The electric filed in V/m is 1.58 * 10^6 V/m
What is the electric field?We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.
Thus;
E = V/d
E = electric field
V = voltage
d = distance of separation
E = 24 * 10^3 V/1.52 * 10^-2 m
E = 1.58 * 10^6 V/m
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Liquids are used in thermometer but not the solid.
Solids have very small intermolecular space and when heated, do not expand as per the container volume. Basically, their expansion is not good.
Liquids, on the other hand, have a definite volume and their expansion is ideal. Also, they fill the container only to the desired level, unlike gases.
A 60 year old person has a threshold of hearing of 79.0 dB for a sound with frequency f=10,000 Hz. By what factor must the intensity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.
43.DB(decibels)
⇉ Unit conversion
4.3(bels)
4.951NP
4.3(decades)
14.28(octaves)
Interpretations
⇉ Power quantity level
Basic unit dimensions
⇉ ( logarithmic quantity ).
They provided the intensity in decibels for the problem, but they are unsure by what factor to increase it (I) to make the sound loud enough for the elderly person to hear.
Neglect f entirely.
The following equation must be used to convert decibels (dB) to I:
I=(10^(dB/10))*10^-12
Divide the elder person's dB by the younger person's dB after doing this for each dB.
1.8372093023.
Decibels are used to measure sound (dB). A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper. Your hearing may begin to be harmed if exposed to noise over 70 dB for an extended period of time. Your ears can suffer instant damage from loud noise above 120 dB.
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quations (3.4) and (3.5) are equivalent expressions for Lagrange's equations.
Exercise 3.1 Using the Nielsen form, determine the equation of motion
for a mass m connected to a spring of constant k.
Exercise 3.2 Using the Nielsen form, determine the equations of motion
for a planet in orbit around the Sun. (Answer: mr - mro² = - GMm and
mrö +2mr00.)
3.2 Hamilton's principle
echanical system composed of N particles can be described by n = 3N
inande
Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the
Explanation:
A phone with a mass of 0.2 kg is dropped from a height of 30 m what is it’s speed when it hits the ground?the acceleration of gravity is 9.8ms
Answer:
24.2 m/s
Explanation:
Mass is irrelevant in this situation....
Displacement: ( to find time)
x = xo + vo t - 1/2 at^2
30= 0 + 0 - 1/2 (9.8)t^2
t = 2.47 seconds
Velocity:
vf = a t = 9.8 (2.473) = 24.2 m/s
Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person wearing a g-suit can withstand accelerations up to about 9g (88.2 m/s2) without losing consciousness.
(a) If a human centrifuge has a radius of 4.00 m, what angular speed (in rad/s) results in a centripetal acceleration of 9g?
_______rad/s
(b) What linear speed (in m/s) would a person in the centrifuge have at this acceleration?
________m/s
(a) The angular speed of the human centrifuge is 4.7 rad/s.
(b) The linear speed (in m/s) a person in the centrifuge would have at this is 18.8 m/s.
Angular speed of the human centrifuge
a = ω²r
where;
a is centripetal acceleration = 9gr is radius of human centrifugeω = √(a/r)
ω = √(9g/r)
ω = √(9 x 9.8 / 4)
ω = 4.7 rad/s
Linear speed of the centrifugev = ωr
where;
v is linear speedv = (4.7 rad/s) x 4 m
v = 18.8 m/s
Thus, the angular speed of the human centrifuge is 4.7 rad/s. The linear speed (in m/s) a person in the centrifuge would have at this is 18.8 m/s.
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In a truck-loading station at a post office, a small 0.200 kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (Figure 1). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.
A). What is the coefficient of kinetic friction on the horizontal surface?
b How much work is done on the package by friction as it slides down the circular arc from A to B ?
The coefficient of kinetic friction on the horizontal surface, μ = 0.39.
The work done on the package by friction as it slides down the circular arc from A to B, W = -4.176.
What is the frictional force acting on the package moving on the track?
Frictional force is the force which opposes the motion of an object over another.
Frictional force acts at the surface of contact of the objects.
The frictional force, F is related to the coefficient of kinetic friction on the horizontal surface, μ, and the normal reaction or weight, mg, of the package as follows:
F = μmg
Work done against friction, W = F * d
W = μmgd
The Kinetic energy of the package is equal to the work done against friction.
Kinetic energy of the package = mv²/2
μmgd = mv²/2
μ = v²/2gd
μ = 4.8²/(2 * 9.81 * 3)
μ = 0.39
b. Work done on the package by friction as it slides down the circular arc from A to B is given as follows:
W = mv²/2 - mgh
W = v²/2 - gh
W = 4.8²/2 - 9.81 * 1.6
W = -4.176
In conclusion, the work done by friction is lost in the form of heat.
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Which one of the following is a source of mechanical form of energy? A. Coal B.Wind C.Battery D.Magnet
wind (b), also some others are:
- steam
- gas
- liquid fules
Which type of wave can travel in a vacuum?
Answer:
Electromagnetic waves
density of a substance ratio
Answer:
Density of a substance is the ratio of mass of the substance to its volume.
Answer:
Density of a substance is the ratio of mass of the substance to its volume.
light has a frequency of about 6 x1014 Hz.
Answer:
The visible light frequency is 400 THz to 700 THz, approximately. A THz is a Terahertz, which is a unit of frequency equal to one trillion Hertz.
The tires of a car make 68 revolutions as the car reduces its speed uniformly from 85.0 km/h to 56.0 km/h. The tires have a diameter of 0.80 m.
a)What was the angular acceleration of the tires?
Express your answer using two significant figures.
b)If the car continues to decelerate at this rate, how much more time is required for it to stop?
Express your answer to two significant figures and include the appropriate units.
c)If the car continues to decelerate at this rate, how far does it go? Find the total distance.
Express your answer to three significant figures and include the appropriate units.
a) The wheel of the car decelerates at an angular acceleration of - 0.577 radians per square second.
b) The wheel of the car requires a time of 51.149 seconds to stop.
c) The wheel of the car travels a distance of 683.225 meters before stopping.
How to analyze a decelerating rotating wheel
a) In this case we have a rotating wheel that decelerates at constant rate. The angular acceleration, in radians per square second, of the tires is determined by the following formula:
α = [ω'² - ω²] / (2 · θ) (1)
Where:
ω - Initial angular velocity, in radians per second.ω' - Final angular velocity, in radians per second.θ - Change in the angular displacement, in radians.The initial and final angular velocities, in radians per second, are now determined:
ω' = v' / R (2)
ω = v / R (3)
Where:
v - Initial linear velocity, in meters per second.v' - Final linear velocity, in meters per second.R - Radius of the tire, in meters.If we know that R = 0.80 m, v = 23.611 m / s, v' = 15.556 m / s and θ ≈ 427.257 radians, then the angular acceleration of the tire is:
ω' = (15.556 m / s) / (0.80 m)
ω' = 19.445 rad / s
ω = (23.611 m / s) / (0.80 m)
ω = 29.513 rad / s
α = [(19.445 rad / s)² - (29.513 rad / s)²] / [2 · (427.257 rad)]
α = - 0.577 rad / s²
b) The time required to stop the car, in seconds, is determined by the following expression:
t = (ω' - ω) / α (4)
t = (0 rad / s - 29.513 rad / s) / (- 0.577 rad / s²)
t = 51.149 s.
c) First, we find the change in angular displacement of the tire:
θ = [ω'² - ω²] / (2 · α) (5)
θ = [(0 rad / s)² - (29.513 rad / s)²] / [2 · (- 0.577 rad / s²)]
θ = 754.781 rad
Lastly, the distance traveled by the vehicle is:
s = R · θ (6)
s = (0.80 m) · (754.781 rad)
s = 683.825 m
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progress of science has not caused any ill effect true or false?
Is it possible for an object to be in equilibrium (no net force), if only one force is acting on it ?
1. What type of waves can only travel through a medium?
Answer:
A mechanical wave can only travel through a medium.
*photo attached* The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 1.01 m, respectively. The respective rotational speeds are 444 rev/min and 4,130 rev/min. Calculate the speeds of the tips of both rotors.
main rotor ______m/s
tail rotor _______m/s
Compare these speeds with the speed of sound, 343 m/s.
vmain rotor = _______ vsound
vtail rotor = _______ vsound
(a) The speeds of the tips of both rotors; main rotor 178.3 m/s and tail rotor 218.4 m/s.
(b) The speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
Linear speed of main motor and tail rotorv = ωr
where;
ω is the angular speed (rad/s)r is radius (m)v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)
v(main rotor) = 178.3 m/s
v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)
v(tail rotor) = 218.4 m/s
Speed of the rotors with respect to speed of sound% speed (main motor) = 178.3/343 = 0.52 = 52 %
% speed (tail motor) = 218.4/343 = 0.64 = 64 %
Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
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A bead of mass 15g is sliding on a wire. It has a speed of 2.0m/s at A, and it stops as it reaches the point C. The length of the wire from A to C is 250cm. How large an average friction force opposed the motion of the bead?
Answer:
.12 N
Explanation:
Velocity goes from 2 m/s to 0 m/s so average velocity = 1 m/s
.250 m / 1 m/s = .25 second to stop
change in velocity / change in time = accel
2 m/s / .25 sec = 8 m/s^2
F = ma
= .015 kg * 8 m/s^2 = .12 N
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and
400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the
steam is cooled. Determine the compression work if the final state is
a) 1.0 MPa and 250°C and
b) 500 kPa.
c) Also determine the temperature at the final state in part b.
A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,
The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.
An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.
Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.
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Peter has started a workout where he jumps rope, does push-ups, kicks a ball through a pattern of cones, does sit-ups, and then repeats the cycle. This is an example of:
Specificity training
Overload training
Circuit training
Interval training
The workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.
What is circuit training?A workout is any activity that requires much physical or mental effort, or produces strain.
There are several approaches that can be employed to achieve workout procedures and they are as follows:
Circuit trainingInterval trainingSpecificity trainingCircuit training is a combination of six or more exercises performed with short recovery periods between them for either a set number of repetitions or a prescribed amount of time.
On the other hand, interval training is similar to circuit training but differs in the sense that a period of rest is allowed in between the exercises.
Therefore, the workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.
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In thicker materials the particles can move more easily, therefore the resistance has to ____________.
A. Increase
B. Decrease
C. Stay the same
D. Not enough info
In thicker materials the particles can move more easily, therefore the resistance has to decrease.
Hence Option (b) is correct.
Resistivity, which is sometimes represented by the Greek letter rho, is quantitatively equal to a specimen's resistance R times its cross-sectional area A times its length.
ρ = R × A/ l
R = ρ × l/ A
It means the Resistance of the material is inversely proportional to the area of the material.
As the area of the material increases, the resistance decreases and vice versa.
Hence, In thicker particles, there is a greater area available and that's why resistance has to decrease so that particles can move more easily.
Hence, Option (b) is correct.
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Consider the ballistic pendulum collision. The projectile, of mass m, is fired into a large block of mass M. (Figure 1)
a) Derive a formula for the fraction of the magnitude of kinetic energy lost
Express your answer in terms of the variables m and M .
b)Evaluate the fraction for m = 18.0 g and M = 380 g .
Express your answer using three significant figures.
The fraction of the magnitude of the kinetic energy lost is [tex]\frac{(change) KE }{KE} = 1 - \frac{m}{m + M }[/tex]. = 0.955.
using the law of conservation of momentum,
[tex]mv=(m+M)V[/tex]
[tex]V= m ( \frac{1}{M+m} ) v\\[/tex]
kinetic energy lost,
Δ[tex]KE=KE_{i} -KE_{f}[/tex]
(see image )
now, for the other part
the fraction of the kinetic energy lost,
ΔKE/ KE = [tex]1- m (\frac{1}{m+M} )[/tex]
ΔKE/ KE = [tex]1 - 18 ( \frac{1}{18 + 380 } )[/tex]
ΔKE/ KE = 0.955.
what is kinetic energy ?
The energy that an object has as a result of motion is known as kinetic energy. It is stated as the effort needed to transfer a bulk body from rest to the given velocity. The body holds onto the kinetic energy it gained during its acceleration until its speed changes.
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How does the suns energy affect the climate of an area
Answer:
the earth receives the suns energy as radiation which is able to travel without the need of matter from point A to point B and this radiation is also in the form of heat.
Explanation:
Sun's energy fall on the earth in form of radiation and heats up the atmosphere and produces unequal heating due to its spherical nature and impacts climate of different areas.
The Sun is the only source of energy for the Earth's climate system. Fundamental to atmospheric composition is the warming of the atmosphere caused by solar radiation, which also causes global wind patterns and aids in the development of clouds, storms, and rainfall.
When the sun is at its maximum (peak in sunspots, prominences, and flares), it produces more ultraviolet (UV) radiation. More ozone is produced as a result of UV light, warming the stratosphere. The lower atmosphere and Earth's surface warm slightly as a result of the increased solar radiation.
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A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calculate the angle of refraction in the glass.
Answer:
Approximately [tex]28^{\circ}[/tex].
Explanation:
The refractive index of the air [tex]n_{\text{air}}[/tex] is approximately [tex]1.00[/tex].
Let [tex]n_\text{glass}[/tex] denote the refractive index of the glass block, and let [tex]\theta _{\text{glass}}[/tex] denote the angle of refraction in the glass. Let [tex]\theta_\text{air}[/tex] denote the angle at which the light enters the glass block from the air.
By Snell's Law:
[tex]n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})[/tex].
Rearrange the Snell's Law equation to obtain:
[tex]\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}[/tex].
Hence:
[tex]\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}[/tex].
In other words, the angle of refraction in the glass would be approximately [tex]28^{\circ}[/tex].
A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of
39 N · m.
If the mechanic grasps the wrench as shown in the figure below, determine the magnitude (in N) of the force she must exert on the wrench.
______N
The magnitude (in N) of the force she must exert on the wrench is 150.1 N.
Force exerted by the wrench
The force exerted by the wrench is calculated using torque formula as follows;
torque, τ = F x r x sinθ
where;
F is the applied forcer is the perpendicular distance if force appliedF = τ /(r sinθ)
F = (39) / (0.3 sin 60)
F = 150.1 N
Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.
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Using the Left Hand Rule, if motion is down and the current is toward you, which way is the field?
In the left-hand rule, the field is represented by the forefinger and it is perpendicular to the motion.
Fleming’s Left-Hand Rule:A force perpendicular to the field's direction and the direction of the current flow is experienced by a current-carrying conductor when it is exposed to an external magnetic field. According to Fleming's Left Hand Rule, if the thumb, forefinger, and middle finger are arranged in a straight line on the left hand, the thumb will point in the direction of the force experienced by the conductor, and the forefinger will point in the direction of the magnetic field, and the middle finger will point in the direction of the electric current. This rule is employed to determine the magnetic force's direction within an electric motor.
Fleming’s Left-Hand Rule are essential rules applicable in magnetism and electromagnetism. They were created by John Ambrose Fleming in the late 19th century as an easy method of determining the direction of motion in an electric motor.
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the meaninn of dynamics
Answer:
Dynamics is just a nice word used in physics denoting a branch of physics, related to the study of forces. Usually these forces are not in mechanical equilibrium, else the branch would be statics.
Is there any change in mass of substance after it changes its state? Explain with an example
Answer:
No
Explanation:
When a substance changes its state , there will not be any change in the mass of substance . For example , if we change 100 grams of water (liquid) to ice(solid) , the mass of ice will be same i.e., 100 grams . This shows there is no change in the mass even if a substance changes its state.
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An object is thrown upward with initial velocity of 30m/s at angle of 30 degree to the horizontal. calculate the components of the initial velocity.
Answer:
Vo (vertical) = Vo sin 30 = 30 m/s / 2 = 15 m/s
Vo (horizontal) = Vo cos 30 = .866 30 m/s = 26 m/s
a man carries a hand bag by hanging on his hand and moves horizontally where the bag does not move up or down. What is the work done on the bag? The man gets tired after sometime of the movement. Why
The work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.
What is the work done?In Physics, we define the work done as the product of the of force and distance. Hence, we generally define the work done as that which occurs when the force applied moves a distance in the direction of the force. This implies that the work done is not a vector but a scalar quantity.
In relation to the man and the bag, the work done is the product of the force that the man applies and the displacement of the bag. As such, the reason for the tiredness of the man is that the internal energy that he possesses is transferred to moving the bag.
Thus, the work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.
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If a skydiver jumps out of a plane horizontally (in other words with no initial vertical velocity), then what will her vertical speed be after having fallen a vertical distance of 50.8m if you neglect air resistance over that distance?
The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
Time of motion of the girl
The time of motion of the girl is calculated as follows;
h = vt + ¹/₂gt²
where;
v is initial vertical velocity = 0t is time of motiong is acceleration due to gravitySubstitute the given parameters and solve for time of motion;
50.8 = 0 + ¹/₂(9.8)t²
2(50.8) = 9.8t²
101.6 = 9.8t²
t² = 101.6/9.8
t² = 10.367
t = √10.367
t = 3.22 seconds
Final vertical velocity of the skydivervf = vi + gt
where;
vi is the initial vertical velocity = 0
vf = 0 + 9.8(3.22)
vf = 31.56 m/s
Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
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A uniform plank of length 2.00 m and mass 30.23 kg is supported by three ropes. A 700 N person is a distance, of 0.55 m from the left end. A) find the magnitude of the tension in the vertical rope on the left end. Give your answers in newtons. B) find the magnitude of the tension in the rope in the right end. Give your answers in newtons C) find the magnitude of the tension in the horizontal rope on the left end. Give your answers in newtons
a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N
b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N
What is tension force?Tension force refers to a pulling force that is exerted by a string or cable about an axis.
a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:
Taking moment about the vertical axis
T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°
Solving for T₁ by taking the left end as the pivot;
T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0
T₁ * 1.285 = 681.5563
T₁ = 530.4 N
Therefore;
T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°
T₂ = 655.62 N
b. From calculation, the magnitude of the tension in the rope in the right end is T₁.
T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:
Taking moments about the left end in the horizontal direction;
T₃ = T₁ * cos 40°
T₃ = 530.4 N * cos 40°
T₃ = 406.3 N
In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.
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