Answer: The answer is 225 joules
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 7.2 m/s, the marine biologist measures the number of clicks occurring per second to be at a frequency of 2674 Hz. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
Answer:
12.64968 Hz
Explanation:
v = Velocity of sound in seawater = 1522 m/s
u = Velocity of dolphin = 7.2 m/s
f' = Actual frequency = 2674 Hz
From Doppler effect we get the relation
[tex]f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz[/tex]
The frequency that will be received is 2661.35032 Hz
The difference in the frequency will be
[tex]2674-2661.35032=12.64968\ Hz[/tex]
Final answer:
The difference between the observed frequency and the actual frequency of clicks emitted by the dolphin, calculated using the Doppler Effect, is approximately 12.7 Hz.
Explanation:
The question involves calculating the observed frequency of sound due to the Doppler Effect, which occurs when the source of the sound is moving relative to the observer. In this case, the dolphin is the source of sound waves (clicks), moving away from the marine biologist, who acts as the observer.
The formula for the Doppler Effect for a source moving away from the observer is given by:
f' = f * (v / (v + vs))
Where:
f' is the observed frequency
f is the actual frequency emitted by the source
v is the speed of sound in the medium (1522 m/s in seawater)
vs is the speed of the source relative to the medium (7.2 m/s)
The marine biologist measures the observed frequency (f') as 2674 Hz. We can rearrange the formula to solve for the actual frequency (f):
f = f' * (v + vs) / v
f = 2674 Hz * (1522 m/s + 7.2 m/s) / 1522 m/s
After performing the calculation:
f ≈ 2686.7 Hz
The difference in frequency is then:
Δf = f - f'
Δf ≈ 2686.7 Hz - 2674 Hz
Δf ≈ 12.7 Hz
Therefore, the difference between the observed frequency and the actual frequency of clicks emitted by the dolphin is approximately 12.7 Hz.
The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered what very significant fact about the spectra produced by hot gases, such as elements heated in a flame?
The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.
Explanation:A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.
Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.
The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a 3.00-m^2 window that is 0.635 cm thick (1/4 in) if the temperatures of the inner and outer surfaces are 5.00ºC and −10.0ºC, respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation. (answer in ×10^{3} W)
Answer:
[tex]5.9527559\times 10^3\ W[/tex]
Explanation:
Q = Heat
t = Thickness = d = 0.635 cm
[tex]k_g[/tex] = Heat conduction coefficient of glass = 0.84 W/m °C (general value)
[tex]\Delta T[/tex] = Change in temperature
A = Area = 3 m²
Power is given by
[tex]P=\frac{dQ}{dt}=\frac{kA\Delta T}{d}\\\Rightarrow P=\frac{k_gA\Delta T}{d}\\\Rightarrow P=\frac{0.84\times 3(5-(-10))}{0.00635}\\\Rightarrow P=5.9527559\times 10^3\ W[/tex]
The rate of conduction in watts through the window is [tex]5.9527559\times 10^3\ W[/tex]
The rate of heat conduction out of a window is calculated by using Fourier's law. Substituting the given values into Fourier's law yields a heat transfer rate of around 1.97 × 10⁴ W. This represents a significant amount of heat loss through the windows.
Explanation:The heat conduction rate can be calculated using Fourier's law, which states that the rate of heat transfer per time, or heat flux, is proportional to the temperature gradient. The formula for this law is Q = k*A*(T1-T2)/d, where Q is the heat transfer rate in watts, k is the thermal conductivity, A is the surface area in square meters, T1 and T2 are the temperatures of the two surfaces, and d is the thickness of the material. Given that glass has a thermal conductivity of about 0.84 W/mºC, when we substitute these values into Fourier's law, we obtain:
Q = 0.84 W/mºC * 3.00 m² * (5.00ºC - (-10.0ºC))/0.635 cm, whereby this yields around 1.97 × 10⁴ W.
This indicates that a significant amount of heat is being lost through the windows, contributing to the chilling of the air in the room.
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Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A)-0 B)-370JC) -190 J D-1800 J
Answer:
the network done by friction on a box that moves in a complete circle is 185.7 joules
Explanation:
Step one
Given
Radius of circle =1.82m
Circumference of the circle =2*pi*r
=2*3.142*1.82=11.43
Hence distance =11.43m
Coefficient of friction u=0.25
Weight of box =65N
We know that work =force*distance
But the limiting force =u*weight
Hence the net work done by friction
Wd=0.25*65*11.43
Wd=185.7 joules
The work done by friction on a box moving in a complete circle is zero because the force of friction is always perpendicular to the direction of the box’s displacement.
Explanation:The subject of this question is the work done by friction on a box moving in a circle. When an object moves in a circle, it is the force of friction that prevents the object from sliding off the circular path and helps maintain a curved path. However, it is important to note that friction does no work when an object moves in a complete circle because the force of friction is at every point perpendicular to the direction of the box’s velocity.
To calculate work, we generally use the formula W = F * d * cos(θ), where F is the force (which is the kinetic friction force in this case), d is the displacement of the box (the path covered), and θ is the angle between the force and the direction of displacement. As the box moves in a complete circle, the angle is always 90 degrees. The cosine of 90 degrees is 0, hence the work done by kinetic friction in this case would be zero (0).
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The shear modulus of steel is 8.1 × 1010 N/m2. A steel nail of radius 7.5 × 10–4 m projects 0.040 m horizontally outward from a wall. A man hangs a wet raincoat of weight 28.5 N from the end of the nail. Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?
Answer:
[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
Explanation:
given,
shear modulus of steel = 8.1 × 10¹⁰ N/m²
radius of steel nail = 7.5 × 10⁻⁴ m
Projection outward = 0.040 m
Weight of wet raincoat = 28.5 N
Vertical deflection of other end of the nail = ?
we know
[tex]G = \dfrac{\dfrac{F}{A}}{\dfrac{\Delta y}{L}}[/tex]
[tex]\Delta\ y = \dfrac{FL}{AG}[/tex]
A = π r²
A = π x (7.5 x 10⁻⁴)²
A = 1.767 x 10⁻⁶ m²
[tex]\Delta\ y =\dfrac{28.5 \times 0.04}{1.767\times 10^{-6}\times 8.1\times 10^{10}}[/tex]
[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
Thus, the vertical deflection of the other end of the nail is[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
The shear modulus of steel, and applying the necessary formulas, we calculated the vertical deflection due to the weight of the raincoat. This means that the vertical deflection of the other end of the nail is approximately 7.968 μm.
To find the vertical deflection of the steel nail when a weight is hung from it, we can use the relationship between shear deformation, shear modulus, and applied force. The steps to solve the problem are as follows:
Step 1: Identify the given values.
Shear modulus of steel (G) = 8.1 × 10¹⁰ N/m² Radius of the nail (r) = 7.5 × 10⁻⁴ m Length of the nail projecting from the wall (L) = 0.040 m Weight of the raincoat (F) = 28.5 NStep 2: Calculate the cross-sectional area (A) of the nail.
The area can be computed using the formula for the area of a circle:
[tex]A = rac{ ext{π} imes r^2}{1}[/tex]
Substituting the value for radius:
[tex]A = ext{π} imes (7.5 × 10^{-4} ext{ m})^2[/tex]
[tex]A \approx 1.7671 × 10^{-6} ext{ m}^2[/tex]
Step 3: Calculate the shear stress (τ) in the nail.
Shear stress is defined as the force applied per unit area. Thus:
[tex]τ = rac{F}{A}[/tex]
Substituting the force and area:
[tex]τ = rac{28.5 ext{ N}}{1.7671 × 10^{-6} ext{ m}^2}[/tex]
[tex]τ \approx 1.6168 × 10^{7} ext{ N/m}^2[/tex]
Step 4: Calculate the angle of deflection (θ) in radians.
The shear strain (γ) is related to shear stress (τ) via the shear modulus (G):
[tex]γ = rac{τ}{G}[/tex]
Substituting in the values:
[tex]γ = rac{1.6168 × 10^{7} ext{ N/m}^2}{8.1 × 10^{10} ext{ N/m}^2}[/tex]
[tex]γ \approx 0.0001992[/tex]
Step 5: Calculate the vertical deflection (δ) at the end of the nail.
Deflection under shear can be computed using the formula:
[tex]δ = γ imes L[/tex]
So substituting in our calculated shear strain and the length:
[tex]δ = 0.0001992 imes 0.040 ext{ m}[/tex]
[tex]δ \approx 7.968 × 10^{-6} ext{ m}[/tex]
The science that uses principles of physics to understand the human machine is called
Answer:
Biomechanics
Explanation:
Biomechanics is a branch of biophysics that uses principles of physics to study and understand movement in organisms using the laws of mechanics. One of the applications of Biomechanics is its usage in sport and exercise activities. Biomechanics is used to study body movement of athletes during sport or exercise activities to improve performance of athletes and make the activities better and safer.
Two 2.0 kg bodies, A and B, collide. The velocities before the collision are
v→A=(15i^+30j^) m/s
and
v→B=(−10i^+5.0j^) m/s
. After the collision,
v→A=(−5.0i^+20j^) m/s
. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
Answer:
Explanation:
Given
mass of body A [tex]m_a=2 kg[/tex]
mass of body [tex]m_b=2 kg[/tex]
Velocity before Collision is [tex]u_a=15\hat{i}+30\hat{j}[/tex]
[tex]u_b=-10\hat{i}+5\hat{j} m/s[/tex]
after collision [tex]v_a=-5\hat{i}+20\hat{j} m/s[/tex]
let [tex]v_b_x[/tex] and [tex]v_b_y[/tex] velocity of B after collision in x and y direction
conserving momentum in x direction
[tex]m_a\times 15+m_b\times (-10)=m_a\times (-5)+m_b\times (v_b_x)[/tex]
as [tex]m_a=m_b[/tex] thus
[tex]15-10=-5+v_b_x[/tex]
[tex]v_b_x=10 m/s[/tex]
Conserving momentum in Y direction
[tex]m_a\times 30+m_b\times 5=m_a\times 20+m_b\times (v_b_y)[/tex]
[tex]30+5=20+v_b_y[/tex]
[tex]v_b_y=15 m/s[/tex]
thus velocity of B after collision is
[tex]v_b=10\hat{i}+15\hat{j}[/tex]
(b)Change in total Kinetic Energy
Initial Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{15^2+30^2})^2=1125 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+5^2})^2=125 J[/tex]
Total initial Kinetic Energy =1250 J
Final Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{5^2+20^2})^2=425 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+15^2})^2=325 J[/tex]
Final Kinetic Energy [tex]=425+325=750 J[/tex]
Change [tex]\Delta K.E.=1250-750=500 J[/tex]
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ above the horizontal. The block is displaced by 19.9 m, and the coefficient of kinetic friction is 0.209.
a. Find the work done by the 156 N force. The acceleration of gravity is 9.8 m/s^2
b. Find the magnitude of the work done by the orce of friction
c. What is the sign of the work done by the frictional force?
d. Find the work done by the normal force
Answer:
a) W = 2635.56 J
b) Wf = 423.27 J
c) c) The Sign of the work done by the frictional force (Wf) is negative (-)
d) W=0
Explanation:
Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .
The formula for calculate the work is :
W = F*d*cosα
Where:
W : work in Joules (J)
F : force in Newtons (N)
d: displacement in meters (m)
α :angle that form the force (F) and displacement (d)
Known data
m = 18.8 kg : mass of the block
F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal
μk= 0.209 : coefficient of kinetic friction between the cart and the surface
g = 9.8 m/s²: acceleration due to gravity
d = 19.9 m : displacement of the block
Forces acting on the block
We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.
W: Weight of the cart : In vertical direction downaward
N : Normal force : In vertical direction the upaward
F : Force applied to the block
f : Friction force: In horizontal direction
Calculated of the weight of the block
W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N
x-y components of the force F
Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N
Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n
Calculated of the Normal force
Newton's second law for the block in y direction :
∑Fy = m*ay ay = 0
N-W+Fy= 0
N-184.24+82,44= 0
N = 184.24-82,44
N = 101.8 N
Calculated of the kinetic friction force (fk):
fk = μk*N = (0.209)*( 101.8)
fk = 21.27 N
a) Work done by the F=156N.
W = (Fx) *d *cosα
W = (132.44 )*(19.9)(cos0°) (N*m)
W = 2635.56 J
b) Work done by the force of friction
Wf = (fk) *d *cos(180°)
Wf = (21.27 )*(19.9) (-1) (N*m)
Wf = - 423.27 J
Wf = 423.27 J :magnitude
c) The Sign of the work done by the frictional force is negative (-)
d) Work done by the Normal force
W = (N) *d *cos(90°)
W = (101.8 )*(19.9) (0) (N*m)
W = 0
The work done by the 156 N force is 2630.77 J. The magnitude of the work done by the friction force is 743.14 J. The work done by the normal force is zero.
Explanation:a. The work done by a force is calculated by multiplying the force applied by the displacement in the direction of the force. In this case, the force is applied at an angle of 31.9 degrees above the horizontal, so we need to find the component of the force in the horizontal direction.
Fx = F * cos(θ) = 156 N * cos(31.9°) = 132.3 N
The work done is given by W = Fx * d = 132.3 N * 19.9 m = 2630.77 J
b. The magnitude of the work done by the force of friction can be calculated using the formula:
Wfriction = friction force * displacement = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the displacement. Substituting the given values:
Wfriction = 0.209 * 18.8 kg * 9.8 m/s2 * 19.9 m = 743.14 J
c. The work done by the frictional force is negative, indicating that it acts against the motion of the block.
d. The work done by the normal force is zero because the displacement of the block is perpendicular to the direction of the normal force.
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The sun radiates energy. Does the earth similarly radiate energy? If so, why can’t we see the radiant energy from the earth?
1. Yes, the earth radiates energy, but only in the sense that the energy is reflected sunlight, which we can’t see from the earth’s surface.
2. No, the earth does not similarly radiate energy.
3. Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the absolute temperature of the radiator, the wavelength of the radiation is far too long for us to see.
4. Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the temperature of the radiator in degrees Kelvin, the frequency of the radiation from the earth is in the ultraviolet range (so that.
Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the absolute temperature of the radiator, the wavelength of the radiation is far too long for us to see.
Answer: Option C
Explanation:
Radiations or light rays are the basic name for electromagnetic energy packets travelling through space. It goes extremely quick (multiple times around the earth in one second) and can go through a vacuum. It needn't bother with material to travel in.
It has numerous structures, including visible light, infrared (IR), bright (UV), X-beams, microwaves, and radio waves. These are a no different structure of energy, just with various frequencies and measures of energy. Various frequencies of radiation communicate with an issue in an unexpected way, which causes them to appear to be more changed to us than they truly are.
A jet plane at take-off can produce a sound of intensity 9.70 W/m2 at 30.3 m away. But you prefer the tranquil sound of normal conversation, which is 1.0μW/m2. Assume that the plane behaves like a point source of the sound.
a. What is the closest distance you should live from the airport runway to preserve your peace of mind?
b. What intensity from the jet does your friend experience if she lives twice as far from the runway as you do?
c. What power of sound does the jet produce at take-off?
Assume that the plane behaves like a point source of the sound.The answers are a. Closest distance is approximately 29820 meters. b. Your friend experiences an intensity of 0.25 μW/m². c. The jet produces approximately 112 kW of sound power at take-off.
Given the sound intensity[tex]\(I = 9.70 \text{ W/m}^2\)[/tex]at a distance [tex]\(d = 30.3 \text{ m}\)[/tex], and the desired intensity[tex]\(I_{desired} = 1.0 \mu \text{W/m}^2 = 1.0 \times 10^{-6} \text{ W/m}^2\),[/tex]we can use the inverse square law for sound intensity:
[tex]\[ I \propto \frac{1}{d^2} \][/tex]
(a) To find the closest distance [tex]\(d_{desired}\)[/tex]:
[tex]\[ \frac{I}{I_{desired}} = \left( \frac{d_{desired}}{d} \right)^2 \]\[ \frac{9.70}{1.0 \times 10^{-6}} = \left( \frac{d_{desired}}{30.3} \right)^2 \]\[ d_{desired} \approx 30.3 \times \sqrt{9.70 \times 10^6} \approx 29820 \text{ m} \][/tex]
(b) The intensity at twice the distance[tex]\(2d_{desired}\)[/tex]:
[tex]\[ I_{friend} = \frac{I_{desired}}{4} = 0.25 \times 1.0 \mu \text{W/m}^2 = 0.25 \mu \text{W/m}^2 \][/tex]
(c) The power of the jet can be found using:
[tex]\[ I = \frac{P}{4\pi d^2} \]\[ P = I \times 4\pi d^2 \]\[ P = 9.70 \times 4\pi \times (30.3)^2 \approx 1.12 \times 10^5 \text{ W} \][/tex]
What would be the wavelength of the fundamental and first two overtones produced by an organ pipe of length L that is closed at one end and open at the other?
A) L, ½ L, ¼ L
B) ½ L, ¼ L, 1/6 L
C) 2L, L, ½ L
D) 4L, 2L, L
E) 4L, 4/3 L, 4/5 L
Answer:
option E
Explanation:
given,
organ pipe of length L
using formula ,
[tex]L = (2n - 1)\dfrac{\lambda}{4}[/tex]
n is the number of nodes
[tex]\lambda= \dfrac{4L}{(2n - 1)}[/tex]
now at n = 1
[tex]\lambda_1= \dfrac{4L}{(2(1) - 1)}[/tex]
[tex]\lambda_1= 4L[/tex]
now at n = 2
[tex]\lambda_2= \dfrac{4L}{(2(2) - 1)}[/tex]
[tex]\lambda_2= \dfrac{4}{3}L[/tex]
now at n = 2
[tex]\lambda_3= \dfrac{4L}{(2(3) - 1)}[/tex]
[tex]\lambda_3= \dfrac{4}{5}L[/tex]
The correct answer is option E
Answer:
The correct answer is option E
Explanation:
Most cells cannot harness heat to perform work because
Answer:
temperature is usually uniform throughout a cell or temperature is usually uniform to do work.
Explanation:
Heat (thermal energy) is a kinetic energy. It s connected with the random movement of the atoms or molecules. The temperature is usually uniform throughout a cell, so most of the cells cannot harness heat to perform work.
Cells are unable to harness heat to perform work due to the second law of thermodynamics, which results in energy being lost in a form that is unusable, often as heat, during transfers and transformations. This heat energy is essentially lost to the cell for performing work. And as entropy increases, less energy becomes available for work.
Explanation:Most cells cannot harness heat to perform work because of the second law of thermodynamics. This law states that all energy transfers and transformations are never completely efficient, with some amount of energy being lost in a form that is unusable, commonly as heat energy. Strictly speaking, heat energy is defined as the energy transferred from one system to another that is not doing work.
For example, during cellular metabolic reactions, a portion of the energy is lost as heat energy. Despite it contributing to maintaining the body temperature of warm-blooded creatures, it is essentially lost to the cell for performing work. This demonstrates how the second law of thermodynamics makes the tasks of cells obtaining, transforming, and using energy to do work more difficult than they appear.
Taking into account the entropy, less and less energy in the universe is available to do work as entropy increases. Eventually, as all fuels are exhausted and temperatures equalize, it will become impossible for heat engines to function, or for work to be done. Hence, cells can't harness heat to perform work due to the inefficiencies described by the second law of thermodynamics.
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A copper-constantan thermocouple generates a voltage of 4.75 x 10-3 volts when the temperature of the hot junction is 110 °C and the reference junction is kept at 0 °C. If the voltage is proportional to the difference in temperature between the junctions, what is the temperature in degrees Celsius of the hot junction when the voltage is 1.76 x 10-3 volts?
Answer:
The temperature of the hot junction is 40.7°C.
Explanation:
Given that,
Voltage [tex]V=4.75\times10^{-3}\ volts[/tex]
Voltage [tex]V'=1.76\times10^{-3}\ volt[/tex]
Temperature of hot junction = 110°C
If the voltage is proportional to the difference in temperature between the junctions,
We need to calculate the temperature of the hot junction
Using formula of temperature
[tex]\dfrac{V}{V'}=\dfrac{\Delta T}{\Delta T}[/tex]
[tex]\dfrac{V}{V'}=\dfrac{T_{2}-T_{1}}{T_{2}-T_{1}}[/tex]
Here,T₁=0°C
[tex]\dfrac{V}{V'}=\dfrac{110}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{4.75\times10^{-3}}{1.76\times10^{-3}}=\dfrac{110}{T_{2}}[/tex]
[tex]T_{2}=\dfrac{110\times1.76\times10^{-3}}{4.75\times10^{-3}}[/tex]
[tex]T_{2}=40.7^{\circ}C[/tex]
Hence, The temperature of the hot junction is 40.7°C.
Final answer:
To determine the temperature of the hot junction in a copper-constantan thermocouple when the voltage is 1.76 x 10-3 volts, we can set up a proportion between the voltages and the temperature difference.
Explanation:
To determine the temperature of the hot junction when the voltage is 1.76 x 10-3 volts, we can use the proportionality between voltage and temperature difference. We know that the voltage generated by the thermocouple is directly proportional to the difference in temperature between the junctions. So, we can set up a proportion:
(4.75 x 10-3 volts) / (110 °C - 0 °C) = (1.76 x 10-3 volts) / (x °C - 0 °C)
By cross-multiplying and solving for x, we can find the temperature of the hot junction. The result is x = 41.5 °C.
Suppose a star with radius 8.50 x 10⁸ m has a peak wavelength of 685 nm in the spectrum of its emitted radiation. (a) Find the energy of a photon with this wavelength. (b) What is surface temperature of the star? (c) At what rate is energy emitted from the star in the form of radiation? Assume the star is a blackbody (e = 1). (d) Using the answer to part (a), estimate the rate at which photons leave the surface of the star.
Answer:
a) Energy of Photon = 2.9 * 10^ -19 J
What is the initial drag force for a golf ball with the following conditions? Frontal area: 14 cm2 (remember to convert to m2), air density: 1.204 kg/m3, initial velocity: 50 m/s, coefficient of drag: 0.3
Answer:
F= 0.6321 N
Explanation:
Given
Area ,A= 14 cm²
Density ,ρ = 1.204 kg/m³
Velocity ,v= 50 m/s
Drag coefficient ,C=0.3
The drag force on the golf given as
[tex]F=\dfrac{1}{2}\rho CAv^2[/tex]
Now by putting the values
[tex]F=\dfrac{1}{2}\rho CAv^2[/tex]
[tex]F=\dfrac{1}{2}\times 0.3\times 1.204 \times (14\times 10^{-4})\times 50^2\ N[/tex]
F= 0.6321 N
Therefore force due to drag on the golf is 0.6321 N
The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.19 m apart. One of the spheres had a mass of 188 kg, while the mass of the other sphere was 0.93 kg.What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
The ratio is found by dividing the gravitational force, determined by the universal law of gravitation, by the weight of the lighter sphere, which is the product of its mass and the acceleration due to gravity.
Explanation:The ratio of the gravitational force between the spheres to the weight of the lighter sphere can be found by dividing the gravitational force by the weight of the lighter sphere. The gravitational force (F) between two objects can be found using the universal law of gravitation: F = G * (m1*m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. Plugging in the given values, we can find the gravitational force. The weight of the lighter sphere can be found by multiplying its mass by the acceleration due to gravity (9.8 m/s^2). Divide the former by the latter to get the answer.
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The ratio of the gravitational force to weight can be calculated using Newton's law of gravitation and the definition of weight. The gravitational constant G used in the calculations was accurately measured by Henry Cavendish in 1798. This allows the measurement of minute gravitational attractions, pivotal in understanding the strength of gravitation.
Explanation:The ratio of the gravitational force between the two spheres to the weight of the lighter sphere is obtained through the gravitational law and the definition of weight. The universal gravitational constant G was accurately measured by English scientist Henry Cavendish in 1798 using an incredibly sensitive balance. The gravitational force (F) among the two spheres is computed through: F = G * m1 * m2 / r², where m1 and m2 are the masses of the spheres and r is the distance between the centers of the spheres, and G is the gravitational constant (6.67 × 10−11 N·m²/kg²).
Weight (W) is calculated as the mass (m) of an object multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s² on Earth. Thus, the weight of the lighter sphere is: W = m * g. Therefore, the ratio of the gravitational force to the weight of the lighter sphere is calculated by dividing the gravitational force by the weight of the lighter sphere. The tiny gravitational attraction between ordinary-sized masses measured by Cavendish is significant in the world of physics, as it determines the strength of one of the nature's fundamental forces.
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Nuclear fusion differs from nuclear fission because nuclear fusion reactions
Answer:
Nuclear fusion reactions produce more energy.( basic difference)
Explanation:
Nuclear fusion reactions are different from nuclear fusion reactions :- In nuclear fission one heavy and unstable nuclear breaks into two parts(smaller) and produce energy( very high energy than normal reaction). whereas in nuclear fusion two small and unstable nuclei fuse and produce very high amount of energy( even higher than nuclear fusion).
Answer:
they are different because nuclear fission is one heavy and unstable nuclear breaks into two parts, or half, and produce very high energy. nuclear fusion two small fuse and produce very high amount of energy
Explanation:
During metamorphism, what is the major effect of chemically active fluids?
a. increase the pressures in deeply buried, regional-metamorphic zones
b. aid in the movement of dissolved silicate ions and facilitate growth of the mineral grains
c. prevent partial melting so solid rocks can undergo very high temperature regional metamorphism
d. facilitate the formation of schistosity and gneissic banding in hornfels and slates
Final answer:
Chemically active fluids during metamorphism help in the transfer and movement of ions, facilitating the growth and reorganization of mineral grains in the rock.
Explanation:
During metamorphism, chemically active fluids primarily aid in the movement of dissolved silicate ions and facilitate growth of the mineral grains. These fluids can enhance the transfer of ions, which allows for the reorganization of atoms and the growth of new mineral structures within the rock. The presence of chemically active fluids during metamorphism can therefore play a critical role in the transformation of a rock's mineral composition and texture.
A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum". The ballistics pendulum consistsof a block of wood, which has a mass of M = 5.00 kg., attached to two stringswhich are L = 125 cm long. When the block is struck by the bullet the blockswings backward until the angle between the ballistics pendulum and thevertical reaches a maximum angle ofa= 38.0o.a. What will be the maximum gravitational energy contained in the ballisticspendulum when it reaches the maximum angle?b. What was the velocity of the block of wood immediately after being struck bythe bullet?c. What was the velocity of the bullet immediately before it strikes the block ofwood?d. How much work was done by the bullet as it lodged in the block of wood?e. What will be the recoil velocity of the rifle?f. How much energy was released when the bullet was fired?
Answer:
Part a)
[tex]U = 13 J[/tex]
Part b)
[tex]v = 2.28 m/s[/tex]
Part c)
[tex]v = 177.66 m/s[/tex]
Part d)
[tex]W = 1012.7 J[/tex]
Part e)
[tex]v = 2.1 m/s[/tex]
Part f)
[tex]E = 1037.2 J[/tex]
Explanation:
Part a)
As we know that the maximum angle deflected by the pendulum is
[tex]\theta = 38^o[/tex]
so the maximum height reached by the pendulum is given as
[tex]h = L(1 - cos\theta)[/tex]
so we will have
[tex]h = L(1 - cos38)[/tex]
[tex]h = 1.25(1 - cos38)[/tex]
[tex]h = 0.265 m[/tex]
now gravitational potential energy of the pendulum is given as
[tex]U = mgh[/tex]
[tex]U = 5(9.81)(0.265)[/tex]
[tex]U = 13 J[/tex]
Part b)
As we know that there is no energy loss while moving upwards after being stuck
so here we can use mechanical energy conservation law
so we have
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.81)(0.265)}[/tex]
[tex]v = 2.28 m/s[/tex]
Part c)
now by momentum conservation we can say
[tex]mv = (M + m) v_f[/tex]
[tex]0.065 v = (5 + 0.065)2.28[/tex]
[tex]v = 177.66 m/s[/tex]
Part d)
Work done by the bullet is equal to the change in kinetic energy of the system
so we have
[tex]W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2[/tex]
[tex]W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2[/tex]
[tex]W = 1012.7 J[/tex]
Part e)
recoil speed of the gun can be calculated by momentum conservation
so we will have
[tex]0 = mv_1 + Mv_2[/tex]
[tex]0 = 0.065(177.6) + 5.50 v[/tex]
[tex]v = 2.1 m/s[/tex]
Part f)
Total energy released in the process of shooting of gun
[tex]E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2[/tex]
[tex]E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)[/tex]
[tex]E = 1037.2 J[/tex]
A child bounces in a harness suspended from a door frame by three elastic bands. If each elastic band stretches 0.300 m while supporting a 7.15-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight.
Answer:
The force constant for each elastic band is 77.93 N/m
Explanation:
Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:
[tex] F_{e}= kx [/tex] (1)
With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:
[tex] \sum F=0 [/tex] (2)
[tex]\frac{W}{3}-F_{e}=0\Rightarrow F_{e}=\frac{W}{3} [/tex] (3)
Using (1) on (3):
[tex] kx=\frac{W}{3}\Rightarrow k=\frac{mg}{3x} [/tex](4)
[tex] k=\frac{7.15\,kg*9.81\,\frac{m}{s^{2}}}{3*0.300\,m}\simeq\mathbf{77.93\frac{N}{m}} [/tex]
The force constant of each elastic band is 77.87 N/m.
To find the force constant for each elastic band, we use Hooke's law
Hooke's lawHooke's law states that the force applied to an elastic material is directly proportional to its extension provided the elastic limit is not exceeded
W = ke............ Equation 1Where:
W = Weight of the child on one elastic bandk = Force constante = extensionMake k the subject of the equation
k = W/e.............. Equation 2From the question,
Given:
W = (7.15×9.8)/3 = 23.36 Ne = 0.3 mSubstitute these values into equation 2
k = 23.36/0.3k = 77.87 N/mHence, The force constant of each elastic band is 77.87 N/m.
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How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What would be it's stopping distance?
The stopping distance is 45.0 m
Explanation:
First of all, we find the acceleration of the truck, by using Newton's second law:
[tex]F=ma[/tex]
where
[tex]F=-1.26\cdot 10^4 N[/tex] is the net force on the truck
[tex]m=2.3\cdot 10^3 kg[/tex] is the mass of the truck
a is its acceleration
Solving for a,
[tex]a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2[/tex]
where the negative sign means the acceleration is opposite to the direction of motion.
Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity of the truck
u = 22.2 m/s is the initial velocity
[tex]a=-5.48 m/s^2[/tex] is the acceleration
s is the stopping distance
And solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m[/tex]
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Air conditioners are rated by their coefficient of performance at 61 ∘ F inside temperature and 99 ∘ F outside temperature. An efficient but realistic air conditioner has a coefficient of performance of 3.2. What is the maximum possible coefficient performance?
Answer:
the maximum possible coefficient performance is 13.7
Explanation:
inside temperature, [tex]T_{C}[/tex] = 61 F = 289.26 K
outside temperature, [tex]T_{H}[/tex] = 99 F = 310.37 K
coefficient of performance, COP (real) = 3.2
according to Carnot's theorem, the coefficient of performance is
[tex]COP_{max}[/tex] = [tex]\frac{T_{C} }{T_{H}-x_{C} }[/tex]
where
[tex]T_{C}[/tex] is cold temperature
[tex]T_{H}[/tex] is hot temperature
thus,
[tex]COP_{max}[/tex] = [tex]\frac{289.26}{310.37-289.26}[/tex]
= 13.7
The mechanical vibrations triggered by sound waves are transduced into neural impulses by
Sound waves cause the eardrum to vibrate, which then send these vibrations to the middle ear and then to the cochlea of the inner ear. Herein, the vibrations are converted by hair cells on the basilar membrane into neural impulses which are transmitted to the brain and perceived as sound.
Explanation:The mechanical vibrations, also referred to as sound waves, reach the outer ear and are transferred to the ear canal where they cause the tympanic membrane, or eardrum, to vibrate. These vibrations are passed on to the three bones of the middle ear, namely the malleus, incus, and stapes. The stapes then transmits these vibrations to a structure known as the oval window, which is the outermost structure of the inner ear. Herein lies the cochlea, a spiral-shaped structure filled with fluid, and the vibrations from the oval window create pressure waves within this fluid. Within the cochlea, a structure known as the basilar membrane, which hosts receptor hair cells, is mechanically stimulated by these pressure waves. When the hair cells 'bend', this initiates a process of mechanical transduction, wherein the mechanical vibrations from sound waves are transformed, or transduced, into electrical signals known as neural impulses. These impulses are then transmitted to the brain via the cochlear nerve, whereby they are perceived as sound.
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The mechanical vibrations of sound waves are transduced into neural impulses by the hair cells in the inner ear.
The mechanical vibrations of sound waves are transduced into neural impulses by specialized cells in the inner ear called hair cells. These hair cells are located within the cochlea, a spiral-shaped organ responsible for auditory processing. When sound waves reach the inner ear, they cause the hair cells to bend. This bending motion activates ion channels on the hair cell membrane, leading to the generation of electrical signals or neural impulses.
These signals are then transmitted through the auditory nerve to the brain, where they are interpreted as sound. The hair cells' ability to transduce mechanical vibrations into neural impulses is a crucial step in the auditory process, allowing us to perceive and understand various sounds in our environment, from music to speech and other auditory stimuli. This complex process enables our sense of hearing and is essential for our communication and interaction with the world around us.
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A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the wall, the ball is moving to the right at 20 m/s. What is the impulse of the net force on the ball during its collision with the wall?
Answer:
Impulse, J = 20 m/s
Explanation:
Given that,
Mass of the ball, m = 0.4 kg
Initial speed of the ball, u = -30 m/s (left)
Final speed of the ball after hitting, v = 20 m/s (right)
Let J is the impulse of the net force on the ball during its collision with the wall. The change in momentum of an object is equal to the impulse imparted to it. It is given by :
[tex]J=m(v-u)[/tex]
[tex]J=0.4\ kg(-30-20)\ m/s[/tex]
J = -20 m/s
So, the magnitude of impulse of the net force on the ball during its collision with the wall is 20 m/s.
The impulse on the ball is 20 Ns.
What is impulse?This can be defined as the product of force and time. The impulse of a force acting on a body is also equal to the change in momentum of the body
To calculate the impulse of the net force on the ball during the collision, we use the formula below.
Formula:
I = m(v-u)............ Equation 1Where:
I = Impulse on the ballv = Final velocity of the ballu = Initial velocity of the ballm = mass of the ball.From the question,
Given:
m = 0.40 kgv = 20 m/s u = -30 m/s (to the right)Substitute these values into equation 1
I = 0.40[20-(-30)]I = 0.40(20+30)I = 0.40(50)I = 20 Ns.Hence, the impulse on the ball is 20 Ns.
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s) A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of the centripetal force acting on the body. (Be sure to correctly state the units of the force
Answer:
Centripetal force acting on the body = 9.47 N
Explanation:
Mass of body, m = 2 kg
Radius, r = 3 m
It makes one revolution in 5 seconds.
Period, T = 5 s
[tex]\texttt{Angular velocity, }\omega =\frac{2\pi }{T}=\frac{2\pi }{5}=1.256rad/s[/tex]
Centripetal force, F = mrω²
F = 2 x 3 x 1.256² = 9.47 N
Centripetal force acting on the body = 9.47 N
While a car is stopped at a traffic light in a storm, raindrops strike the roof of the car. The area of the roof is 5.0 m2. Each raindrop has a mass of 3.7 ✕ 10−4 kg and speed of 2.5 m/s before impact and is at rest after the impact. If, on average at a given time, 150 raindrops strike each square meter, what is the impulse of the rain striking the car?
Answer:
J = 0.693 N.s
Explanation:
The impulse of one single drop is given by:
J1 = m*(Vf - Vo) where Vf = 0
[tex]J1 = -9.25*10^{-4}N.s[/tex]
The magnitude of the total impulse will be:
Jt = J1 * 150 * 5
Jt = 0.693 N.s
Technician A says that all FWD vehicles use adjustable front wheel bearings. Technician B says that most FWD vehicles use sealed non-adjustable front wheel bearings. Which technician is correct?
Answer:
Explanation:
Technician B only.
Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
Answer:
t = 83.93 nm
Explanation:
given,
blue light wavelength (λ)= 470 nm
refractive index of oil (μ)= 1.40
minimum thickness of an oil slick on water = ?
using constructive interference formula
now,
[tex]2 \mu t = (m + \dfrac{1}{2})\lambda[/tex]
where, t is the thickness of the oil slick
m = 0,1,2
for minimum thickness m = 0
[tex]2\times 1.40\times t = (0 + 0.5)\times 470[/tex]
[tex]2.8\times t = 235[/tex]
[tex]t = \dfrac{235}{2.8}[/tex]
t = 83.93 nm
minimum thickness of an oil slick on water = t = 83.93 nm
The minimum thickness of an oil slick should be 83.93 nm
Important information:Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40
Calculation of the minimum thickness:[tex]2\times 1.40 \times t= (0 + 0.5) \times 470\\\\28 \times t = 235[/tex]
t = 83.93 nm
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Blocks I and II, each with a mass of 1.0 kg, are hung from the ceiling of an elevator by ropes 1 and 2.What is the force exerted by rope 1 on block I when the elevator is traveling upward at a constant speed of 2.0 m/s?What is the force exerted by rope 1 on block II when the elevator is stationary?
When a body in the elevator is hung then-
If the elevator is at rest the only its weight is acting on it,
a.If the elevator is moving upward with constant velocity the wight of the body will not change.
b.If the elevator is moving upward with constant acceleration the weight of the body will increase.
c.If the elevator is moving downward with constant velocity the weight of the body will remain the same.
d.If the elevator is moving downward with constant acceleration the weight of the body will decrease.
Answer:
Part a)
[tex]T = 9.8 N[/tex]
Part b)
[tex]T = 9.8 N[/tex]
Part c)
If elevator is accelerating upwards then we have
So tension will be more than the weight
Now if the elevator is moving with uniform velocity
tension will be equal to the weight of the block
Explanation:
Part a)
As we know that rope 1 is connected to block 1 at its lower end
So here we have elevator is moving at constant velocity in upward direction
so we have
[tex]T - mg = 0[/tex]
[tex]T = mg[/tex]
[tex]T = 1\times 9.8[/tex]
[tex]T = 9.8 N[/tex]
Part b)
now for block II we have
[tex]T - mg = ma[/tex]
[tex]T - mg = 0[/tex]
[tex]T = 1 \times 9.8[/tex]
[tex]T = 9.8 N[/tex]
Part c)
If elevator is accelerating upwards then we have
[tex]T - mg = ma[/tex]
[tex]T = mg + ma[/tex]
So tension will be more than the weight
Now if the elevator is moving with uniform velocity
[tex]T - mg = 0[/tex]
[tex]T = mg[/tex]
so tension will be equal to the weight of the block
Methane gas absorbs red light, and methane clouds reflect blue light, giving Uranus and Neptune their distinctive blue colors. Why do Uranus and Neptune have methane clouds, but Jupiter and Saturn do not
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation:
Jupiter and Saturn have high densities meaning any methane on them is dragged down to their hot surface which prevents methane from ever forming clouds.
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation: