Answer:
64.57
Explanation:
We are given that
For decreasing measured sound intensity level=36.2 dB
We have to find the factor by which the distance increase.
Let initial distance=x
Final distance=x'
According to question
[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]
[tex]36.2=10log(\frac{x'}{x})^2[/tex]
[tex]36.2=10\times 2log\frac{x'}{x}[/tex]
[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]
[tex]\frac{x'}{x}=10^{1.81}[/tex]
[tex]x'=10^{1.81}x=64.57x[/tex]
Hence, the distance is increases by factor of 64.57
Between this and the next assignment, we want to get a better under- standing of how light interacts with the eye. Here are two questions to get us started, focused on diffraction (i.e., the spreading of light when it passes through a narrow opening). A. To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright
Answer:
θ₁ = 3.35 10⁻⁴ rad , θ₂ = 8.39 10⁻⁵ rad
Explanation:
This is a diffraction problem for a slit that is described by the expression
sin θ = m λ
the resolution is obtained from the angle between the central maximum and the first minimum corresponding to m = 1
sin θ = λ / a
as in these experiments the angle is very small we can approximate the sine to its angle
θ = λ / a
In this case, the circular openings are explicit, so the system must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ / D
where D is the diameter of the opening
let's apply this expression to our case
indicates that the wavelength is λ = 550 nm = 550 10⁻⁹ m
the case of a lot of light D = 2 mm = 2 10⁻³ m
θ₁ = 1.22 550 10-9 / 2 10⁻³
θ₁ = 3.35 10⁻⁴ rad
For the low light case D = 8 mm = 8 10⁻³
θ₂ = 1.22 550 10-9 / 8 10⁻³
θ₂ = 8.39 10⁻⁵ rad
What is energy?
O
A. A form of sound
O
B. The ability to do work
O
c. The number of atoms in an object
O
D. The size of an object
A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 28cm before coming to rest at its lowest point. It then continues to oscillate vertically.
A. What is the spring constant?
B. What is the amplitude of the oscillation?
C. What is the frequency of the oscillation?
Explanation:
Given that,
Length of the spring, l = 50 cm
Mass, m = 330 g = 0.33 kg
(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :
[tex]kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m[/tex]
(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.
(C) The frequency of the oscillation is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz[/tex]
cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 15.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy.
Answer:
a). 675J
b). 337.5J
c). 1012.5J
Explanation:
M = 6.0kg
V = 15.0m/s
a). Translational energy
E = ½ *mv²
E = ½ * 6 * 15²
E = 675J
b). Rotational kinetic energy K.E(rot) = Iw²
But moment of inertia of a cylinder (I) = ½Mr²
I = ½mr²
V = wr, r = v / w
K.E(rot) = ¼ mv²
K.E(rot) = ¼* 6 * 15²
K.E(rot) = 337.5J
Total energy of the system = K.E(rot) + Translational energy = 337.5 + 675
T.E = 1012.5J
1. A 500-g block is placed on a level, frictionless surface and attached to an ideal spring. At t = 0 the block moves through the equilibrium position with speed vo in the –x direction, as shown below. At t = sec, the block reaches its maximum displacement of 40 cm to the left of equilibrium. a. Determine the value of each of the following quantities. Show all work. • period: • spring constant: b. Using x(t) = A cos(t + o) as the solution to the differential equation of motion: i. Determine the form of the function v(t) that represents the velocity of the block. ii. Evaluate all constant parameters (A, , and o) so as to completely describe both the position and velocity of the block as functions of time. c. Consider the following statement about the situation described above. "It takes the first seconds for the block to travel 40 cm, so the initial speed vo can be found by dividing 40 cm by seconds." Do you agree or disagree with this statement? If so, explain why you agree. If not, explain why you disagree and calculate the initial speed vo of the block.
a. The period (T) of the motion is [tex]\(2\pi\)[/tex] seconds, and the spring constant (k) is [tex]\(m\omega^2\) where \(m\) is the mass, and \(\omega\)[/tex] is the angular frequency. For the given problem, [tex]\(T = 2\pi\) seconds and \(k = \frac{m}{\pi^2}\).[/tex]
b. i. The velocity function [tex]\(v(t)\) is given by \(v(t) = -A\omega\sin(\omega t + \phi_0)\).[/tex]
ii. Evaluating the constant parameters:
[tex]\(A = 0.4\ m\) (maximum displacement),[/tex]
[tex]\(\omega = \frac{2\pi}{T} = 1\ \text{rad/s}\),[/tex]
[tex]\(\phi_0 = -\frac{\pi}{2}\) (phase angle).[/tex]
c. Disagree. The block doesn't travel 40 cm in the first [tex]\(\pi\)[/tex] seconds. The correct initial speed [tex]\(v_0\) is found using \(v_0 = A\omega\), so \(v_0 = 0.4\ m \times 1\ \text{rad/s} = 0.4\ \text{m/s}\).[/tex]
Explanation:a. The period of the motion [tex](\(T\))[/tex] is the time taken for one complete oscillation. For simple harmonic motion, [tex]\(T = 2\pi/\omega\), where \(\omega\)[/tex] is the angular frequency. In this case, [tex]\(T = 2\pi\)[/tex] seconds. The spring constant [tex](\(k\))[/tex] is related to the angular frequency by [tex]\(k = m\omega^2\), where \(m\)[/tex] is the mass. Substituting [tex]\(T\) into the formula for \(\omega\), we find \(k = \frac{m}{\pi^2}\).[/tex]
b. i. The velocity function [tex]\(v(t)\)[/tex] is the derivative of the displacement function [tex]\(x(t)\). For \(x(t) = A\cos(\omega t + \phi_0)\), \(v(t) = -A\omega\sin(\omega t + \phi_0)\).[/tex]
ii. To fully describe the motion, we find [tex]\(A\), \(\omega\), and \(\phi_0\). \(A\)[/tex] is the amplitude, given as 0.4 m. [tex]\(\omega\)[/tex] is calculated from the period, resulting in 1 rad/s. [tex]\(\phi_0\)[/tex] is the phase angle, set to [tex]\(-\frac{\pi}{2}\)[/tex] to match the initial condition.
c. The statement is incorrect. The block does not travel 40 cm in the first [tex]\(\pi\)[/tex] seconds. The initial speed [tex](\(v_0\))[/tex] can be found using [tex]\(v_0 = A\omega\), which yields \(0.4\ m \times 1\ \text{rad/s} = 0.4\ \text{m/s}\).[/tex]
1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.
Find the given attachments
Projectile motion is a combination of which two types of motion?
Answer:horizontal motion and vertical motion
Explanation:projectile motion is a combination of both vertical and horizontal motion
Answer:
C. Horizontal and Vertical
Explanation:
edge
Monochromatic light is incident on a pair of slits that are separated by 0.230 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.) (a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.57 cm, find the wavelength of the incident light
Answer:
The wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
Explanation:
The physicist Thomas Young established, through his double slit experiment, a relation between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
[tex]\Lambda x = L\frac{\lambda}{d} [/tex] (1)
Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.
The values for this particular case are:
[tex]L = 2.60m[/tex]
[tex]d = 0.230mm[/tex]
[tex]\Lambda x = 1.57cm[/tex]
Then, [tex]\lambda[/tex] can be isolated from equation 1
[tex]\lambda = \frac{d \Lambda x}{L}[/tex] (2)
However, before equation 2 can be used, it is necessary to express [tex]\Lambda x[/tex] and d in units of meters.
[tex]\Lambda x= 1.57cm \cdot \frac{1m}{100cm}[/tex] ⇒ [tex]0.0157m[/tex]
[tex]d = 0.230mm \cdot \frac{1m}{1000mm}[/tex] ⇒ [tex]2.3x10^{-4}m[/tex]
Finally, equation 2 can be used.
[tex]\lambda = \frac{(2.3x10^{-4}m)(0.0157m}{(2.60m)}[/tex]
[tex]\lambda = 1.38x10^{-6}m[/tex]
Hence, the wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
What caused Hurricane Sandy to weaken on October 30th and October 31st?
Answer:
It made landfall.
Explanation:
On land there is more friction than in open water, so it slowed down the hurricane.
Electrons are ejected from a metallic surface with speeds of up to 4.60x105 m/s when light with a wave length of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface? (
Answer:
The solution to the question above is explained below:
Explanation:
(a) The work function of the surface is:
The work function of a metal, Φ, it's the minimum amount of energy required to remove electron from the conduction band and remove it to outside the metal. It is typically exhibited in units of eV (electron volts) or J (Joules). Work Function, is the minimum thermodynamic work.
hf = hc /λ = φ + Kmax = φ + 1 /2 mev ²max
where, φ=work function of a metal
eV=electron volts
J= Joules
λ=the Plank constant 6.63 x 10∧-34 J s
f = the frequency of the incident light in hertz
∧= signifies raised to the power in the solution
Kmax= the maximum kinetic energy of the emitted electrons in joules
φ= ( hc /λ -1/2mev ²max= 6.63 × 10∧-34Js × 3.00 × 10∧8 m/s ÷ 625 × 10∧-9) -
(1/2 × 9.11 × 10∧-31 kg × 4.60 × 10∧5 m/s) = 2.21 × 10∧-19 J
ans= 1.38 eV
(b) The cutoff frequency for this surface is:
The cutoff frequency is the minimum frequency that is required for the emission of electrons from a metallic surface at which energy flowing through the metallic surface begins to be reduced rather than passing through.
Light at the cutoff frequency only barely supplies enough energy to overcome the work function. The value of the cutoff frequency is in unit hertz (Hz)
hfcut = φ
fcut = φ /h
ans= 334 THz
Which of the following statements are true concerning the reflection of light?
Check all that apply.
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
e. For specular reflection, the angle of incidence is less than the angle of reflection.
Answer:
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
Explanation:
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc
b. The reflection of light from a smooth surface is called specular reflection.
This is correct
c. The reflection of light from a rough surface is called diffuse reflection.
This is correct
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection. The only difference between this type of reflection and specular reflection, is that the normal for diffuse reflection is not parallel to each due to the rough surface in which the light incidents.
For specular reflection, the angle of incidence is less than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection
The correct statements are statement 2 and statement 3.The true statements are: the reflection of light from a smooth surface is called specular reflection, and the reflection of light from a rough surface is called diffuse reflection. The others are incorrect as the angle of incidence always equals the angle of reflection.
To answer the question about the reflection of light, let's analyze each statement:
The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.In summary, the true statements concerning the reflection of light are:
The reflection of light from a smooth surface is called specular reflection.The reflection of light from a rough surface is called diffuse reflection.
A coil formed by wrapping 65 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 µT to 600 µT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire?
Final answer:
To find the total length of the wire in a 65-turn square coil subjected to a changing magnetic field, apply Faraday's law of induction to calculate the magnetic flux change and then determine the side length of the coil. Multiply the coil's perimeter by the number of turns to obtain the total length.
Explanation:
The student's question pertains to electromagnetic induction in a coil exposed to a changing magnetic field. To solve for the total length of the wire, we can use Faraday's law of induction, which states that the induced emf (electromotive force) in a coil is proportional to the rate of change of magnetic flux through the coil. Given an induced emf of 80.0 mV and a change in magnetic field from 200 µT to 600 µT over 0.400 s, the change in magnetic flux φ can be calculated.
Since the coil is square and positioned at a 30.0° angle to the magnetic field, the effective area A for inducing emf can be determined using the cosine of the angle and the side length of the square, assuming all sides are equal. With the number of turns (N) being 65, we can apply Faraday's law to find the magnetic flux and subsequently the side length of the square. The total length of the wire is simply the perimeter of the square (4 times the side length) multiplied by the number of turns (N).
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius RRR which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2cm2. How large does the radius RRR of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/mmV/m at the receiver? For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius RRR refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal
Complete Question
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.
How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.
Give your answer in centimeters, to two significant figures.
Answer:
The radius of the dish is [tex]R = 18cm[/tex]
Explanation:
From the question we are told that
The radius of the orbit is = [tex]R = 35,000km = 35,000 *10^3 m[/tex]
The power output of the power is [tex]P = 1 kW = 1000W[/tex]
The electric vector amplitude is given as [tex]E = 0.1 mV/m = 0.1 *10^{-3}V/m[/tex]
The area of thereciever is [tex]A_R = 5cm^2[/tex]
Generally the intensity of the dish is mathematically represented as
[tex]I = \frac{P}{A}[/tex]
Where A is the area orbit which is a sphere so this is obtained as
[tex]A = 4 \pi r^2[/tex]
[tex]= (4 * 3.142 * (35,000 *10^3)^2)[/tex]
[tex]=1.5395*10^{16} m^2[/tex]
Then substituting into the equation for intensity
[tex]I_s = \frac{1000}{1.5395*10^{16}}[/tex]
[tex]= 6.5*10^ {-14}W/m2[/tex]
Now the intensity received by the dish can be mathematically evaluated as
[tex]I_d = \frac{1}{2} * c \epsilon_o E_D ^2[/tex]
Where c is thesped of light with a constant value [tex]c = 3.0*10^8 m/s[/tex]
[tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]8.85*10^{-12} N/m[/tex]
[tex]E_D[/tex] is the electric filed on the dish
So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish
Now making the eletric field intensity the subject of the formula
[tex]E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }[/tex]
substituting values
[tex]E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }[/tex]
[tex]= 7*10^{-6} V/m[/tex]
The incident power on the dish is what is been reflected to the receiver
[tex]P_D = P_R[/tex]
Where [tex]P_D[/tex] is the power incident on the dish which is mathematically represented as
[tex]P_D = I_d A_d[/tex]
[tex]= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)[/tex]
And [tex]P_R[/tex] is the power incident on the dish which is mathematically represented as
[tex]P_R = I_R A_R[/tex]
[tex]= \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]
Now equating the two
[tex]\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]
Making R the subject we have
[tex]R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }[/tex]
Substituting values
[tex]R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }[/tex]
[tex]R = 18cm[/tex]
Please select that whether below statements are correct or not?
1. Resistivity is the property that measures a material's ability to provide "obstacles" to the flow of electrons caused by an external electric field. Such a flow of electrons is called an electric current. Resistivity rho is defined as the ratio of the magnitude of the electric field E to the magnitude of the current density J: rho=EJ.
2. Resistance is a measure of an object's ability to provide "obstacles" to electric current. The resistance R of a conductor (often, a metal wire of some sort) is defined as the ratio of the voltage V between the ends of the wire to the current I through the wire: R=VI.
3. Ohm's law is not a fundamental law of physics; it is valid under certain conditions (mostly, metal conductors in a narrow range of temperatures). Still, Ohm's law is a very useful tool, since many circuits operate under these conditions.
Answer:
Statements 1, 2 and 3 are all correct.
The human circulatory system is closed-that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart’s four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. If the aorta (diameter da) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?
(a) √da (b) da/√2 (c) 2da
(d) da/2
Answer:
(b) da/√2
Explanation:
Detailed explanation and calculation is shown in the image below
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.8 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 140 rad/s2.
What is the moment of inertia of the boxer's forearm?
Answer:
0.366kgm²
Explanation:
F = 1.8*10³N
r = 2.85cm = 0.0285m
α = 140rad/s²
Torque = applied force * distance
τ = r * F
τ = 0.0285 * 1.8*10³
τ = 51.3N.m
but τ = I * α
I = τ / α
I = 51.3 / 140
I = 0.366kgm²
Newton's Law of Universal Gravitation states that the force F between two masses, m1 and m2, is given below, where G is a constant and d is the distance between the masses. Find an equation that gives an instantaneous rate of change of F with respect to d. (Assume m1 and m2 represent moving points.)
The instantaneous rate of change of the force between two masses with respect to distance is given by the derivative of Newton's Law of Universal Gravitation, resulting in [tex]dF/dd = -2G(m1*m2)/d^3.[/tex]
Explanation:The question is asking for the instantaneous rate of change of the force between the two masses with respect to the distance. This can be found by taking the derivative of Newton's Law of Universal Gravitation.
The gravitational force, F, is defined by the equation[tex]F = G(m1*m2)/d^2.[/tex]Taking the derivative of this function with respect to d, gives us [tex]dF/dd = -2G(m1*m2)/d^3.[/tex] This equation represents how the force changes with a small change in distance.
Learn more about Instantaneous Rate of Change here:https://brainly.com/question/31402058
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Two vectors have magnitudes of 1.8 m and 2.4 m. How are these vectors arranged, so their sum is 0.6 m?
Answer:
Perpendicular
Explanation:
write the angle between them as a
cosine theoreme
[tex]0.6^{2} =1.8^{2} +2.4^{2}-2*1.8*2.4*cos(a)[/tex]
cos(a)=(3.24+5.76-0.6)/(2*1.8*2.4)
cos(a)=1
a=90°
To find the arrangement of two vectors with magnitudes of 1.8 m and 2.4 m so that their sum is 0.6 m, we need to consider vector addition. The magnitude of the resultant vector is 4.2 m. One possible arrangement is to align the vectors in opposite directions, cancelling out each other's effects and resulting in a net sum of zero.
Explanation:To find the arrangement of two vectors with magnitudes of 1.8 m and 2.4 m, so that their sum is 0.6 m, we need to understand vector addition. The sum of two vectors can be found by aligning the vectors head to tail and drawing a line connecting the head of the first vector to the tail of the second vector. The resulting vector, called the resultant vector, is the sum of the two vectors.
In this case, let's call the first vector A with magnitude 1.8 m and the second vector B with magnitude 2.4 m. To find the arrangement where their sum is 0.6 m, we can set up the equation: A + B = 0.6. Since the magnitudes are given, we can rearrange the equation to solve for the magnitude of the resultant vector: |A| + |B| = |0.6|. Substituting the given values, we get: 1.8 + 2.4 = 0.6. Simplifying the equation, we find that the magnitude of the resultant vector is 4.2 m.
Now, to find the arrangement, we need to consider the direction of the vectors. Since the sum of the two vectors is not zero, their directions cannot be directly opposite. Therefore, we need to arrange them in a way that their individual directions cancel out each other to result in a net sum of zero. One possible arrangement is to align the vectors such that they form a straight line in opposite directions. In this case, A and B would have the same magnitude but opposite directions, cancelling out each other's effects and resulting in a net sum of zero.
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The total mass of the sled, man, and rock is 96.5 kg. The mass of the rock is 0.300 kg and the man can throw it with a speed of 17.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving).
Answer: 0.5 m/s
Explanation:
Given
Speed of the sled, v = 0.55 m/s
Total mass, m = 96.5 kg
Mass of the rock, m1 = 0.3 kg
Speed of the rock, v1 = 17.5 m/s
To solve this, we would use the law of conservation of momentum
Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns
When the man throws the rock forward
rock:
m1 = 0.300 kg
V1 = 17.5 m/s, in the same direction of the sled with the man
m2 = 96.5 kg - 0.300 kg = 96.2 kg
v2 = ?
Law of conservation of momentum states that the momentum is equal before and after the throw.
momentum before throw = momentum after throw
53.08 = 0.300 * 17.5 + 96.2 * v2
53.08 = 5.25 + 96.2 * v2
v2 = [53.08 - 5.25 ] / 96.2
v2 = 47.83 / 96.2
v2 = 0.497 ~= 0.50 m/s
A couple of soccer balls of equal mass are kicked off the ground at the same speed, butat different angles. Soccer ball A is kicked off at an angle slightly above the horizontal,whereas ball B is kicked slightly below the vertical.(a) How does the initial kinetic energy of ball A compare to the initial kinetic energyof ball B?(b) How does the change in gravitational potential energy from the ground to thehighest point for ball A compare to the change in gravitational potential energyfrom the ground to the highest point for ball B?(c) If the energy in part (a) di ers from the energy in part (b), explain why there is adi erence between the two energies.
Answer:
Remember that Kinetic energy is a scalar quantity and it only depends on the speed and not necessary not the angle
Thus,Since the masses and the speed are same for both A and B, the initial kinetic energy of A and B are same.
b]
The difference or variation in gravitational potential energy is again a scalar quantity and so as long as the initial speed is same, the change in gravitational potential energy will also be the same [though they may not occur at the same horizontal position].
therefore, from the ground to the highest point of both A and B, both will have same potential energy.
Also The energy in part (a) differs from part (b),
In part (a) energy mention is kinetic energy that depends on mass and velocity of particle whereas in part (b) energy is potential energy that depends on mass and the position with reference of ground. Potential energy is a state function but kinetic energy is not.
Answer: they are the same
Explanation:
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive direction and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
Answer:
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
[tex]my'' + \zeta y' + ky=0[/tex]
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
[tex]w - kL=0[/tex]
[tex]mg - kL=0[/tex]
[tex]4 - k\frac{1}{6}=0[/tex]
[tex]k = 4\times 6 =24[/tex]
The damping constant can be found by
[tex]F = -\zeta y'[/tex]
[tex]6 = 3\zeta[/tex]
[tex]\zeta = \frac{6}{3} = 2[/tex]
Finally, the mass m can be found by
[tex]w = 4[/tex]
[tex]mg=4[/tex]
[tex]m = \frac{4}{g}[/tex]
Where g is approximately 32 ft/s²
[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]
Therefore, the required differential equation is
[tex]my'' + \zeta y' + ky=0[/tex]
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
The initial position is
[tex]y(0) = \frac{1}{2}[/tex]
The initial velocity is
[tex]y'(0) = 0[/tex]
We formulated the initial value problem for a damped spring-mass system:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0
with initial conditions x(0) = 0.5 ft and dx/dt(0) = 0 ft/s.
Let's break down the problem with the given data:
The mass weighs 4 lb.The spring stretches 2 inches under this weight.Additional displacement given is 6 inches in the positive direction.Viscous resistance is 6 lb when the velocity is 3 ft/s.First, find the spring constant k:
The weight of the mass stretches the spring by 2 inches (0.1667 feet).
The force exerted by the weight = 4 lb = mg
The spring force F = kx
So,
k = F/x k= 4 lb / 0.1667 ft k ≈ 24 lb/ftThe general form of the second-order differential equation governing the motion of the spring-mass-damper system is:
m*d²x/dt² + c*dx/dt + k*x = 0The viscous resistance given is 6 lb at 3 ft/s. Therefore, the damping coefficient c:
c = 6 lb / 3 ft/s c = 2 lb·s/ftThe initial conditions are displacement 6 inches (0.5 feet) and initial velocity 0:
x(0) = 0.5 ftCombining these elements, the initial value problem is:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
56.5 m/s²
Explanation:
From the law of conservation of momentum,
mu+m'u' = mv+m'v'........................ Equation 1
Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.
From the question,
The golf ball is at rest, Hence u' = 0 m/s
mu = mv+m'v'
Make v' the subject of the equation
v' = (mu-mv)/m'........................... Equation 2
Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.
Substitute into equation 2
v' = (0.152×44.8+0.152×27.7)/0.046
v' = (6.8096-4.2104)/0.046
v' = 2.5992/0.046
v' = 56.5 m/s²
In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to yourself. If it takes 5 seconds for the sound of the thunder to reach you, then the lightning bolt was 1 mile away from you. Sound travels at a speed of 331 meters/second. How accurate is the rule of thumb? Express your answer as a percent error.
Answer:
2.837% less than actual value.
Explanation:
Based on given information let's calculate our value.
S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.
1mile = 1609.34meters.
percentage error is.
[tex]\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%[/tex]
negative indicates less than actual value.
The rule of thumb that states the lightning bolt was one mile away for every five seconds between seeing the flash and hearing the thunder is not very accurate. The actual distance the lightning bolt would be approximately 3.19 miles away, which results in a percent error of approximately 219%. The rule tends to underestimate the distance to the lightning bolt.
Explanation:The "rule of thumb that many people follow" during a thunder and lightning storm is based on the fact that light travels much faster than sound. When you see a flash of lightning, the sound wave created by the thunder associated with the lightning bolt takes more time to reach the observer than the light from the flash.
According to the rule of thumb, we estimate one mile per five seconds because sound travels at a speed of approximately 331 meters per second. However, to calculate the actual distance in miles, you would multiply the time (in seconds) by the speed of sound and convert to miles (1 meter = 0.00062137 miles). This gives an actual distance of about 0.00063741 miles/second. Therefore, for every second delay between the lightning and the thunder, the lightning bolt would be about 0.00063741 miles away.
So, if we see a flash of lightning and hear the thunder 5 seconds later, according to the accurate calculation, the lightning bolt was just over 3.19 miles away (5 seconds * 0.00063741 miles/second). That's a sizeable difference from the one-mile estimate given by the rule of thumb. To find the percent error, we subtract the accepted value from the experimental value, divide by the accepted value, and multiply by 100. That gives us a percent error of approximately 219%. So the rule of thumb is not particularly accurate.
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In a ballistics test, a 26 g bullet traveling horizontally at 1000 m/s goes through a 35-cm-thick 400 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.
How long is the bullet in the target?
Answer: 3.60*10^-4 s
Explanation:
Given
Mass of bullet, m = 26 g = 0.026 kg
Initial speed of bullet, u = 1000 m/s
Length of target, s = 35 cm = 0.35 m
Mass of target, M = 400 kg
Final speed of bullet, v = 950 m/s
Using equation of motion
v² = u² + 2as, making a subject of formula, we have
a = (v² - u²) / (2*s)
a = 950² - 1000² / 2 * 0.35
a = 902500 - 1000000 / 0.7
a = -97500 / 0.7
The acceleration = - 1.39*10^5 m/s² ( it is worthy of note that the acceleration is negative)
now, using another equation of motion, we have
v = u + a*t
we know our a, so we make t subject of formula
time t = (v-u) / a
t = (950 - 1000) / -1.39*10^5
t = -50 / -1.39*10^5
t = 3.60*10^-4 s
A firework shell is launched vertically upward from the ground with an initial speed of 44m/s. when the shell is 65 m high on the way up it explodes into two wequal mass halves, one half is observed to continue to rise straight up to a heigh of 120 m. How high does the other half go?
Answer:
[tex]h = 83.093\,m[/tex]
Explanation:
The speed of the firework shell just before the explosion is:
[tex]v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}[/tex]
[tex]v \approx 25.712\,\frac{m}{s}[/tex]
After the explosion, the initial speed of one of the mass halves is:
[tex]v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s[/tex]
[tex]v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s[/tex]
[tex]v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}[/tex]
[tex]v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}[/tex]
[tex]v_{o} \approx 32.845\,\frac{m}{s}[/tex]
The initial speed of the other mass half is determined from the Principle of Momentum Conservation:
[tex]m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v[/tex]
[tex]25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v[/tex]
[tex]v = 18.838\,\frac{m}{s}[/tex]
The height reached by this half is:
[tex]h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}[/tex]
[tex]h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]h = 83.093\,m[/tex]
Answer:
The other half goes 17.4m high
Explanation:
Pls see calculation in the attached file
The shock absorbers in an old car with mass 1160 kg are completely worn out. When a 970 N person climbs slowly into the car, it deforms 3.0 cm. The car is now towed down the road (with the person inside). The car hits a bump, and starts oscillating up and down with an amplitude of 6.4 cm.
Model the car and person as a single body on a spring and find the period and frequency of oscillations.
Answer:
[tex]f = 0.806\,hz[/tex], [tex]T = 1.241\,s[/tex]
Explanation:
The problem can be modelled as a vertical mass-spring system exhibiting a simple harmonic motion. The spring constant is:
[tex]k = \frac{970\,N}{0.03\,m}[/tex]
[tex]k = 32333.333\,\frac{N}{m}[/tex]
The angular frequency is:
[tex]\omega = \sqrt{\frac{32333.333\,\frac{N}{m} }{1258.879\,kg} }[/tex]
[tex]\omega = 5.068\,\frac{rad}{s}[/tex]
The frequency and period of oscillations are, respectively:
[tex]f = \frac{5.068\,\frac{rad}{s} }{2\pi}[/tex]
[tex]f = 0.806\,hz[/tex]
[tex]T = \frac{1}{0.806\,hz}[/tex]
[tex]T = 1.241\,s[/tex]
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.A) At what point in the cycle does the penny first lose contact with the piston?midpoint moving uplowest pointhighest pointmidpoint moving downB) What is the maximum frequency for which the penny just barely remains in place for the full cycle?Express your answer with the appropriate units.
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
In part (a), we are asked to determine the point at which the penny first loses contact with the piston.
In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
The question is based on Simple Harmonic Motion: The penny first loses contact with the piston at its highest point (option c) in the motion. The maximum frequency for which the penny just barely remains in place for the full cycle is approximately 12.5 Hz.
Explanation:The penny will first lose contact with the piston at its highest point, which is option c) highest point in the cycle.
To determine the maximum frequency for which the penny just barely remains in place for the full cycle, we need to consider the condition for the penny to stay in contact with the piston. At the highest point of the motion, the penny experiences an upward gravitational force and a downward centripetal force due to the circular motion.
When the centripetal force becomes greater than or equal to the gravitational force, the penny will lose contact with the piston. This occurs when
mvmax2/r = mg
Solving for the maximum velocity using vmax = 2πfA, where A is the amplitude, and substituting into the equation above, we find that
fmax = g / (4π2A)
Substituting the given values, with a=4.0 cm = 0.04 m and g=9.8 m/s2:
fmax = 9.8 / (4π2(0.04)) ≈ 12.5 Hz
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A 1500 kg car carrying four 90 kg people travels over a "washboard" dirt road with corrugations 3.7 m apart. The car bounces with maximum amplitude when its speed is 20 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?
Answer:
Car body rise on its suspension by 0.0309 m
Explanation:
We have given mass of the car m = 1500 kg
Mass of each person = 90 kg
Speed of the car [tex]v=20km/hr=20\times \frac{5}{18}=5.555m/sec[/tex]
Distance traveled by car d = 3.7 m
So time period [tex]T=\frac{distance}{speed}=\frac{4}{5.55}=0.72sec[/tex]
Frequency [tex]f=\frac{1}{T}=\frac{1}{0.72}=1.388Hz[/tex]
Angular frequency is [tex]\omega =2\pi f=2\times 3.14\times 1.388=8.722rad/sec[/tex]
Angular frequency is equal to [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]8.722 =\sqrt{\frac{k}{1500}}[/tex]
k = 114109.92 N/m
Now weight of total persons will be equal to spring force
[tex]4mg=kx[/tex]
[tex]4\times 90\times 9.8=114109.92\times x[/tex]
x = 0.0309 m
A howler monkey is the loudest land animal and under some circumstances, can be heard up to a distance of 5.0km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0*10^-12 W/m^2. The acoustic power emitted by the howler is clostest to:
A) 0.31mW
B) 1.1mW
C) 3.2mW
D) 11mW
Answer:
Power emitted will be 0.314 mW
So option (A) will be correct option
Explanation:
We have given threshold hearing [tex]I=10^{-12}W/m^2[/tex]
Distance is given r = 5 km =5000 m
We have to find the power emitted
Power emitted is equal to
[tex]P=I\times A[/tex]
[tex]=10^{-12}\times 4\pi r^2[/tex]
[tex]=10^{-12}\times 4\times 3.14\times (5000)^2[/tex]
=[tex]314\times 10^{-6}watt=0.314mW[/tex]
So power emitted will be 0.314 mW
So option (A) will be correct option.
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.40. The thickness t of the air layer can be varied. Light with wavelength λ in air is at normal incidence onto the top of the air film. There is constructive interference between the light reflected at the top and bottom surfaces of the air film when its thickness is 700 nm. For the same wavelength of light the next larger thickness for which there is constructive interference is 980 nm.
a. What is the wavelength λ of the light when it is traveling in air?
b. What is the smallest thickness t of the air film for which there is constructive interference for this wavelength of light?
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= [tex](m+ \frac{1}{2} \lambda)[/tex]
replacing ;
Δx = 2t ; we have:
2t = [tex](m+ \frac{1}{2} \lambda)[/tex]
Given that thickness t = 700 nm
Then
2× 700 = [tex](m+ \frac{1}{2} \lambda)[/tex] --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = [tex](m+ \frac{1}{2} \lambda)[/tex] ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness [tex]t_{min} ; \ \ \ m =0[/tex]
∴ [tex]2t_{min} =\frac{\lambda}{2}[/tex]
[tex]t_{min} =\frac{\lambda}{4}[/tex]
[tex]t_{min} =\frac{560}{4}[/tex]
[tex]t_{min} =140 \ \ nm[/tex]
Thus, the smallest thickness t of the air film = 140 nm
Answer:
1.4x10^7m & 98nm
Explanation:
Pls the calculation is in the attached file