eyepiece (ocular) to magnify this image further into a virtual image. The correct statements include that the eyepiece forms a virtual image, and the image from the objective lens should be within the eyepiece's focal length for further magnification.
Explanation:To understand how a compound microscope works, we analyze the function of its two lenses: the objective lens and the eyepiece. Initially, an object just beyond the objective lens's focal length (fobj) creates a real, inverted image that is larger than the object. This initial image acts as the object for the eyepiece lens, or ocular, which is placed within its own focal length (feye) to further magnify the image, effectively functioning as a magnifying glass. The resulting image is a virtual, larger, and further magnified version of the initial image, making it more comfortable for viewing as the eye is relaxed when looking at distant objects.
If we consider the statements provided, the correct statements are as follows:
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A 20 kg box on a horizontal frictionless surface is moving to the right at a speed of 4.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then oscillates back and forth. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero. ____ Positive ____ Negative ____ Zero
Answer:
Work done by the spring is negative
Explanation:
We can answer this question by thinking what is the force acting on the box.
In fact, the force acting on the box is the restoring force of the spring, which is given by Hooke's Law:
[tex]F=-kx[/tex]
where
k is the spring constant
x is the displacement of the box with respect to the equilibrium position of the spring
The negative sign in the equation indicates that the direction of the force is always opposite to the direction of the displacement: so, whether the spring is compressed or stretched, the force applied by the spring on the box is towards the equilibrium position.
The work done by the restoring force is also given by
[tex]W=Fx cos \theta[/tex]
where
F is the restoring force
x is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and the displacement
Here we know that the force is always opposite to the displacement, so
[tex]\theta=180^{\circ} \rightarrow cos \theta =-1[/tex]
Which means that the work done by the spring is always negative, since the direction of the restoring force is always opposite to the direction of motion.
The work done by the spring on the box as it compresses is negative, due to the force applied by the spring being in the opposite direction of the box's displacement.
Explanation:For the given problem, the work done by the spring on the box is negative. This is because the spring is exerting a force in the opposite direction to that of the box's motion, leading to a decrease in the kinetic energy of the box. According to the work-energy principle, if the force applied by the spring opposes the direction of motion, the work done is negative. When the spring compresses, it stores elastic potential energy, which is a conversion of the box's kinetic energy. The spring force acts to the left (assuming the right is the positive direction), while the displacement of the box due to that force is to the right, resulting in negative work done by the spring.