Answer:
The answer is 2000 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 500 N
distance = 4 m
We have
workdone = 500 × 4
We have the final answer as
2000 JHope this helps you
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is [tex] v = 350 \ m/s [/tex]
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by [tex]k = 60 cm = 0.6 \ m[/tex]
The frequency of both speakers is [tex]f = 700 \ Hz[/tex]
Generally the distance of the listener to the first speaker is mathematically represented as
[tex]L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]
[tex]L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]
[tex]L_1 = 5.39 \ m [/tex]
Generally the distance of the listener to second speaker at its new position is
[tex]L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]
[tex]L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]
[tex]L_2 = 5.64 \ m [/tex]
Generally the path difference between the speakers is mathematically represented as
[tex]pD = L_2 - L_1 = \frac{n * \lambda}{2}[/tex]
Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{v}{f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * \frac{v}{f}}{2}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=> [tex] 5.64 - 5.39 = \frac{1 * v}{2*700}[/tex]
=> [tex] v = 350 \ m/s [/tex]
The speed of sound in air is 350 m/s
Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,
L = √[(5 m)² + (4/2 m)²]
= √[25 m² + (2 m)²]
= √[25 m² + 4 m²]
= √29 m² = 5.39 m.
Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now
L' = √[(5 m)² + (4/2 + 0.6 m)²]
= √[25 m² + (2 m + 0.6 m)²]
= √[25 m² + (2.6 m)²]
= √[25 m² + 6.76 m²]
= √31.76 m²
= 5.64 m.
Now, the path difference when we first have destructive interference is
ΔL = L' - L
= 5.64 - 5.39
= 0.25
Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.
Now, since we have destructive interference for the first time, n = 0.
So, ΔL = (n + 1/2)λ
ΔL = (0 + 1/2)v/f
ΔL = v/2f
Making v subject of the formula, we have
v = 2fΔL
Substituting the values of the variables into the equation, we have
v = 2fΔL
v = 2 × 700 Hz × 0.25 m
v = 350 m/s
So, the speed of sound in air is 350 m/s
Learn more about interference of sound here:
https://brainly.com/question/1346741
Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. The energies of atoms are quantized.
Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. FALSE. The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. FALSE. The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. FALSE. During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. TRUE. The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. TRUE. In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
The reason why atoms emit only specific wavelengths is because the energy levels in atoms are quantized.
Max Plank introduced the idea of quantization of energy in the early 1900s. He introduced the idea that energy can only take on certain specific values. This idea was later extended to atoms by Neils Bohr.
The following statements explain why atoms only emit certain wavelengths of light when they are excited;
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are quantized.Learn more: https://brainly.com/question/24381583
To obtain your Class E learner's license, you'll need to _____.
A. pass a vision and hearing test
B. pass a literacy test
C. submit proof of employment
D. submit proof of insurance
Answer:
This answer was wrong
Explanation:
I took the test and I missed this question. So it is not answer B: pass a literacy test.
It will need an A. pass a vision and hearing test
Class E license:It is the standard driver's license for people that drive personal vehicles. It permits for drive a noncommercial vehicle that weighs less than 26,001 pounds.So the vision and hearing test should be required.Learn more about the insurance here: https://brainly.com/question/989103?referrer=searchResults
What is the definition of inertia
Answer: a tendency to do nothing or to remain unchanged.
Explanation:
For example, if you roll a ball, it will continue rolling unless friction or something else stops it by force.
Answer:
the property of matter by which it retains its state of rest or its velocity along a straight line so long as it is not acted upon by an external force. an analogous property of a force: electric inertia.
or a tendency to do nothing or to remain unchanged.
Write an expression for the magnitude of six ball's velocity, in terms of the angles given in the problem and the magnitude of the eight ball's initial velocity, v.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The expression for the velocity of the sixth ball is
[tex]v_2 = \frac{u_1 sin\theta }{sin \phi cos\theta + cos \phi sin\theta}[/tex]
Explanation:
From the question we are told that
The mass of the 8 ball is [tex]m_1 = 0.5 \ kg[/tex]
The initial velocity of the 8 ball is [tex]u_1 = 3.7 \ m/s[/tex]
The initial velocity of the 6 ball is [tex]u_2 = 0 \ m/s[/tex]
The mass of the 6 ball is [tex]m_2 = 0.5 \ kg[/tex]
Generally from the law of momentum conservation is
[tex]m_1 *u_1 + m_2 * u_2 = m_1 * v_1 +m_2 *v_2[/tex]
Now along the y axis
[tex]u_1 =u_y_1 = 0 \ m/ s[/tex]
[tex]u_2 =u_y_2= 0 \ m/ s[/tex]
[tex]v_1 =v_y_1= v_1 sin \theta \ m/ s[/tex]
[tex]v_2 =v_y_2= v_2 sin \phi \ m/ s[/tex]
So
[tex]m_1 *0 + m_2 * 0 = m_1 * (v_1 sin \theta) +m_2 *(v_2 sin \phi )[/tex]
=> [tex]v_1 = \frac{v_2 sin \phi }{ sin \theta}[/tex]
Now along the x axis
[tex]u_1 =u_x_1 = u_1 \ m/ s[/tex]
[tex]u_2 =u_2_2= 0 \ m/ s[/tex]
[tex]v_1 =v_x_1= v_1 cos \theta \ m/ s[/tex]
[tex]v_2 =v_x_2= v_2 cos \phi \ m/ s[/tex]
So
[tex]m_1 *u_1 + m_2 * 0 = m_1 *(v_1 cos \theta) +m_2 *(v_2 cos \phi)[/tex]
=>[tex]m_1 *u_1 = m_1 *(v_1 cos \theta) +m_2 *(v_2 cos \phi)[/tex]
substituting for [tex]v_1[/tex] in the above equation
=>[tex]m_1 *u_1 = m_1 *([ \frac{v_2 sin \phi }{ sin \theta}] cos \theta) +m_2 *(v_2 cos \phi)[/tex]
=>[tex]m_1 *u_1 = m_1 *([ \frac{v_2 sin \phi }{ sin \theta}] cos \theta) +m_2 *(v_2 cos \phi)[/tex]
given that [tex]m_1 = m_2 = m[/tex]
=>[tex]m *u_1 = m *([ \frac{v_2 sin \phi }{ sin \theta}] cos \theta) +m *(v_2 cos \phi)[/tex]
=>[tex]m *u_1 = m [([ \frac{v_2 sin \phi }{ sin \theta}] cos \theta) + (v_2 cos \phi)][/tex]
=>[tex]u_1 = ([ \frac{v_2 sin \phi }{ sin \theta}] cos \theta) + (v_2 cos \phi)[/tex]
=>[tex]u_1 = v_2 sin\phi cot \theta + v_2 cos\phi [/tex]
[tex]v_2 = \frac{u_1 }{sin\phi cot\theta + cos\phi }[/tex]
converting [tex]cot \phi[/tex] back to [tex]\frac{cos \theta}{sin\theta }[/tex]
So
[tex]v_2 = \frac{u_1 }{sin\phi [\frac{cos \theta}{sin\theta }] + cos\phi }[/tex]
multiply through by [tex]\frac{sin \theta}{sin \theta }[/tex]
[tex] \frac{sin \theta}{sin \theta } *v_2 = \frac{sin \theta}{sin \theta } * \frac{u_1 }{sin\phi [\frac{cos \theta}{sin\theta }] + cos\phi }[/tex]
=> [tex]v_2 = \frac{u_1 sin\theta }{sin \phi cos\theta + cos \phi sin\theta}[/tex]
A student must determine the relationship between the inertial mass of an object, the net force exerted on the object, and the object’s acceleration. The student uses the following procedure. The object is known to have an inertial mass of 1.0kg .
Step 1: Place the object on a horizontal surface such that frictional forces can be considered to be negligible.
Step 2: Attach a force probe to the object.
Step 3: Hang a motion detector above the object so that the front of the motion detector is pointed toward the object and is perpendicular to the direction that the object can travel along the surface.
Step 4: Use the force probe to pull the object across the horizontal surface with a constant force as the force probe measures force exerted on the object. At the same time, use the motion detector to record the velocity of the object as a function of time.
Step 5: Repeat the experiment so that the object is pulled with a different constant force.
Can the student determine the relationship using this experimental procedure?
Answer choices:
A) Yes, because Newton’s second law of motion must be used to determine the acceleration of the object.
B) Yes, because the net force exerted on the object and its change in velocity per unit of time are measured.
C) No, because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed.
D) No, because knowing the net force exerted on the object and its change in velocity per unit of time is not sufficient to determine the relationship.
Answer:
C
Explanation:
In order to obtain data about the object’s velocity as a function of time, the object must move either toward or away from the motion detector.
The student cannot determine the relationship using this experiment ; ( C ) Because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed
Given that the aim of the experiment is to determine the relationship between mass of inertia of object, net force exerted and acceleration of the object we have to first obtain the object's velocity (i.e. Distance travelled by object / time ). and
To obtain velocity of object with respect to time, the object must move in either direction ( back or front ) from its parallel position away from the position of the detector ( motion )
Hence we can conclude that The student cannot determine the relationship using this experimental procedures.Because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed
Learn more : https://brainly.com/question/20591341
Please help with this science question
Answer:
Community A
Explanation:
Community A most likely use hydroelectric power for generating most of their electricity.
Hydroelectric power utilizes the energy from water to produce electricity.
A water body is dammed, then channeled for driving a turbine which in turn generates electricity by induction.
The water used is a renewable resource. This method leads to destruction of the habitat of the fish and other aquatic life in that environment.During a thunderstorm the electric field at a certain point in the earth's atmosphere is 1.07 105 N/C, directed upward. Find the acceleration of a small piece of ice of mass 1.08 10-4 g, carrying a charge of 1.05 10-11 C.
Answer:
The acceleration of a small piece of ice is 10.40 m/s².
Explanation:
The electric force is given by:
[tex]F = Eq[/tex]
Where:
E is the electric field = 1.07x10⁵ N/C
q is the charge = 1.05x10⁻¹¹ C
The electric force is equal to Newton's second law:
[tex] Eq = ma [/tex]
Where:
m is the mass = 1.08x10⁻⁴ g = 1.08x10⁻⁷ kg
a is the acceleration
Hence, the acceleration is:
[tex] a = \frac{Eq}{m} = \frac{1.07 \cdot 10^{5} N/C*1.05 \cdot 10^{-11} C}{1.08 \cdot 10^{-7} kg} = 10.40 m/s^{2} [/tex]
Therefore, the acceleration of a small piece of ice is 10.40 m/s².
I hope it helps you!
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
Answer:
(a) 24.56 N
(b) 142.28 N
Explanation:
(a)
The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.
The centripetal function is generated from either scenario by Equation:
⇒ [tex]Fc = \frac{mv^2}{r}[/tex]
On putting the values, we get
⇒ [tex]=\frac{12\times 1.33^2}{0.864}[/tex]
⇒ [tex]=24.56 \ N[/tex]
(b)
Use T to denote whatever arm stress we can get at the bottom including its circle:
⇒ [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]
⇒ [tex]T = mg + Fc[/tex]
⇒ [tex]=12\times 9.81+24.56[/tex]
⇒ [tex]=142.28 \ N[/tex]
g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at the other. The cylinder has a radius of 0.0200 m and the safety valve is simply a 0.00750-m radius circular opening at one end, sealed with a disk. The disk is held in place by a spring with a spring constant of 950 N/m that has been compressed 0.0085 m from its natural length. Determine the magnitude of the minimum force that must be exerted on the piston in order to open the safety valve.
Answer:
58.32 N
Explanation:
Area of a circle = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the circle.
The cylinder has a radius of 0.02 m, its area is;
[tex]A_{1}[/tex] = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](0.02)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x 0.0004
= 1.2571 x [tex]10^{-3}[/tex]
Area of the cylinder is 0.0013 [tex]m^{2}[/tex].
The safety valve has a radius of 0.0075 m, its area is;
[tex]A_{2}[/tex] = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](0.0075)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x 5.625 x [tex]10^{-5}[/tex]
= 1.7679 x [tex]10^{-4}[/tex]
Area of the valve is 0.00018 [tex]m^{2}[/tex].
From Hooke's law, the force on the safety valve can be determined by;
F = ke
[tex]F_{2}[/tex] = 950 x 0.0085
= 8.075 N
Minimum force, [tex]F_{1}[/tex], required can be determined by;
[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
[tex]\frac{F_{1} }{0.0013}[/tex] = [tex]\frac{8.075}{0.00018}[/tex]
[tex]F_{1}[/tex] = [tex]\frac{0.0013 *8.075}{0.00018}[/tex]
= 58.32
The minimum force that must be exerted on the piston is 58.32 N.
Which kind of energy do flowing electrons have?
A. Electrical
B Elastic
C. Nuclear
D. Gravitational
