Answer:106.6N
Explanation:
Centripetal force is a force that cause a body to move along a circular path in a field. The body possesses a mass and velocity(since it's moving) and also radius alone the circular path on which it moves.
Given mass (m) = 2kg, velocity (v) = 8m/s and radius (r) = 1.2m
Centripetal acceleration(a)= v²/r
Centripetal force = ma = mv²/r
Substituting this values in the formula above we have;
Centripetal force = 2×8²/1.2m
Centripetal acceleration = 128/1.2
= 106.6N
A charge of −−3.00 nC is placed at the origin of an xyxy-coordinate system, and a charge of 2.00 nC is placed on the yy-axis at y=y= 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point x=x= 3.00 cm, y=y= 4.00 cm, find the xx- and yy-components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.
Answer:
Explanation:
a )
Force on 5 nC due to -3 nC
= 9 x 10⁹x 5 x 3 x 10⁻¹⁸ / ( 5 x 10⁻²)²
= 5.4 x 10⁻⁵ N
X component = 5.4 x 10⁻⁵ cosθ
= -5.4 x 10⁻⁵ x 3/5
= -3.24 x 10⁻⁵ N
y component
= -5.4 x 10⁻⁵ x 4/5
= -4.32 x 10⁻⁵ N
Force on 5 nC due to 2 nC
= 9 x 10⁹x 5 x 2 x 10⁻¹⁸ / ( 3 x 10⁻²)²
= 10 x 10⁻⁵ N
x component = 10 x 10⁻⁵ N
Total x component = (10-3.24)x10⁻⁵ N
= 6.76 x 10⁻⁵ N
y component = -4.32 x 10⁻⁵ N
magnitude = √( 6.76² + 4.32²) x 10⁻⁵ N
= 8 .02 N
DIRECTION =
Tan θ = -4.32 / 6.76
θ = 32.5
with negative x -direction , south west.
One can find the total forces exerted on the third charge in x and y directions using Coulomb's Law. The total force endured by the third charge is calculated by computing the vector sum of the individual forces. The direction of this force is determined using trigonometric principles.
Explanation:Using Coulomb’s law, we can determine the forces exerted by each individual charge on the third charge. First, we need to find the distance between each charge and the third charge. We have (x12 = 3 cm, y12 = 4 cm) for the first charge and (x23 = 0 cm, y23 = 0 cm) for the second charge.
Now, the forces in x and y directions can be calculated using the formula: F = k * |q1 * q2| / r², where k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
The components of the total force experienced by the third charge due to the first and the second charge can be calculated by adding the individual forces in both x and y directions. For total force magnitude use this formula: Ftotal = sqrt((Fx)² + (Fy)²).
To find the direction of the total force, you can use this formula: Tan θ = Fy / Fx. The direction of the force will be in the quadrant of the combined forces.
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The drawings show three situations in which a positively charged particle is moving through a uniform magnetic field B with a velocity v. For each situation, what is the direction of the magnetic force F exerted on the particle?
Answer:
Case A: - x axis
Case B: - x axis
Case C: no force as the direction of v and B are parallel.
Explanation:
Solution:
- The direction of Force exerted by the magnetic field B with a moving charge q with a velocity of v is given by an expression:
F = q*(v x B)
- Where F: force vector exerted by the field
v: velocity vector of charged particle q.
B: magnetic field direction
- The cross product of v with B:
We will use right hand rule in which our fingers curl from v to B, the direction of the thumb denotes the direction of applied force. Hence,
Case A: - x axis
Case B: - x axis
Case C: no force as the direction of v and B are parallel.
Final answer:
The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). The force is perpendicular to the plane formed by the particle's velocity and the magnetic field. The direction of the force depends on the charge of the particle and is opposite for positive and negative charges.
Explanation:
The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). According to RHR-1, if you point your right thumb in the direction of the particle's velocity (v) and your right fingers in the direction of the magnetic field (B), then your right palm will point in the direction of the magnetic force (F) acting on the particle.
For example, if the particle is moving perpendicular to the magnetic field (B), as shown in the figure, and the particle has a positive charge, then the force (F) will be directed outward from the plane formed by v and B.
It's important to note that the direction of the magnetic force on a negatively charged particle will be in the opposite direction to that on a positively charged particle.
Points A and B are in a region of uniform electric field. Point A is at the origin and point B is on the x-axis at x = 0.150 m. The electric potential at point A is 200 v and the electric potential at point B is 500 v. What are the magnitude and direction of the uniform field in this region?
Answer: 2000 v/m, from B to A.
Explanation: if point A is at the origin (x=0m) and point B is at the point x= 0.150m, the distance between both points (d) = 0.150 - 0 = 0.150m
Point A is at a 200v potential and point B is at a potential of 500v.
Difference in potential produces a voltage (v) = 500 - 200 = 300v.
The relationship between voltage, electric field intensity and distance is given by the formulae below
v=Ed
Where v = voltage = 300v, electric field =?, d = 0.150m
300 = E×0.150
E = 300/0.150
E = 2000 v/m.
Since point B is at higher potential than A, it implies that if there is an electron in this field, it will move from B to A thus making the direction of field be from B to A.
The electric field's magnitude is 2000v/m and its direction is from B to A along the negative x-axis.
Explanation:The electric field (E) in a region of space is defined as the electric force per unit charge. The electric field direction is taken to be the direction of the force it would exert on a positive test charge. The electric potential at a point in the field is the work done in bringing unit positive charge from infinity to that point. In your case, the electric potential difference, ΔV between point A and point B is: ΔV = Vb - Va = 500v - 200v = 300v. The separation between A and B, Δx is 0.150m. The magnitude of the electric field E can be calculated by the equation E = - ΔV / Δx = - 300v / 0.150m = - 2000v/m. The negative sign indicates that the field direction is from B to A along the negative x-axis.
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When we initially detect physical stimuli, such as odors, light, and sound, we call this Group of answer choices a.perception. b.sensation. c.absolute threshold. d.difference threshold.
Answer: B. Sensation
Explanation:
Sensation is input about the physical world obtained by our sensory receptors, and perception is the process by which the brain selects, organizes, and interprets these sensations. In other words, senses are the physiological basis of perception. Perception of the same senses may vary from one person to another because each person’s brain interprets stimuli differently based on that individual’s learning, memory, emotions, and expectations.
The sensitivity of a given sensory system to the relevant stimuli can be expressed as an absolute threshold. Absolute threshold refers to the minimum amount of stimulus energy that must be present for the stimulus to be detected 50% of the time.
Sometimes, we are more interested in how much difference in stimuli is required to detect a difference between them. This is known as the just noticeable difference (jnd) or difference threshold.
Final answer:
The initial detection of physical stimuli such as odors, light, and sound is known as sensation, which is distinct from perception, the interpretation of these stimuli.
Explanation:
When we initially detect physical stimuli such as odors, light, and sound, the process is referred to as sensation. Sensation occurs when sensory receptors detect sensory stimuli, involving the conversion of physical energy like light or sound waves into a form of energy that the brain can understand, which is electrical stimulation. On the other hand, perception involves the organization, interpretation, and conscious experience of those sensations. It is during the perception process that we can identify specific objects, sounds, or smells and comprehend what they mean or represent in our environment. While absolute threshold refers to the minimum amount of stimulus energy required to be detected about 50% of the time, it is not the initial detection of stimuli itself.
A park merry-go-round consists of a 255 kg circular wooden platform 4.10 m in diameter.?
Four children running alongside push tangentially along the platform's circumference until, starting from rest, the merry-go-round reaches a steady speed of one complete revolution every 2.8 s.
(a) If each child exerts a force of 26 N, how far does each child run?
(b) What is the angular acceleration of the merry-go-round?
(c) How much work does each child do?
(d) What is the kinetic energy of the merry-go-round?
Answer: a) 12.9 m b) 0.40 rad/s² c) 335.4 J d) 1349.06 J
Explanation:
a)
Distance traveled= R∅
where ∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α
To find α we have
Sum of torque= Iα
F x r x number of children = m*r²/2 * α
26 x [tex]\frac{4.1}{2}[/tex] * 4 = (255) * (4.10/2)² /2 *α
Solving we get,
α= [tex]\frac{213.2}{535.82}[/tex]
α= 0.40
So,
∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α
Becomes
∅= 6.29
Now,
Distance traveled= 4.1/2 * 6.29
Distance traveled= 12.9 m
b)
Angular acceleration= 0.40
c)
Work done= F x distance traveled
Work done= 26 x 12.9
Work done= 335.4 J
d)
Kinetic energy= [tex]\frac{1}{2}* I * (\frac{2\pi }{2.8})^2[/tex]
Putting values we get,
Kinetic energy= 267.91 * 5.03
Kinetic energy= 1349.06 J
The merry-go-round problem involves calculating the distance each child runs, the angular acceleration, the work done by each child, and the kinetic energy of the system, using principles of rotational dynamics and energy.
Explanation:We are given a merry-go-round scenario with children exerting force to set it in motion. Let's break down the problem step by step.
(a) Distance run by each child:
Determine the circumference of the merry-go-round: Circumference = π × diameter = 3.14159 × 4.10 m.Find the number of revolutions: One revolution takes 2.8 seconds, so the number of revolutions is the time spent pushing divided by 2.8 s.Distance run by each child is the product of the number of revolutions and the circumference.(b) Angular acceleration:
Calculate angular velocity (omega), given by omega = 2π / T, where T is the period (2.8 s).Use the kinematic equation for rotational motion: omega = omega_initial + alpha × t, where omega_initial = 0 (starting from rest) and t is the time. Solve for alpha (angular acceleration).(c) Work done by each child:
Work done equals force times distance: Work = force × distance run by the child.(d) Kinetic energy of the merry-go-round:
Find moment of inertia (I) for the merry-go-round: I = (1/2) × mass × radius^2.Calculate kinetic energy: KE = (1/2) × I × omega^2.
Which seismic wave is characterized by alternating compression-expansion parallel to the direction of wave movement?
Answer: Acoustic or sound wave
Explanation:
Acoustic wave is a type of wave energy that travels through a medium by adiabatic compression and decompression, they have acoustic velocity which is determined by the type and nature of medium they travel through. They are mechanical and longitudinal waves with characteristic features such as amplitude, period, frequency and wavelength.
Two parallel conducting plates that are in deep space are brought to a potential difference of 3000 V, and a small pellet of mass 2.60 mg carrying a charge of 6.00 × 10-7 C accelerates from rest from the positive plate. With what speed will it reach the other plate?
Answer:
qU = Ek = ½mv²
6.00 * 10^-7 * 3000 = ½ * 0.0026 * v²
v² = 0.7222167
v = 0.8498 m/s
Examine the symbols below:
Four schematics are shown. Symbol A shows two dots and a line draw from one not connected to the other. Symbol B shows two dots and a line draw from one connected to the other. Symbol C shows vertical lines in the pattern long, short, long, and short with a plus and minus symbol on it. Symbol D shows a dot in the center of two brackets with short horizontal dashes connected to them.
Which image represents an open switch in a circuit?
Symbol B
Symbol A
Symbol C
Symbol D
Answer: Symbol A
Explanation:
The four symbols described here represent:
- Symbol A shows two dots and a line draw from one not connected to the other. --> this is an open switch. A switch is component of a circuit that is used to open/close the circuit in order to interrupt/allow the flow of current through the circuit. In this case, the switch is open, since the line does not connect the second dot.
- Symbol B shows two dots and a line draw from one connected to the other. --> this is the symbol used to represent the switch when it is closed, so it is a closed switch.
- Symbol C shows vertical lines in the pattern long, short, long, and short with a plus and minus symbol on it. --> this symbol represents a battery, which consists of two or more cells and provides the electromotive force that pushes the electrons along the circuit.
Therefore, the correct symbol representing the open switch is
Symbol A
For this case we have that by definition, the switch is a control component that opens or closes a circuit.
It is necessary to emphasize that if the switch is open then the flow of electrons from one point to another is not allowed. Thus, it is said that we are in the presence of an open circuit.
The correct option is symbol A, an open switch is shown graphically in the attached image.
Answer:
Option B
"Electromagnetic radiation travels or propagates through space in the form of a wave but can interact with matter as a particle of energy called a photon. This dual nature is referred to as":____________
Answer:
"Electromagnetic radiation travels or propagates through space in the form of a wave but can interact with matter as a particle of energy called a photon. This dual nature is referred to as" Wave particle duality (answer)
Explanation:
Electromagnetic radiation is created when an atomic particle, such as an electron, is accelerated by an electric field, causing it to move, and it is a form of energy that spreads as both magnetic and electrical waves that travels in vessels of energy called photons.
All types of electromagnetic radiation travel at the speed of light, radiation can be also described in terms of particles of energy, called photons. Electromagnetic radiation is generated when an electrical charge is accelerated. The acceleration produces oscillating electric and magnetic fields. Electromagnetic radiation ranges from gamma rays with very short wavelength to long radio waves. Electromagnetic radiation has the dual nature: its exhibits wave properties and photon properties.
Waves are characterized by frequency, wavelength, speed and phase, where as, a photon is the basic unit of all light. Wave particle duality can be explained as when an entity exhibits a wavelike and a particlelike properties though these properties never appear simultaneously.
When you try to stretch a bungee cord a distance x, it resists with an opposing force of the form b x2 , where b is a constant. If b is measured to be 6 N/m2 , how much work does it take to stretch the bungee cord a distance of 1 meterA.2JB.3JC.0.5JD. 1J
When a bungee cord is stretched a distance x, it exerts an opposing force given by the formula F = bx2. Integrated over the distance of stretch, this results in a work done of 2 Joules for a stretch distance of 1 meter and force constant b = 6 N/m^2.
Explanation:The work done in stretching an elastic object like a bungee cord is given by the integral of the force over the distance of stretch. Given the force exerted by the bungee cord is F = bx^2, where b = 6 N/m^2 and the work W is defined as the integral from 0 to x of F dx, the formula to calculate work done becomes W = ∫from 0 to 1 (bx^2) dx. Solving this integral gives W = (b/3)x^3 evaluated from 0 to 1, which simplifies to W = (6/3) (1) = 2 Joules. So, the correct answer is 2 Joules.
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A car traveling at 60 mph has how much more energy than a car going at 15 mph?
KE increases by a factor of____.
1. 1/16
2. 4
3. 1/4
4. 16
4. KE increases by a factor of 16 is the answer
Explanation:
Kinetic energy = (1/2)mv² = 0.5 mv²
where
m = mass, and v = velocity.
So at 15 mph,
K E = 0.5 m (15) ² = 112.5 m
And at 60 mph,
K E = 0.5 m (60)² = 1800 m
m is the mass, and not meters.
So, 1800 m/112. 5 m = 16
16 times the Kinetic Energy.
a 8.0 * 104 kg spaceship is at rest in deep space. it's thrusters provide a force of 1200 kn. the spaceship fires its thrusters for 20 s, then coasts for 12 km. how long does it take the spaceship to coast this distance?
Answer:
40 seconds
Explanation:
Thrusting speed = V
Where V=at
a= F/m
Therefore V = Ft/m
F = 1200kN = 1200000N
t = 20s
m = 8.0 * 104 kg
V = 1200000 * 20 / 80000
V = 300m/s
Time for the spaceship to coast a distance of 12km
Distance = Vt
t = 12000/300 = 40seconds
The time taken for taking the spaceship to coast this distance is 40 seconds
The calculation is as follows:
Thrusting speed = V
Where
V=at
And,
[tex]a= F\div m[/tex]
Therefore [tex]V = Ft\div m[/tex]
Now
F = 1200kN
= 1200000N
t = 20s
[tex]m = 8.0 \times 104 kg[/tex]
Now
[tex]V = 1200000 \times 20 \div 80000[/tex]
V = 300m/s
Time for the spaceship to coast a distance of 12km
Distance = Vt
[tex]t = 12000\div 300[/tex]
= 40seconds
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What frequencies (in Hz) will a 1.85 m long tube produce in the audible range (20 Hz - 20,000 Hz) at 18.0°C for the following cases?
Answer:
The lowest frequency is 45.01 Hz.
The second lowest frequency is 135.03 hz.
The highest frequency is 19993.4 Hz.
Explanation:
Given that,
Length = 1.85 m
Range of frequency = 20 hz - 20000 Hz
Temperature = 18.0°C
Suppose, the tube is closed at one end lowest frequency , second lowest frequency and highest frequency
We need to calculate the velocity of sound
Using formula of sound velocity
[tex]v=v_{0}+0.61\times T[/tex]
Put the value into the formula
[tex]v=332+0.061\times18[/tex]
[tex]v=333.09\ m/s[/tex]
(a). For closed end,
We need to calculate the lowest and second lowest frequency
Using formula of frequency
[tyex]f_{n}=(2n+1)\times\dfrac{v}{4l}[/tex]
If n =0
[tex]f_{0}=\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{0}=\dfrac{333.09}{4\times1.85}[/tex]
[tex]f_{0}=45.01\ Hz[/tex]
If n = 1
[tex]f_{1}=\dfrac{3v}{4l}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{3\times333.09}{4\times1.85}[/tex]
[tex]f_{1}=135.03\ Hz[/tex]
Now, The maximum audible range is 20000 Hz.
We need to calculate the value of n
Using formula of frequency
[tex]f_{n}=(2n+1)\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]20000=(2n+1)\times45.01[/tex]
[tex]20000=2n\times45.01+45.01[/tex]
[tex]n=\dfrac{20000-45.01}{2\times45.01}[/tex]
[tex]n=221.6[/tex]
We need to calculate the maximum frequency
Using formula of frequency
[tex]f_{n}=(2n+1)\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{221.6}=(2\times221.6+1)\times45.01[/tex]
[tex]f_{221.6}=19993.4\ Hz[/tex]
Hence, The lowest frequency is 45.01 Hz.
The second lowest frequency is 135.03 hz.
The highest frequency is 19993.4 Hz.
Final answer:
To find the resonant frequencies of a 1.85 m long tube, calculate the speed of sound at 18.0°C, and determine the wavelengths for both an open and closed tube. Use these to calculate the fundamental frequencies, which are 92.5 Hz for the open tube and 46.3 Hz for the closed at one end tube, with higher harmonics also possible in the audible range.
Explanation:
The question involves calculating the resonant frequencies of a tube at a certain temperature, which relates to the physics concept of standing waves in air columns. Specifically, a 1.85 m long tube will produce different frequencies based on whether it's closed at one end or open at both ends, due to the formation of nodes and antinodes. At 18.0°C, the speed of sound in air can be calculated using the formula v=331.4 + 0.6T, where T is the temperature in degrees Celsius. This gives 331.4 + 0.6(18.0) = 342.2 m/s for the speed of sound. The wavelength λ for the fundamental frequency (for a tube open at both ends) is 2L, where L is the length of the tube. Thus, λ = 2(1.85 m) = 3.70 m. The frequency can then be calculated using f = v/λ, resulting in approximately 92.5 Hz. For a tube closed at one end, the fundamental frequency has a wavelength of 4L, because only quarter-wavelengths can fit in the tube, making the fundamental frequency approximately 46.3 Hz. Higher harmonics can also be calculated for both cases, but they will depend on the number of nodes and antinodes that can fit within the tube for closed and open situations respectively.
The vibrations along a transverse wave move in a direction _________.
Answer: perpendicular to it oscillations.
Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.
By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.
Examples of transverse waves includes wave in a string, water wave and light.
Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.
If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.
Since the oscillations is perpendicular to the direction of wave, it is a transverse wave
In a transverse wave, the vibrations move in a direction perpendicular to the direction of wave travel. This distinguishes them from other types of waves. An example is the movement of water waves.
Explanation:In a transverse wave, the vibrations or disturbances move in a direction perpendicular (at right angles) to the direction of the wave's travel. This is the key characteristic that separates transverse waves from other types of waves. A common example of transverse waves would be waves in a body of water where the water particles move up and down causing the wave to move forward.
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anna litical is practicing a centripetal force demonstration at home. she fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when its half- full of water and when it is quater- full of water. In which case is more force required to stop spin the bucket circle.
Answer:
half filled bucket requires more force to stop
Explanation:
When spinning a bucket half filled it is clear that is has greater mass of water than the quarter filled bucket.
While revolving any mass tied about a fixed point we have a centripetal force acting on the bucket which makes it take the circular path during the motion.
This is centripetal force is given as:
[tex]F_c=m.\frac{v^2}{r}[/tex]
where:
[tex]m=[/tex] mass of the revolving body
[tex]v=[/tex] tangential velocity
[tex]r=[/tex] radius of revolution
From the above equation we observe that centripetal force is directly proportional to mass and square of the velocity but inversely proportional to the radius of the revolution which is same as the length of the rope between the hand and the bucket (more precisely the distance between the center of revolution and the center of mass of the revolving body). While this force acts inward to the circular path and not along the tangential direction.The revolving mass has to be brought to rest in this case the momentum of the heavier mass will be greater and from the Newton's second law of motion we have the the rate of change in momentum directly proportional to the force applied.
Mathematically:
[tex]F=\frac{d}{dt}(m.v)[/tex]
here the mass is constant so,
[tex]F=m.\frac{d}{dt} v[/tex]
Therefore if the length of the rope, and the speed of revolution is same in both the case then the half filled bucket whose mass is greater than the quarter filled bucket will require more force to stop the circular motion of the bucket.
Answer: so there is potiental enrgy need to stop spinning the bucket
Explanation:The Force of and bucket outcome is determed by the force of that i uesd to
Frequency division multiplexing: A. operates by statistically time slicing the signal B. operates by dividing the signal into different frequencies C. uses a codec that divides signals into different channels D. operates by time slicing the signal operates by light dividing the signal
Frequency division multiplexing operates by dividing the signal into different frequencies
Explanation:
The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.
Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.
Frequency Division Multiplexing operates by dividing a signal into different frequencies to enable multiple transmissions at the same time. This method is used in FM radio and television broadcasts, as well as cell phone conversations and computer data transmissions.
Explanation:Frequency Division Multiplexing (FDM) operates by dividing the signal into different frequencies. This can be seen in how FM (Frequency Modulation) radio signals are used. In FM radio transmission, the information is carried by varying the frequency of the carrier wave and keeping its amplitude constant. This forms different channels on which multiple signals can be transmitted simultaneously without interfering with each other.
Another example is how television broadcasts. Since a vast amount of visual and audio information needs to be carried, each channel requires a larger range of frequencies, these fall under VHF and UHF (high to ultra-high frequencies).
Cell phone conversations and computer data are also transmitted using a similar approach by converting the signal into a sequence of binary ones and zeros. This binary sequence can be transmitted via different frequencies, allowing multiple data transmissions to take place at the same time.
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What is the velocity of an electron that has a de Broglie wavelength approximately the length of a chemical bond? Assume this length to be 1.2 * 10-10m.
Answer : The velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]
Explanation :
According to de-Broglie, the expression for wavelength is,
[tex]\lambda=\frac{h}{p}[/tex]
and,
[tex]p=mv[/tex]
where,
p = momentum, m = mass, v = velocity
So, the formula will be:
[tex]\lambda=\frac{h}{mv}[/tex] .............(1)
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\lambda[/tex] = wavelength = [tex]1.2\times 10^{-10}m[/tex]
m = mass of electron = [tex]9.11\times 10^{-31}kg[/tex]
v = velocity of electron = ?
Now put all the given values in formula 1, we get:
[tex]1.2\times 10^{-10}m=\frac{6.626\times 10^{-34}Js}{(9.11\times 10^{-31}kg)\times v}[/tex]
[tex]v=6.1\times 10^{6}m/s[/tex]
Thus, the velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]
The area under the curve of the net external force vs time graph is equal to __________ or ________.
Impulse delivered
or
Change in momentum.
Final answer:
The area under the net external force vs time graph is equal to impulse or change in momentum, representing the impulse-momentum theorem.
Explanation:
The area under the curve of the net external force vs time graph is equal to impulse or change in momentum. This relationship arises from Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum. In graphical terms, when force is plotted on the y-axis and time on the x-axis, the area under the curve represents the impulse, which is the product of the force and the time interval over which it acts. This impulse is equivalent to the change in momentum of the object according to the impulse-momentum theorem.
Apply Newton's first law to each of the following situations. In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? A book slides across the table and comes to a stop.
Answer:
kinetic frictional force opposes the relative motion between the surfaces in contact of the book and the table.
Explanation:
When a book slides on across the table and comes to stop then there must be force acting on it which hinder its state of uniform motion.According to the Newton's first law of motion every body continues to be in the state of rest or in uniform motion until acted upon by any external force.Here while the book slides on the table there acts a force of friction between the table surface and the surface of the book which is in contact to the top of the table while the relative motion between the surfaces there acts a kinetic frictional force which opposes the relative motion between the two.The height of an object dropped from the top of a 64-foot building is given by h(t)=-16t^2+64. How long will it take the object to hit the ground?
Answer:
1.86 s
Explanation:
Given the expression
h(t) = -16t²+ 64...................... Equation 1
Where h = height of the object, t = time it will take the object to hit the ground.
Given: h = 64 foot.
We have to concert from foot to meters
If 1 foot = 0.3048 meters
Then, 64 foot = 0.3048×64 = 19.51 meters.
We substitute the value of h into equation
119.51 = -16t²+64
-16t² = 199.51-64
-16t² = 55.51
t² = 55.51/-16
t² = 3.469
t = √3.469
t = 1.86 s.
Hence it will take the object 1.86 s to hit the ground.
Humans have existed for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as 1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existed?
Answer:
8.64 seconds
Explanation:
According to the question
[tex]10^{10}\ years=1\ day[/tex]
[tex]1\ year=\dfrac{1}{10^{10}}\ day[/tex]
Humans have lived for [tex]10^6[/tex] years
In one universe day
[tex]10^6\ years=10^6\times \dfrac{1}{10^{10}}=10^{-4}\ day[/tex]
In seconds
[tex]10^{-4}\times 24\times 60\times 60=8.64\ seconds[/tex]
In one universe day humans have lived for 8.64 seconds
Humans have been around for 8.64 'universe seconds'.
The question asks for a conversion of human existence into 'universe seconds' given that humans have been around for approximately 106 years and the universe is approximately 10^10 years old. To answer this, one must conceptualise a 'universe day' as representing the entire 10^10 years of the universe's existence and break it down into 'universe seconds' in the same way that a normal day is broken down into normal seconds.
A regular day has 86,400 seconds (24 hours × 60 minutes per hour × 60 seconds per minute). As humans have existed for 10^6 years in the context of the universe's 10^10 years, we can calculate the fraction of the universe's existence that humans have been present for:
Human existence fraction = (Human years) / (Universe years) = 10^6 / 10^10 = 1 / 10^4
Now, we multiply this fraction by the total number of seconds in a day to find the equivalent 'universe seconds' of human existence:
Universe seconds of human existence = 1 / 10^4 × 86,400
Universe seconds of human existence = 8.64
Therefore, humans have existed for 8.64 'universe seconds' when defining the age of the universe as 1 'universe day'.
If an object which weighs 100 lbs on the Earth's surface were placed on a planet with 3 times the radius of the Earth and with 5 times the Earth's mass, how much would that object weigh? Enter answer to nearest 0.1 lbs.
The object's weight on the other planet is determined by the force of gravity on that planet, which depends on the planet's mass and radius. The object's weight can be found by plugging these values into the formula for gravitational force, once the actual mass of the object is obtained by dividing its weight on Earth by the Earth's gravitational acceleration.
Explanation:To find the weight of the object on the other planet, we need to calculate the gravitational pull on that planet. The force of gravity is given by the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two objects (which in this case is the radius of the planet).
On Earth, the object weighs 100 lbs. This is its mass times the gravity of Earth, which is roughly 9.8 m/s^2. So we can find the mass of the object by dividing the weight (100 lbs) by the acceleration due to gravity (9.8 m/s^2).
The planet in question is stated to have 3 times the Earth's radius and 5 times its mass. So we substitute these values into the formula along with the mass of the object we calculated, and solve for F, the force, which will be the weight of the object on the other planet.
Learn more about Force of gravity here:https://brainly.com/question/12753714
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the New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.12 HzHz , and the acceleration of the top of the building can reach 2.3 %% of the free-fall acceleration, enough to cause discomfort for occupants. what is______________
Answer:
The question is incomplete or some details are missing. The last part of the question says ; What is the total distance, side to side, that the top of the building moves during such an oscillation? Express your answer to two significant figures and include the appropriate units.
Total distance (x) = 0.3972m
Explanation:
The detailed steps is as shown in the attachment
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?
Answer:
1keff=1k1+1k2
see further explanation
Explanation:for clarification
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?
From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.
Now the spring is in series combination
F[tex]\alpha[/tex]e
F=ke
k=f/e.........*
where k is the force constant or the constant of proportionality
k=f/e
[tex]f_{eff} =f_{1} +f_{2}[/tex]............................1
also for effective force constant
divide all through by extension
1) Total force is
Ft=F1+F2
Ft=k1e1+k2e2
F = k(e1+e2) 2)
Since force on the 2 springs is the same, so
k1e1=k2e2
e1=F/k1 and e2=F/k2,
and e1+e2=F/keq
Substituting e1 and e2, you get
1/keq=1/k1+1/k2
Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.
An 80-kg skater is coasting at a velocity of 6 m/s. She sees a small child in her way and picks him up as she skates by. Her velocity decreases to 1.12 m/s. What is the mass of the child she picked up?
Answer:
348.57 kg
Explanation:
From the law of conservation of momentum,
Total momentum before the small child was picked = Total momentum after the small child was picked
mu + m'u' = V(m+m')................... Equation 1
Where m = mass of the skater, u = initial velocity of the skater, m' = mass of the small child, u' = initial velocity of the small child, V = common velocity after the child was picked.
Note: Assuming the small child was stationary before he was picked, the u' = 0 m/s.
making m' the subject of the equation
m' = (mu-mV)/V........................... Equation 2
Given: m = 80 kg, u = 6 m/s, V = 1.12 m/s.
Substitute into equation 2
m' = [(80×6)-(80×1.12)]/1.12
m' = (480-89.6)/1.12
m' = 390.4/1.12
m' = 348.57 kg.
Thus the mass of the small child = 348.57 kg
What tension must a 42.1 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 2.85 m/s?
Answer:
Tension in the string will be 19.293 N
Explanation:
We have given length of the string r = 42.1 cm = 0.421 m
Mass of the stone m = 1000 gram
We know that 1000 gram = 1 kg
Velocity in the circular path v = 2.85 m/sec
We have to find the tension in the string
Tension in the string will be equal to centripetal force
So tension [tex]T=\frac{mv^2}{r}[/tex], here m is mass, v is velocity and r is length of the string
So tension in the string [tex]T=\frac{1\times 2.85^2}{0.421}=19.293N[/tex]
So tension in the string will be equal to 19.293 N
An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and its volume is 1.08×1012km3. What is the earth’s average density?
Final answer:
To find the Earth's average density, we divide its mass (5.94×10²⁴ kg) by its volume converted to cubic meters (1.08×10²⁴ m³), resulting in a density of approximately 5.5 kg/m³.
Explanation:
An object's average density is defined as the ratio of its mass to its volume (rho = M/V). To calculate the Earth's average density, we use the given mass of the Earth (5.94×10²⁴ kg) and the given volume of the Earth (1.08×10¹² km³).
It's important to convert the volume from cubic kilometers to cubic meters to ensure consistency in units since density is typically expressed in kilograms per cubic meter (kg/m³). There are 1,000,000,000,000 (1×10¹²) cubic meters in one cubic kilometer, so the volume in cubic meters is 1.08×10¹² km³ × 1×1012 m³/km³ = 1.08×1024 m³.
Now, dividing the mass by the volume in consistent units gives:
Density = Mass / Volume = 5.94×10²⁴ kg / 1.08×10²⁴ m³ = 5.5 kg/m³
Therefore, The average density of Earth is roughly 5.5 kg/m³.
A ball is moving across a level platform 1.6m above the floor. After rolling off the ball hits the floor 20 from the base of the platform. What is the velocity of the ball as it left the platform? Remember that the platform is level and that the ball is moving horizontally when it leaves the platform.
Answer:
The correct answer is
35.01 m/s
Explanation:
To solve the question, we note the given variables thus
Height of platform S = 1.6 m
Distance from the platform the ball landed = 20 m
S = ut + 0.5gt² therefore 1.6 = 0.5 × 9.81 × t² or t = 0.57 s
Distance = 20 m = velocity × time
Therefore velocity = Distance / time = 20 m/0.57 s = 35 m/s
Answer:
Explanation:
When the skydiver descends to a certain height above the ground, she deploys her parachute to ensure a safe landing. Usually the parachute is deployed when the skydiver reaches an altitude of about 900 m (3000 ft). Immediately after deploying the parachute, does the skydiver have a nonzero acceleration?
Final answer:
After deploying the parachute, a skydiver does experience nonzero acceleration as the increased drag causes rapid deceleration until reaching a new lower terminal velocity.
Explanation:
Immediately after deploying her parachute, a skydiver does indeed have a nonzero acceleration. When a parachute opens, it rapidly increases the area exposed to air resistance, causing a significant increase in drag. As a result, the force of air resistance acting on the skydiver greatly exceeds the force of gravity, which causes a rapid deceleration of the skydiver. The magnitude of this deceleration is determined by the net force acting on the skydiver, which now includes the strong opposing force from the parachute's drag. This deceleration continues until the skydiver reaches a new, much lower terminal velocity with the parachute open.
At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during this 2.4 s interval?
m/s2
a) in the initial direction of motion
b) opposite the initial direction of motion
c) direction changes continuously
Answer:
(a) (18m/s/t₁)m/s²
(b) -12.5m/s²
(c) -20mls²
Explanation:
(a) Let t₁ be the initial time
a = v-u/t
acc = (18m/s/t₁)m/s²
(b) acc = -30m/s/2.4
= -12.5m/s²
(c)The particle was at a speed of 18m/s in the positive x-direction and later after 2.4s ≡Δt, it was at speed of -30m/s in the negative x-direction.
so this imply that the velocity was first v₁ =18m/s and later v₂ = -30m/s.
The average acceleration is then:
Aavg = Δv
Δt
= v₂-v₁/Δt
= -30-18/2.4 = -20mls²