To solve this problem it is necessary to apply the concepts presented in Kepler's third law in which the period is described, as well as the theorems developed for acceleration based on gravity.
Acceleration in gravitational terms can be expressed as
[tex]a = \frac{GM}{r^2}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Earth
r = Distance
At the same time the Period by Kepler's law the Period is described as
[tex]T^2 = \frac{4\pi^2r^3}{GM}[/tex]
That is equal to
[tex]T^2 = \frac{4\pi^2r}{\frac{GM}{r^2}}[/tex]
Using the equation of acceleration,
[tex]T^2 = \frac{4\pi^2r}{a}[/tex]
Re-arrange to find a,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
Our values are given as
[tex]r = 1.5*10^{11}m[/tex]
[tex]T = 365days(\frac{86400s}{1days}) = 31536000s[/tex]
Replacing we have,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
[tex]a = \frac{4\pi^2(1.5*10^{11})}{(31536000)^2}[/tex]
[tex]a = 0.005954m/s^2[/tex]
Therefore the magnitude of the acceleration of the earth is [tex]0.005954m/s^2[/tex]
The magnitude of the acceleration of the Earth in its orbit around the sun, assuming a circular orbit, is approximately 0.0059 m/s².
Explanation:The concept at play here is centripetal acceleration, which is the rate of change of tangential velocity and points toward the center of the circle around which Earth, or any object, orbits. For Earth orbiting the sun, we can use the formula ac = v²/r (where v is velocity, r is radius), but given that v isn't directly stated, one must compute it first by v = 2πr/T (where T is the period of the orbit in seconds).
So, the velocity of the Earth in its orbit is approximated as 2π(1.50x10¹¹ m)/(365.25x24x60x60 s) = 29,785.6 m/s. To get the acceleration, plug this computed v value into the ac equation: ac = (29,785.6 m/s)²/(1.50 x 10¹¹ m) = 0.0059 m/s².
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A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential is â150 V. What is the change in the protonâs electric potential? What is the change in the potential energy of the proton? What is the work done on the proton?
Answer:
(a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Explanation:
Given that,
Speed [tex]v= 3.50\times10^{5}\ m/s[/tex]
Initial potential V=100 V
Final potential = 150 V
(a). We need to calculate the change in the protons electric potential
Potential energy of the proton is
[tex]U=qV=eV[/tex]
Using conservation of energy
[tex]K_{i}+U_{i}=K_{f}+U_{f}[/tex]
[tex]\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}[/tex]
[tex]]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})[/tex]
[tex]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V[/tex]
[tex]\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}[/tex]
Put the value into the formula
[tex]\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}[/tex]
[tex]\Delta V=639.2=0.639\ kV[/tex]
(b). We need to calculate the change in the potential energy of the proton
Using formula of potential energy
[tex]\Delta U=q\Delta V[/tex]
Put the value into the formula
[tex]\Delta U=1.6\times10^{-19}\times639.2[/tex]
[tex]\Delta U=1.022\times10^{-16}\ J[/tex]
(c). We need to calculate the work done on the proton
Using formula of work done
[tex]\Delta U=-W[/tex]
[tex]W=q(V_{2}-V_{1})[/tex]
[tex]W=-1.6\times10^{-19}(150-100)[/tex]
[tex]W=-8\times10^{-18}\ J[/tex]
Hence, (a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Final answer:
The change in electric potential and potential energy of the proton can be calculated based on the provided potentials. The work done on the proton equals the change in potential energy.
Explanation:
The change in the proton's electric potential: The change in electric potential is the final potential minus the initial potential, thus the change is -150 V - 100 V = -250 V.
The change in potential energy of the proton: The potential energy change equals the charge of the proton times the change in potential, giving -proton charge x change in potential.
The work done on the proton: The work done is equal to the change in the potential energy of the proton.
An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel turns at 1000 rpm when set to low. When the switch is changed from high to low, it takes the wheel 60 sec to slow down. A) ( 5 points) What is the initial angular speed of the high setting in rad/sec B) (5 points) What is the angular acceleration in rad/s2 of the wheel? C) (5 points) What is the angular speed in rad/s 40 seconds after the setting is changed? D) (5 points) How many revolutions did it make as it changes speed?
a) The initial angular speed is 209.3 m/s
b) The angular acceleration is [tex]-1.74 rad/s^2[/tex]
c) The angular speed after 40 s is 139.7 rad/s
d) The wheel makes 1501 revolutions
Explanation:
a)
The initial angular speed of the wheel is
[tex]\omega_i = 2000 rpm[/tex]
which means 2000 revolutions per minute.
We have to convert it into rad/s. Keeping in mind that:
[tex]1 rev = 2\pi rad[/tex]
[tex]1 min = 60 s[/tex]
We find:
[tex]\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s[/tex]
b)
To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.
Using the same procedure used in part a),
[tex]\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s[/tex]
Now we can find the angular acceleration, given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_i = 209.3 rad/s[/tex] is the initial angular speed
[tex]\omega_f = 104.7 rad/s[/tex] is the final angular speed
t = 60 s is the time interval
Substituting,
[tex]\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2[/tex]
c)
To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:
[tex]\omega' = \omega_i + \alpha t[/tex]
where we have
[tex]\omega_i = 209.3 rad/s[/tex]
[tex]\alpha = -1.74 rad/s^2[/tex]
And substituting t = 40 s, we find
[tex]\omega' = 209.3 + (-1.74)(40)=139.7 rad/s[/tex]
d)
The angular displacement of the wheel in a certain time interval t is given by
[tex]\theta=\omega_i t + \frac{1}{2}\alpha t^2[/tex]
where
[tex]\omega_i = 209.3 rad/s[/tex]
[tex]\alpha = -1.74 rad/s^2[/tex]
And substituting t = 60 s, we find:
[tex]\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad[/tex]
So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,
[tex]\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev[/tex]
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A puck of mass 0.5100.510kg is attached to the end of a cord 0.827m long. The puck moves in a horizontal circle without friction. If the cord can withstand a maximum tension of 126N, what is the highest frequency at which the puck can go around the circle without the cord breaking?
Answer:2.74 Hz
Explanation:
Given
mass Puck [tex]m=0.51 kg[/tex]
length of cord [tex]L=0.827 m[/tex]
Maximum Tension in chord [tex]T=126 N[/tex]
as the Puck is moving in a horizontal circle so maximum Tension in the string will be equal to centripetal force
[tex]F_c=m\omega ^2L=T[/tex]
[tex]126=0.51\times (\omega )^2\times 0.827[/tex]
[tex]\omega =\sqrt{298.74}[/tex]
[tex]\omega =17.28 rad/s[/tex]
[tex]\omega =2\pi f[/tex]
[tex]f=\frac{2\pi }{\omega }[/tex]
[tex]f=2.74 Hz[/tex]
To find the highest frequency at which the puck can go around the circle without the cord breaking, we use the formula for tension in a circular motion and solve for velocity. Then we use the velocity to find the frequency. The highest frequency is approximately 2.18 Hz.
To determine the highest frequency at which the puck can go around the circle without the cord breaking, we need to find the maximum tension in the cord.
Since the tension in the cord is equal to the centripetal force required to keep the puck moving in a circle, we can use the formula:
Tension = mass × velocity² / radius
Substituting the given values, we get:
126N = 0.51kg × v² / 0.827m
Now, solving for v, we find:
v² = (126N × 0.827m) / 0.51kg
v² = 204.2 m²/s²
v = √(204.2 m²/s²) = 14.29 m/s
Since the frequency of an object moving in a circle is equal to its velocity divided by the circumference of the circle, we can calculate the highest frequency as:
Frequency = v / (2πr)
Frequency = 14.29 m/s / (2π × 0.827m)
Frequency ≈ 2.18 Hz
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Complete the following statement: The interior of a thermos bottle is silvered to minimize heat transfer due to
A. conduction and convection
B. conduction
C. conduction, convection and radiation
D. conduction and radiation
E. radiation.
Answer:
E. radiation.
Explanation:
As we know that heat transfer due to conduction depends on thermal conductivity of the materials and heat transfer due to convection depends on the velocity of the fluid.But on the other hand heat transfer due to radiation depends on the surface properties like emmisivity .So when bottle is silvered then it will leads to minimize the radiation heat transfer.
Therefore answer is --
E. radiation.
The answer is option C.
The interior of a thermos is silvered to minimize heat transfer by conduction, convection, and radiation. The silvering acts like a mirror, reflecting heat, and the vacuum between the thermos walls almost eliminates conduction and convection.
Explanation:The interior of a thermos bottle is silvered to minimize heat transfer due to C. conduction, convection and radiation.
This is because the silvering on the inner surface of the thermos acts like a mirror, reducing the amount of heat that can be transferred by all three modes: conduction, convection, and especially radiation.
Conduction is the transfer of heat through direct contact of molecules, minimized in a thermos by the vacuum between its double walls. Convection is the transfer of heat in a fluid (like air or liquid) through the motion of the fluid itself, which is also nearly eliminated by the vacuum. Radiation is the transfer of heat through electromagnetic waves, which the silvering reflects back, greatly reducing heat loss this way.
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Sound level B in decibels is defined as
B= 10 log (i/i)
where i = 1 × 10-12 W/m2 . The decibel
scale intensity for busy traffic is 80 dB. Two
people having a loud conversation have a deci-
bel intensity of 70 dB.
What is the approximate combined sound
intensity?
Answer in units of W/m2
Answer:
The approximate combined sound intensity is [tex]I_{T}=1.1\times10^{-4}W/m^{2}[/tex]
Explanation:
The decibel scale intensity for busy traffic is 80 dB. so intensity will be
[tex]10log(\frac{I_{1}}{I_{0}} )=80[/tex], therefore [tex]I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}[/tex]
In the same way for the loud conversation having a decibel intensity of 70 dB.
[tex]10log(\frac{I_{2}}{I_{0}} )=70[/tex], therefore [tex]I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}[/tex]
Finally we add both of them [tex]I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}[/tex], is the approximate combined sound intensity.
The human eye is most sensitive to green light of wavelength 505 nm . Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina.
1.What is the frequency of this photon?
2.How much energy (in joules and eV ) does it deliver to the receptor cells?
Answer:
Answer:
Answer:
1. f = 5.94 x 10^14 Hz
2. 3.94 x 10^-19 J or 2.46 eV
Explanation:
wavelength, λ = 505 nm = 505 x 10^-9 m
speed of light,c = 3 x 10^8 m/s
1. Let the frequency of the light is f.
[tex]f=\frac{c}{\lambda }[/tex]
[tex]f=\frac{3 \times 10^{8}}{505 \times 10^{-9}}[/tex]
f = 5.94 x 10^14 Hz
2. Energy is given by
E = h x f
where, h is the Plank's constant.
E = 6.63 x 10^-34 x 5.94 x 10^14
E = 3.94 x 10^-19 J
Now, we know that 1 eV = 1.6 x 10^-19 J
E = 2.46 eV
Explanation:
Explanation:
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250, respectively. After sliding 77.3 cm along the incline, the block slides across a frictionless horizontal surface and encounters a spring (k = 35.0 N/m).What is the maximum compression of the spring?
Answer:x=23.4 cm
Explanation:
Given
mass of block [tex]m=0.5 kg[/tex]
inclination [tex]\theta =30[/tex]
coefficient of static friction [tex]\mu =0.35[/tex]
coefficient of kinetic friction [tex]\mu _k=0.25[/tex]
distance traveled [tex]d=77.3 cm[/tex]
spring constant [tex]k=35 N/m [/tex]
work done by gravity+work done by friction=Energy stored in Spring
[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]
[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]
[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]
[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]
[tex]x=0.234 m[/tex]
[tex]x=23.4 cm[/tex]
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.60 x 10^4 kg/m^3 at the center and 2100 kg/m^3 at the surface.
What is the acceleration due to gravity at the surface of this planet?
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance
[tex]\rho(r) = ar^2 - br^3[/tex]
r is the radius of the spherical shell
dr is the thickness
volume of shell
[tex]dV = 4 \pi r^2 dr[/tex]
mass of shell
[tex]dM = \rho(r)dV[/tex]
[tex]\rho = \rho_0 - br[/tex]
now,
[tex]dM = (\rho_0 - br)(4 \pi r^2)dr[/tex]
integrating both side
[tex]M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr[/tex]
[tex]M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})[/tex]
[tex]M = \pi R^3(\dfrac{\rho_0}{3}+\rho)[/tex]
we know,
[tex]a = \dfrac{GM}{R^2}[/tex]
[tex]a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}[/tex]
[tex]a =\pi RG(\dfrac{\rho_0}{3}+\rho)[/tex]
[tex]a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)[/tex]
a = 9.94 m/s²
Consider the air moving over the top of the light bulb. The streamlines near the bulb will be squeezed together as the air goes over the top of the bulb. This leads to a region of _________ on the top of the bulb.
Answer:
low pressure
Explanation:
The streamlines of air particles are squeezed together as the air goes over the top of the bulb. Then, by the law of conservation of mass, the velocity of air particles are increased. And since, the velocity is increase the pressure is bound to decrease. Hence, this leads to region of low pressure on the top of the bulb.
A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.
(a) Calculate the half-life of this nuclide.
What fraction of a sample will remain at the end of (b) 4.44 half-lives and (c) 14.6 days?
Explanation:
Given that,
The disintegration constant of the nuclide, [tex]\lambda=0.0178\ h^{-1}[/tex]
(a) The half life of this nuclide is given by :
[tex]t_{1/2}=\dfrac{ln(2)}{\lambda}[/tex]
[tex]t_{1/2}=\dfrac{ln(2)}{0.0178}[/tex]
[tex]t_{1/2}=38.94\ h[/tex]
(b) The decay equation of any radioactive nuclide is given by :
[tex]N=N_oe^{-\lambda t}[/tex]
[tex]\dfrac{N}{N_o}=e^{-\lambda t}[/tex]
Number of remaining sample in 4.44 half lives is :
[tex]t_{1/2}=4.44\times 38.94[/tex]
[tex]t_{1/2}=172.89\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 172.89}[/tex]
[tex]\dfrac{N}{N_o}=0.046[/tex]
(c) Number of remaining sample in 14.6 days is :
[tex]t_{1/2}=14.6\times 24[/tex]
[tex]t_{1/2}=350.4\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 350.4}[/tex]
[tex]\dfrac{N}{N_o}=0.0019[/tex]
Hence, this is the required solution.
Suppose the Earth's magnetic field at the equator has magnitude 0.00005 T and a northerly direction at all points. How fast must a singly ionized uranium atom (m=238u, q=e) move so as to circle the Earth 1.44 km above the equator? Give your answer in meters/second.
Answer:
Velocity will be [tex]v=1.291\times 10^8m/sec[/tex]
Explanation:
We have given magnetic field B = 0.00005 T
Mass m = 238 U
We know that [tex]1u=1.66\times 10^{-27}kg[/tex]
So 238 U [tex]=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg[/tex]
Radius [tex]=R+1.44=6378+1.44=6379.44KM[/tex]
We know that magnetic force is given by
[tex]F=qvB[/tex] which is equal to the centripetal force
So [tex]qvB=\frac{mv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}[/tex]
[tex]v=1.291\times 10^8m/sec[/tex]
A particular AM radio station broadcasts at a frequency of 1020 kilohertz. What is the wavelength of this electromagnetic radiation?
1 m
How much time is required for the radiation to propagate from the broadcasting antenna to a radio 3 km away?
2 s
Explanation:
Given that,
Frequency of the AM radio station, [tex]f=1020\ kHz=1020\times 10^3\ Hz[/tex]
(a) Let [tex]\lambda[/tex] is the wavelength of this electromagnetic radiation. It can be calculated as :
[tex]\lambda=\dfrac{c}{f}[/tex]
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{1020\times 10^3\ Hz}[/tex]
[tex]\lambda=294.11\ m[/tex]
Since the wavelength of one cycle is 294.12 m, then the total number of cycles over a 3 km distance is:
[tex]n=\dfrac{3000}{294.12}=10.19\ cycles[/tex]
Let the period is the duration of one cycle is given by:
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{1020\times 10^3}[/tex]
[tex]T=9.8\times 10^{-7}\ s[/tex]
So, total time required is t as :
[tex]t=n\times T[/tex]
[tex]t=10.19\times 9.8\times 10^{-7}[/tex]
[tex]t=9.98\times 10^{-6}\ s[/tex]
The wavelength of the electromagnetic radiation from the AM radio station is approximately 293.1 meters. It takes approximately 10 milliseconds for the radiation to propagate from the broadcasting antenna to a radio 3 km away.
Explanation:The wavelength of electromagnetic radiation can be calculated using the formula:
wavelength = speed of light / frequency
Given that the frequency of the AM radio station is 1020 kilohertz and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula to find the wavelength:
wavelength = (3 x 10^8 m/s) / (1020 x 10^3 Hz)
Simplifying the expression gives us a wavelength of approximately 293.1 meters.
To calculate the time required for the radiation to propagate from the broadcasting antenna to a radio 3 km away, we can use the formula:
time = distance / speed
Given that the distance is 3 km and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula:
time = (3 km) / (3 x 10^8 m/s)
Converting kilometers to meters gives us a distance of 3000 meters:
time = (3000 m) / (3 x 10^8 m/s)
Simplifying the expression gives us a time of 0.01 seconds or 10 milliseconds.
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A toy balloon, which has a mass of 2.90 g before it is inflated, is filled with helium (with a density of 0.180 kg/m^3) to a volume of 8400 cm^3. What is the minimum mass that should be hung from the balloon to prevent it from rising up into the air? Assume the air has a density of 1.29 kg/m^3.
Answer:
[tex]M=6.4243\ g[/tex]
Explanation:
Given:
mass of deflated balloon, [tex]m_b=2.9\ g=0.0029\ kg[/tex]density of helium, [tex]\rho_h=0.180\ kg.m^{-3}[/tex]volume of inflation, [tex]V=8400\ cm^3=0.0084\ m^3[/tex]density of air, [tex]\rho_a=1.29\ kg.m^{-3}[/tex]To stop this balloon from rising up we need to counter the buoyant force.
mass of balloon after inflation:
[tex]m=m_h+m_b[/tex]
[tex]m=0.0084\times 0.180+0.0029[/tex]
[tex]m=0.004412\ kg[/tex]
Now the density of inflated balloon:
[tex]\rho_b=\frac{m}{V}[/tex]
[tex]\rho_b=\frac{0.004412}{0.0084}[/tex]
[tex]\rho_b=0.5252\ kg.m^{-3}[/tex]
Now the buoyant force on balloon
[tex]F_B=V(\rho_a-\rho_b).g[/tex]
[tex]F_B=0.0084(1.29-0.5252)\times 9.8[/tex]
[tex]F_B=0.063\ N[/tex]
∴Mass to be hung:
[tex]M=\frac{F_B}{g}[/tex]
[tex]M=0.00642432\ kg[/tex]
[tex]M=6.4243\ g[/tex]
Two waves traveling on a string in the same direction both have a frequency of 135 Hz, a wavelength of 2 cm, and an amplitude of 0.04 m. What is the amplitude of the resultant wave if the original waves differ in phase by each of the following values?
(a) p/6 cm(b) p/3 cm
Answer:
The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
Explanation:
Given that,
Frequency = 135 Hz
Wavelength = 2 cm
Amplitude = 0.04 m
We need to calculate the angular frequency
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\times\pi\times135[/tex]
[tex]\omega=848.23\ rad/s[/tex]
As the two waves are identical except in their phase,
The amplitude of the resultant wave is given by
[tex]y+y=A\sin(kx-\omega t)+Asin(kx-\omega t+\phi)[/tex]
[tex]y+y=A[2\sin(kx-\omega t+\dfrac{\phi}{2})\cos\phi\dfrac{\phi}{2}[/tex]
[tex]y'=2A\cos(\dfrac{\phi}{2})\sin(kx-\omega t+\dfrac{\phi}{2})[/tex]
(a). We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{6}[/tex]
The amplitude of the resultant wave is
[tex]A'=2A\cos(\dfrac{\phi}{2})[/tex]
Put the value into the formula
[tex]A'=2\times0.04\cos(\dfrac{\pi}{12})[/tex]
[tex]A'=0.0772\ m[/tex]
(b), We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{3}[/tex]
[tex]A'=2\times0.04\cos(\dfrac{\pi}{6})[/tex]
[tex]A'=0.0692\ m[/tex]
Hence, The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
About once every 30 minutes, a geyser known as Old Faceful projects water 18.0 m straight up into the air. Use g = 9.80 m/s2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m3. What is the speed of the water when it emerges from the ground?
Answer:
Speed of the water that emerge out of the pipe is 18.8 m/s
Explanation:
Since we know that water drops projected upwards to maximum height of 18 m
So here we can use kinematics equations here
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we have
[tex]v_f = 0[/tex]
[tex]d = 18 m[/tex]
[tex]a = -9.80 m/s^2[/tex]
so we will have
[tex]0 - v_i^2 = 2(-9.80)(18)[/tex]
[tex]v_i = 18.8 m/s[/tex]
A large man sits on a four-legged chair with his feet off the floor. The combined mass of the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.600 cm at the bottom, what pressure does each leg exert on the floor?
Answer:
Pressure, [tex]P=2.05\times 10^6\ Pa[/tex]
Explanation:
Given that,
The combined mass of the man and chair is 95.0 kg, m = 95 kg
The radius of the circular leg of the chair, r = 0.6 cm
Area of the 4 legs of the chair, [tex]A=4\times \pi r^2[/tex]
Let P is the pressure each leg exert on the floor. The total force acting per unit area is called pressure exerted. Its expression is given by :
[tex]P=\dfrac{F}{A}[/tex]
[tex]P=\dfrac{mg}{4\times \pi r^2}[/tex]
[tex]P=\dfrac{95\times 9.8}{4\times \pi (0.6\times 10^{-2})^2}[/tex]
[tex]P=2.05\times 10^6\ Pa[/tex]
So, the pressure exerted by each leg on the floor is [tex]2.05\times 10^6\ Pa[/tex]. Hence, this is the required solution.
To calculate the pressure each chair leg exerts on the floor, convert the combined mass of the man and chair into force using the equation F = mg, and then calculate the area of each chair leg using the formula A = πr². Finally, divide the total force by the number of legs and then that value by the area of each leg.
Explanation:The key to solving this problem is recognizing that the pressure exerted by the man and chair on each chair leg is equal to the force of their combined weight divided by the area of contact the chair legs make with the floor. First, convert the combined mass of the man and chair (95.0 kg) into force in newtons, using the equation F = mg, where F is the force in Newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.8 m/s²). In this case, F = 95.0 kg * 9.8 m/s² = 931 N.
Next, calculate the area of each chair leg that is in contact with the floor. Since each leg is circular, its area, A, can be calculated using the formula A = πr², where r is the radius. Remember to convert the radius from centimeters to meters. Therefore, A = π*(0.006 m)² = 0.000113 m².
Finally, divide the total force by four (since the weight is distributed evenly over four legs) and then that value by the area of each leg to find the pressure each leg exerts: P = F/(4*A) = 931 N / 4 / 0.000113 m² = 2063363.4 Pascal. Remember to use the correct units of pressure (Pascal).
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If a 110-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions. Part A How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 2.8 m away? Express your answer using two significant figures.
Answer:
9.7 x 10¹¹ .
Explanation:
2.5 % of 110 W = 2.75 J/s
energy of one photon
= hc / λ
=[tex]\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}[/tex]
= .036 x 10⁻¹⁷ J
No of photons emitted
= 2.75 / .036 x 10⁻¹⁷
= 76.38 x 10¹⁷
Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance
photons passing per unit area of this sphere
= 76.38 x 10¹⁷ / 4π ( 2.8)²
Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing
= [tex]\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2[/tex]
= 9.7 x 10¹¹ .
Your local AM radio station broadcasts at a frequency of f = 1100 kHz. The electric-field component of the signal you receive at your home has the time dependence E(t) = E0 sin(2πft), where the amplitude is E0 = 0.62 N/C. Radio waves travel through air at approximately the speed of light.
a) At what wavelength, in meters, docs this station broadcast?
b) What is the value of the radio wave's electric field, in newtons per coulomb, at your home at a time of t = 3.1 μs?
Final answer:
a) The wavelength of the radio station's broadcast is approximately 272.73 meters. b) At a time of 3.1 μs, the value of the radio wave's electric field is approximately 0.619 N/C.
Explanation:
a) To calculate the wavelength of the radio station's broadcast, we can use the formula λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. Plugging in the given frequency of 1100 kHz (or 1100 x 10^3 Hz), we get: λ = (3 x 10^8 m/s) / (1100 x 10^3 Hz) = 272.73 m
b) To find the value of the radio wave's electric field at a specific time, we can use the given time dependence equation E(t) = E0 sin(2πft), where E0 is the amplitude, f is the frequency, and t is the time. Plugging in the given amplitude of E0 = 0.62 N/C, frequency of 1100 kHz (or 1100 x 10^3 Hz), and time of 3.1 μs (or 3.1 x 10^-6 s), we get: E(t) = 0.62 sin(2π x 1100 x 10^3 x 3.1 x 10^-6) ≈ 0.619 N/C
A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per second, find the rate at which the top of the ladder is sliding down when the top of the ladder is 8 feet from the ground.
Answer:
6.87 ft/s is the rate at which the top of ladder slides down.
Explanation:
Given:
Length of the ladder is, [tex]L=20\ ft[/tex]
Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.
Now, from triangle ABC,
AB² + BC² = AC²
[tex]h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1[/tex]
Differentiating the above equation with respect to time, 't'. This gives,
[tex]\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2[/tex]
In the above equation the term [tex]\frac{dh}{dt}[/tex] is the rate at which top of ladder slides down and [tex]\frac{db}{dt}[/tex] is the rate at which bottom of ladder slides away.
Now, as per question, [tex]h=8\ ft, \frac{db}{dt}=3\ ft/s[/tex]
Plug in [tex]h=8[/tex] in equation (1) and solve for [tex]b[/tex]. This gives,
[tex]8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft[/tex]
Now, plug in all the given values in equation (2) and solve for [tex]\frac{dh}{dt}[/tex]
[tex]8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s[/tex]
Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.
You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 3.1 g, so that the center of mass of the coins is directly over the end of the meter stick. The pivot point is a distance d from the 0-cm end of the meter stick.
Part (a): Determine the distance d = d1, in centimeters, if there is only one coin o the 0 end of the meter stick and the system is in static equilibrium
The distance [tex]\( d_1 \)[/tex] from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, maintaining static equilibrium, is approximately 46.8 cm.
1. To achieve static equilibrium, the torques on both sides of the pivot point must balance out.
2. The torque due to the meter stick with mass ( M ) is [tex]\( M \times g \times \frac{L}{2} \)[/tex], where ( L ) is the length of the meter stick (100 cm) and \( g \) is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).
3. The torque due to the coin on the 0-cm end is [tex]\( m \times g \times d_1 \)[/tex], where [tex]\( m \)[/tex] is the mass of the coin.
4. Since the torques balance out, we have the equation: [tex]\( M \times g \times \frac{L}{2} = m \times g \times d_1 \).[/tex]
5. Rearrange the equation to solve for [tex]\( d_1 \): \( d_1 = \frac{M \times \frac{L}{2}}{m} \).[/tex]
6. Substitute the given values: [tex]\( d_1 = \frac{95 \, \text{g} \times \frac{100}{2} \, \text{cm}}{3.1 \, \text{g}} \).[/tex]
7. Calculate [tex]\( d_1 \)[/tex]to find the distance from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, which is approximately 46.8 cm.
A uniform-density wheel of mass 10 kg and radius 0.36 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 14 N for 0.76 s. (a) What is the final angular speed? radians/s (b) What was the average angular speed? radians/s (c) Through how big an angle did the wheel turn? radians/s (d) How much string came off the wheel?
Answer:
(a) 5.91 rad/s
(b) 2.96 rad/s
(c) 2.25 rad
(d) 0.81 m
Explanation:
The torque generated by tension force from the string is:
T = FR = 14*0.36 = 5.04 Nm
This torque would then create an angular acceleration on the uniform-density wheel with moments of inertia of
[tex] I = 0.5mR^2 = 0.5*10*0.36^2 = 0.648kgm^2[/tex]
[tex]\alpha = \frac{T}{I} = \frac{5.04}{0.648}=7.78rad/s^2[/tex]
(a) The wheel turns for 0.76s, this means the final angular speed is
[tex]\omega_f = t\alpha = 0.76*7.78 = 5.91 rad/s[/tex]
(b) Since the force is constant, the torque is also constant and so is the angular acceleration. This means angular speed is rising at a constant rate. That means the average angular speed is half of the final speed
[tex]\omega_a = 0.5\omega_f = 0.5*5.91 = 2.96 rad/s[/tex]
(c) The total angle that the wheel turns is the average angular speed times time
[tex]\theta = t\omega_a = 2.96*0.76 = 2.25 rad[/tex]
(d) The string length coming off would equal to the distance swept by the wheel
[tex]d = R\theta = 0.36*2.25 = 0.81 m[/tex]
The final angular speed of this uniform-density wheel of mass is equal to 5.91 radians/s.
Given the following data:
Mass = 10 kg.
Radius = 0.36 m.
Initial velocity = 0 m/s (since it's starting from rest).
Force = 14 Newton.
Time = 0.76 seconds.
How to calculate the final angular speed.First of all, we would determine the torque produced due to the tensional force that is acting on the string by using this formula:
[tex]\tau = Fr\\\\\tau = 14 \times 0.36[/tex]
Torque = 5.04 Nm.
Also, we would determine the moment of inertia by using this formula;
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 10 \times 0.36^2\\\\I=5 \times 0.1296[/tex]
I = 0.648 [tex]kgm^2[/tex]
Next, we would determine the angular acceleration by using this formula;
[tex]\tau=\alpha I\\\\\alpha =\frac{\tau}{I} \\\\\alpha =\frac{5.04}{0.648}\\\\\alpha = 7.78 \;rad/s^2[/tex]
Now, we can calculate the final angular speed:
[tex]\omega_f = t\alpha \\\\\omega_f = 0.76 \times 7.78\\\\\omega_f = 5.91 \;rad/s[/tex]
How to calculate the average angular speed.[tex]\omega_A = \frac{1}{2} \omega_f\\\\\omega_A = \frac{1}{2} \times 5.91\\\\\omega_A =2.96\;rad/s[/tex]
How to calculate the angle.[tex]\theta = t\omega_A \\\\\theta = 0.76 \times 2.96[/tex]
Angle = 2.25 rad.
In order to calculate the length of the string that came off the wheel, we would determine the distance swept by the wheel:
[tex]d=r\theta\\\\d=0.36 \times 2.25[/tex]
d = 0.81 meter.
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Two long parallel wires, each carrying a current of 12 A, lie at a distance of 9 cm from each other. What is the magnetic force per unit length exerted by one wire on the other? The magnetic force per unit length exerted by one wire on the other is ×10-4 N/m.
Answer:
Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]
Explanation:
We have given that two long parallel wires each carrying a current of 12 A
So [tex]I_1=I_2=12A[/tex]
Distance between the two conductors d = 9 cm = 0.09 m
We know that magnetic force between two parallel conductors per unit length is given by
[tex]F=\frac{\mu _0I_1I_2}{2\pi d}=\frac{4\times 3.14\times 10^{-7}\times 12\times 12}{2\times 3.14\times 0.09}=3.2\times 10^{-4}N[/tex]
Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]
Cart A, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart B, which has a mass of 0.8 kg and is initially at rest. After the collision the carts stick together, and the center of mass of the two cart system has a kinetic energy of
A. 3.5 J B. 7.2 J C. 2.4 J D 1.2 JE. 4.8 J
Answer:
C. The two cart system has a kinetic energy of 2.4 J.
Explanation:
Hi there!
The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:
initial momentum of the system = final momentum of the system
pA + pB = p (A + B)
mA · vA + mB · vB = (mA + mB) · v
Where:
pA and pB = initial momentum of carts A and B respectively.
p (A +B) = momentum of the two cart system after the collision.
mA and mB = mass of carts A and B respectively.
vA and vB = initial velocity of carts A and B.
v = velocity of the two cart system.
We have the following data:
mA = 0.4 kg
mB = 0.8 kg
vA = 6 m/s
vB = 0 m/s
Solving the equation for v:
mA · vA + mB · vB = (mA + mB) · v
0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v
2.4 kg m/s = 1.2 kg · v
v = 2.4 kg m/s / 1.2 kg
v = 2 m/s
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where m is the mass of the object and v its speed.
Replacing with the data we have obtained:
KE = 1/2 · 1.2 kg · (2 m/s)²
KE = 2.4 J
The two cart system has a kinetic energy of 2.4 J.
After the collision, the two carts stick together and move with a common velocity. The kinetic energy of the two carts after the collision is 2.4 J.
Explanation:First, we need to calculate the initial momentum of cart A and the final momentum of the two carts combined.
The initial momentum of cart A is given by the formula: momentum = mass * velocity.
So, momentum of cart A = 0.4 kg * 6 m/s = 2.4 kg*m/s.
After the collision, the two carts stick together and move with a common velocity.
Using the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Hence, 2.4 kg*m/s = (0.4 kg + 0.8 kg) * final velocity.
Solving for the final velocity, we get: final velocity = 2.4 kg*m/s / 1.2 kg = 2 m/s.
Finally, we can calculate the kinetic energy of the two carts after the collision using the formula: kinetic energy = (1/2) * mass * velocity^2.
Kinetic energy = (1/2) * (0.4 kg + 0.8 kg) * (2 m/s)^2 = 2.4 J.
Therefore, the correct answer is option C. 2.4 J.
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Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, assuming the following: (1) an average fiber diameter of 10 10-3 mm, (2) an average fiber length of 1 mm, (3) a fiber fracture strength of 2.5 GPa, (4) a fiber-matrix bond strength of 10 MPa, (5) a matrix stress at fiber failure of 10.0 MPa, and (6) a matrix tensile strength of 75 MPa.
Answer:
632.5 MPa
Explanation:
[tex]\sigma_{m}[/tex] = Matrix stress at fiber failure = 10 MPa
[tex]V_f[/tex] = Volume fraction of fiber = 0.25
[tex]\sigma_f[/tex] = Fiber fracture strength = 2.5 GPa
The longitudinal strength of a composite is given by
[tex]\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa[/tex]
The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa
Final answer:
The given question seeks to calculate the longitudinal strength of a carbon fiber-epoxy composite, but lacks sufficient detail or formulae for a complete answer. Typically, this computation would involve using materials science models that consider fiber orientation and other stress-strain interactions between components.
Explanation:
To compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite with a 0.25 volume fraction of fibers, we'd need to consider the following given properties: the average fiber diameter, average fiber length, fiber fracture strength, fiber-matrix bond strength, matrix stress at fiber failure, and matrix tensile strength.
Although the exact method for calculating the longitudinal strength would typically involve applying principles from materials science, such as the rule of mixtures, in combination with the given data points, the actual question does not provide enough information or a specific formula to complete the calculation. For a real-life carbon fiber-epoxy composite, the longitudinal strength could be substantially influenced by the alignment of the fibers, bond quality between the fibers and matrix, and the interaction between the stress and strain of the components.
If we had a suitable model or empirical formula, we would proceed by plugging in the given values to determine the longitudinal strength. However, as the provided data from the question is incomplete for this calculation, it is recommended to refer to a textbook or comprehensive resource on composite material mechanics for the detailed step-by-step methodology and equations.
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 6.17 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 2.09 s. Determine the following. (a) force constant of the springs? N/m
(b) mass of the trailer? kg
(c) frequency of the oscillation? Hz
(d) time it takes for the trailer to bounce up and down 10 times? s
a) The spring constant is 12,103 N/m
b) The mass of the trailer 2,678 kg
c) The frequency of oscillation is 0.478 Hz
d) The time taken for 10 oscillations is 20.9 s
Explanation:
a)
When the two children jumps on board of the trailer, the two springs compresses by a certain amount
[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]
Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:
[tex]2mg = k'\Delta x[/tex] (1)
where
m = 76.2 kg is the mass of each children
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]k'[/tex] is the equivalent spring constant of the 2-spring system
For two springs in parallel each with constant k,
[tex]k'=k+k=2k[/tex]
Substituting into (1) and solving for k, we find:
[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]
b)
The period of the oscillating system is given by
[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]
where
And for the system in the problem, we know that
T = 2.09 s is the period of oscillation
m is the mass of the trailer
[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system
Solving the equation for m, we find the mass of the trailer:
[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]
c)
The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
In this problem, we have
T = 2.09 s is the period
Therefore, the frequency of oscillation is
[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]
d)
The period of the system is
T = 2.09 s
And this time is the time it takes for the trailer to complete one oscillation.
In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.
Therefore, the time needed for 10 oscillations is:
[tex]t=10T=10(2.09)=20.9 s[/tex]
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Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years in a circular orbit around its star, which has a measured mass of 3.03×1030 kg . Find the radius of the exoplanet's orbit.
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of about 4ˆ10´3 mm but there are literally billions of them. The average speed of blood through the capillaries is about 5ˆ10´4 m/s. (i) Calculate the effective cross sectional area of the capillaries and (ii) the approximate number of capillaries.
Answer:
(I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
Explanation:
Given that,
Radius of aorta = 10 mm
Speed = 300 mm/s
Radius of capillary [tex]r=4\times10^{-3}\ mm[/tex]
Speed of blood [tex]v=5\times10^{-4}\ m/s[/tex]
(I). We need to calculate the effective cross sectional area of the capillaries
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
Where. v₁ = speed of blood in capillarity
A₂ = area of cross section of aorta
v₂ =speed of blood in aorta
Put the value into the formula
[tex]A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}[/tex]
[tex]A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}[/tex]
[tex]A_{1}=0.188\ m^2[/tex]
(II). We need to calculate the approximate number of capillaries
Using formula of area of cross section
[tex]A_{1}=N\pi r_{c}^2[/tex]
[tex]N=\dfrac{A_{1}}{\pi\times r_{c}^2}[/tex]
Put the value into the formula
[tex]N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}[/tex]
[tex]N=3.74\times10^{9}[/tex]
Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
Which of the following statements is false?
• The energy of electromagnetic radiation increases as its frequency increases.
• An excited atom can return to its ground state by absorbing electromagnetic radiation.
• An electron in the n = 4 state in the hydrogen atom can go to the n = 2 state by emitting electromagnetic radiation at the appropriate frequency.
• The frequency and the wavelength of electromagnetic radiation are inversely proportional to each other.
An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.
Option B
Explanation:
In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.
Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.
The statement 'An excited atom can return to its ground state by absorbing electromagnetic radiation' is false, as the process actually involves emitting radiation. Electromagnetic radiation's energy increases with frequency, and frequency and wavelength have an inverse relationship.
Explanation:The false statement among the ones provided is: An excited atom can return to its ground state by absorbing electromagnetic radiation. An excited atom returns to its ground state by emitting radiation, not absorbing it. When it comes to electromagnetic radiation, increasing frequency indeed results in increasing energy. This is because the energy of electromagnetic radiation is directly proportional to its frequency. Additionally, when an electron in the n = 4 state in the hydrogen atom transitions to the n = 2 state, it emits radiation at an appropriate frequency. Lastly, the frequency and the wavelength of electromagnetic radiation are inversely proportional, meaning as one increases, the other decreases.
A vehicle moves with a velocity, v(t) = exp(0.2t) - 1, 0 ≤ t ≤ 5 s. Peter would like to calculate the displacement of the vehicle as a function of time, x(t), by integrating given velocity over the time from t = 0. Use t = 0.2 s for trapezoidal rule.
Answer:
[tex]x|_0^{0.2}=1.59535[/tex]
Explanation:
Given expression of velocity:
[tex]v(t)=10^{0.2t}-1 ;\ \ 0\leq t\leq 5\ s[/tex]
For getting displacement we need to integrate the above function with respect to t.
Given period of integration:
[tex]t_0=0\ s \to t_f=0.2\ s[/tex]
For trapezoidal rule we break the given interval into two parts of 0.1 s each.
∴take n=2
hence, [tex]\Delta t= 0.1[/tex]
[tex]v(0)=0[/tex]
[tex]v(0.1)=1.0471[/tex]
[tex]v(0.2)=1.0965[/tex]
Now, using trapezoidal rule:
[tex]\int_{0}^{0.2}v(t)\ dt=\Delta x[\frac{1}{2}\times v(0)+v(0.1)+\frac{1}{2}\times v(0.2)][/tex]
[tex]\int_{0}^{0.2}v(t)\ dt=0.1 [\frac{1}{2}\times 0+1.0471+\frac{1}{2}\times 1.0965][/tex]
[tex]x|_0^{0.2}=1.59535[/tex]
Note:Smaller the value of sub-interval better is the accuracy.
A uniform plank is 3.0 m long and has a mass of 10 kg. It is secured at its left end in a horizontal position to be used as a diving platform. To keep the plank in equilibrium, the point of support must supply: a. an upward force and a clockwise torque b. a downward force and a clockwise torque c. an upward force and a counter-clockwise torque d. a downward force and a counter-clockwise torque e. none of these
To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.
By torque it is understood that
[tex]\tau = F*d[/tex]
Where,
F= Force
d = Distance
The value of the given Torque acts from the center of mass causing it to rotate clockwise.
The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.
I enclose a graph that allows us to understand the problem in a more didactic way.
The correct answer is D.