Answer:
The correct answer is:
d) A and B have the same magnitude flux
Explanation:
Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.
The formula of electric flux is:
Φ = E A Cos θ
(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).
Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.
Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).
Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.
Explanation:The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.
Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.
The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?
Answer:
The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
Explanation:
Given that,
Boiling point = 81.0°C
Atmospheric pressure :
Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.
The value of atmospheric pressure is
[tex]P=101325\ Pa[/tex]
Vapor pressure :
Vapor pressure is equal to the atmospheric pressure.
Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
A hot air balloon uses the principle of buoyancy to create lift. By making the air inside the balloon less dense then the surrounding air, the balloon is able to lift objects many times its own weight. A large hot air balloon has a maximum balloon volume of 2280 m3 and the required lift is 2500 N (rough estimate of the weight of the equipment and passengers). Calculate the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is (exactly) 0◦ C and that air is an ideal gas under these conditions. What factor limit the maximum altitude attainable by this method for a given load?
Using the principle of buoyancy and the ideal gas law equation, we can calculate the temperature needed inside a hot air balloon to provide the required lift. The maximum altitude of the balloon is limited by air temperature, pressure, and other environmental factors.
Explanation:The principle of buoyancy used by hot air balloons to create lift is based on Archimedes' Principle, which states that the buoyant or lifting force exerted on a body immersed in a fluid equals the weight of the fluid the body displaces. In order for a hot air balloon to gain lift, the air inside the balloon must be heated to a temperature that makes it less dense than the surrounding cooler air. This difference in air density is what actually provides the uplift.
To calculate the required temperature inside the balloon, we can use the ideal gas law equation P₁V₁ / P₂V₂. Given the values in the question and the atmospheric pressure, we can derive the temperature inside the balloon required to provide the needed lift.
The maximum altitude attainable by a hot air balloon is limited by the ambient air temperature and pressure, which decrease as altitude increases. If the balloon's lift decreases (because it can no longer heat the air inside it sufficiently to provide lift), the balloon will stop ascending. Wind currents and weather conditions can also affect the maximum altitude.
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A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your answer to the nearest whole number. Use g ≈ 9.8 m/s2.)
The muzzle speed of the gun, when fired at an angle of elevation of 30° and reaching a maximum height of 544 m, is approximately 329 m/s.
Explanation:The physics concept here is projectile motion. The muzzle speed of the gun can be calculated using the equation for the maximum height attained by a projectile, which is given by H = (V^2 * sin^2θ) / 2g, where V represents the muzzle speed, θ is the angle of elevation, and g is the acceleration due to gravity. Rearranging for V, and substituting the given values, we get:
V = sqrt((2 * H * g) / sin^2θ) = sqrt((2 * 544 m * 9.8 m/s^2) / sin^2 30°). Since sin 30° = 0.5, this leads to V = sqrt((2 * 544 m * 9.8 m/s^2) / (0.5)^2). The resulting muzzle speed, when calculated and rounded to the nearest whole number, is 329 m/s.
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An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half the circle in 156 s, determine the following. HINT (a) Determine the magnitude of the airplane's displacement during the given time (in m). m (b) Determine the magnitude of the airplane's average velocity during the given time (in m/s). m/s (c) What is the airplane's average speed during the same time interval (in m/s)
The displacement of the airplane is 6760 meters, its average velocity is 43.33 m/s, and its average speed is 135.08 m/s.
Explanation:To solve this problem, we need to apply the principles of physics in kinematics and circular motion. First of all, let's understand that displacement is the shortest distance the airplane covered, which is the diameter of the circle. We multiply the radius by 2 (2*3.38 km) and convert it to meters to give 6760 meters for part (a).
Next, for average velocity, which is displacement over time, we divide 6760 m by 156 s, yielding approximately 43.33 m/s for part (b).
Lastly, for average speed, we need to consider the total distance travelled. In half a circle, this is pi times the diameter. Therefore, the average speed is (3.14 * 6760 m) / 156 = 135.08 m/s for part (c).
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The magnitude of the airplane's displacement is 6760 m. The magnitude of the average velocity is 43.33 m/s, and the average speed is 67.97 m/s.
Let's break down the problem step-by-step to find the required values.
(a) Magnitude of the Airplane's Displacement:
The airplane flies half of the circle, which means the path is a semicircle. The displacement is the straight-line distance between the start and end points of this path, which is the diameter of the circle.
Radius of the circle, [tex]r = 3.38 km = 3380 m[/tex]
Diameter, [tex]d = 2 * r = 2 * 3380 m = 6760 m[/tex]
Therefore, the magnitude of the displacement is 6760 m.
(b) Magnitude of the Average Velocity:
Average velocity is the displacement divided by the time.
Displacement = 6760 m
Time, [tex]t = 156 s[/tex]
Average velocity, [tex]v_{avg} = Displacement / Time = 6760 m / 156 s = 43.33 m/s[/tex]
Therefore, the magnitude of the average velocity is 43.33 m/s.
(c) Airplane's Average Speed:
The average speed is the total distance traveled along the path divided by the time.
The distance traveled in half a circle is the circumference of the semicircle.
[tex]Distance = (\pi * Diameter) / 2 = (\pi * 6760 m) / 2 = 10602.91 m[/tex]
[tex]Average speed = Distance / Time = 10602.91 m / 156 s = 67.97 m/s[/tex]
Therefore, the airplane's average speed is 67.97 m/s.
A 35.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.42 105 N · m2/C. What is the magnitude of the electric field? MN/C
To solve this problem we will apply the concept of Electric Flow, which is understood as the product between the Area and the electric field. For the data defined by the area, we will use the geometric measurement of the area in a circle (By the characteristics of the object) This area will be equivalent to,
[tex]\phi = 35 cm[/tex]
[tex]r = 17.5 cm = 0.175 m[/tex]
[tex]A = \pi r^2 = \pi (0.175)^2 = 0.09621m^2[/tex]
Applying the concept of electric flow we have to
[tex]\Phi = EA[/tex]
Replacing,
[tex]5.42*10^5N \cdot m^2/C = E (0.09621m^2)[/tex]
[tex]E = 5.6335*10^6N/C[/tex]
Therefore the magnitude of the electric field is [tex]5.6335*10^6N/C[/tex]
The total distance treaveled by a car moving in a straight line is as follows: After the first 7.0 minutes it has gone a total of 2.0 miles. After 14.0 minutes it has traveled a total of 4.5 miles. Finally at 21.0 minutes it has traveled a total of 6.0 miles. Find the average speed at: Time
Answer:
Average speed of the car will be 27.582 km/hr
Explanation:
We have given that in first 5 minutes distance traveled by car is 2 miles
After 14 minutes it has travel 4.5 miles
And finally after 21 minutes distance traveled by car is 6 miles
So total time of traveling t = 21 minutes
As we know that 1 hour = 60 minutes
So 21 minutes [tex]=\frac{21}{60}=0.35hour[/tex]
Total distance traveled = 6 miles
As 1 miles = 1.609 km
So 6 miles [tex]=6\times 1.609=9.654km[/tex]
Average speed is equal to the ratio of total distance and total time
So average speed [tex]v=\frac{9.654}{0.35}=27.582km/hr[/tex]
A baseball was hit and reached its maximum height in 3.00s. Find (a) its initial velocity; (b) the height it reaches.[29.4 m/s; 44.1m]
Answer:
u= 29.43 m/s
h=44.14 m
Explanation:
Given that
t= 3 s
We know that acceleration due to gravity ,g = 9.81 m/s² (Downward)
Initial velocity = u
Final velocity ,v= 0 (At maximum height)
We know v = u +a t
v=final velocity
u=initial velocity
a=Acceleration
Now by putting the values in the above equation
0 = u - 9.81 x 3
u= 29.43 m/s
The maximum height h is given as
v² = u ² - 2 g h
0² = 29.43 ² - 2 x 9.81 x h
[tex]h=\dfrac{29.43^2}{2\times 9.81}\ m[/tex]
h=44.14 m
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.09 m/s . Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.85 m/s . It takes you 14.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running?
Answer:
25.52 m
Explanation:
Relative speed between the person and I would be the difference of our speeds
[tex]v_r=2.85-1.09=1.76\ m/s[/tex]
Time taken by me to walk up to the person = 14.5 s
Distance is given by
[tex]Distance=Speed\times Time\\\Rightarrow s=vt\\\Rightarrow s=1.76\times 14.5\\\Rightarrow s=25.52\ m[/tex]
The person was 25.52 m ahead of me when I started running
An electron moving in the y direction at right angles to a magnetic field experiences a magnetic force in the x direction. The direction of the magnetic field is in the_______.
A. x direction
B. x direction
C. y direction
D. z direction
E. z direction
Answer:
The direction of the magnetic field is in the z-direction
Explanation:
Applying right hand rule, which states that when the thumb, index finger and the middle finger are held mutually at right angle to each other, with the index finger pointing in the direction of the moving charge (y- direction), the thumb pointing in the direction of the magnetic force (x-direction) pushing on the moving charge and then the middle finger pointing in the direction of the magnetic field (z-direction).
z direction.
Explanation:Using John Ambrose Fleming's right hand rule which asks to position the middle finger, the thumb and the index finger as follows;
| thumb
|
|
index |
/
/
/
/
middle
With this arrangement shown above, if you point your index finger in the direction in which the charge is moving, and then the middle finger in the direction of the magnetic field, then your thumb will point in the direction of the magnetic force.
For example;
If a charge is moving in the x direction, and the magnetic force is moving in the y direction, then the above figure is rotated to make sure that the index finger is pointing in the direction of the moving charge so as to get the direction of the magnetic field as follows;
y| thumb (magnetic force)
|
(direction of charge) |
index x |
/
/
/
/ z
middle (direction of field)
Therefore, the magnetic field is in the z direction.
Now to the question, if the charge(electron) is moving in the y direction at right angles to a magnetic field and it experiences a magnetic field in the x direction, the diagram above can be rotated to depict this situation as follows;
y| index (direction of charge)
|
|
| x
/ thumb (direction of force)
/
/
/ z
middle (direction of field)
Therefore the magnetic field will move in the z direction.
Note:
i. For every rotation of right-hand, the index finger must always point in the direction of the charge.
ii. Since the question does not specify which direction the electron is moving in (whether positive or negative), the direction of the magnetic field (whether positive or negative) might not be determined either. But in either case, the field will move in the z direction.
In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 10.1 oC. The temperature at the inside surface of the wall is 20.8 oC. The wall is 0.18 m thick and has an area of 8.9 m2. Assume that one kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar's worth of energy to be conducted through the wall
Answer:
17 hours
Explanation:
k = Thermal conductivity of concrete = 1.1 W/m°C
A = Area = [tex]8.9\ m^2[/tex]
l = Thickness = 0.18 m
[tex]\Delta T[/tex] = Change in temperature = 20.8-10.1
Power is given by
[tex]P=\dfrac{kA\Delta T}{L}\\\Rightarrow P=\dfrac{1.1\times 8.9\times (20.8-10.1)}{0.18}\\\Rightarrow P=581.961\ W[/tex]
Time required to produce 1 kWh
[tex]t=\dfrac{3600\times 10^3}{581.961}\\\Rightarrow t=6185.98153485\ s[/tex]
For one dollar
[tex]t=\dfrac{6185.98153485}{0.1}\\\Rightarrow t=61859.8153485\ s\\\Rightarrow t=\dfrac{61859.8153485}{60\times 60}\\\Rightarrow t=17.1832820412\ hours[/tex]
The time taken is 17 hours
You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10−19 C. Answer in units of µC.
Final answer:
Upon removing 12000 electrons from a neutral balloon, it will acquire a positive charge of 1.92 microcoulombs (1.92 µC). This is calculated by multiplying the number of electrons by the elemental charge and converting to the appropriate unit.
Explanation:
When 12000 electrons are removed from a neutral balloon, it obtains a positive charge because electrons carry a negative charge. To determine the charge the balloon now carries, we multiply the number of electrons removed by the elementary charge of an electron.
The charge (Q) on the balloon can be calculated using the formula:
Q = n × e
Where:
n is the number of removed electronse is the elementary charge per electron (1.6 × 10⁻¹⁹ C)Let's plug in the values into the formula:
Q = 12000 × 1.6 × 10⁻¹⁹ C
Q = 1.92 × 10⁻¹⁵ C
Converting coulombs (C) to microcoulombs (µC) by multiplying by 10¶ gives:
Q = 1.92 × 10⁻¹⁵ C × 10¶ µC/C
Q = 1.92 µC
The balloon will carry a charge of 1.92 µC after 12000 electrons have been removed.
If a dog has a mass of 20.1 kg, what is its mass in the following units? Use scientific notation in all of your answers.
Answer:
dog's mass in grams is [tex]20.1\times 10^3 grams[/tex]
dog's mass in milligrams is [tex]20.1\times 10^6 milligrams[/tex]
dog's mass in micrograms is[tex]20.1\times 10^9 micrograms[/tex]
Explanation:
dog has a mass of m= 20.1 kg
dog's mass in grams is given by [tex]20.1\times 1000 grams=20100 gms =20.1\times 10^3 grams[/tex]
dog's mass in milligrams is given by [tex]20.1\times 10^6 milli grams=20100000 milligrams= 20.1\times 10^6 milligrams[/tex]
dog's mass in micrograms is given by
[tex]20.1\times 10^9 micro grams=20100000000 micrograms= 20.1\times 10^9 micrigrams[/tex]
What causes a meteor shower?
Answer: Meteor showers occur when the earth in its orbit around the Sun passes through debris left over from the destruction of comets.
Explanation:A meteor is a particle broken off an asteroid or comet orbiting the Sun, it burns up as it enters the Earth's atmosphere, creating the effect called shooting star. Cosmic debris of meteor is known as meteoroids. These meteoroids, entering Earth's atmosphere, at extremely high speeds on parallel trajectories is an event known as meteor shower.
A good-quality measuring tape can be off by 0.42 cm over a distance of 28 m. What is its percent uncertainty? (Express your answer to the correct number of significant figures and proper units.)
Answer:
0.015%
Explanation:
Data provided in the question:
Length by which the measuring tape can be off, δL = 0.42 cm
Total measured length for which Error of δL is observed, L = 28 m
Now,
we know,
1 m = 100 cm
Thus,
28 m = 28 × 100 = 2800 cm
Percent uncertainty = [δL ÷ L] × 100%
= [0.42 ÷ 2800] × 100%
= 0.015%
One end of a string 5.02 m long is moved up and down with simple harmonic motion at a frequency of 61 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm
Answer:
Explanation:
One end of a string 5.02 m long is moved up and dowBDBBn with simple harmonic motion at a frequency of 61 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm
The wavelength of the waves on the string is calculated using the wave speed (found by distance/time) and the frequency. After performing the calculations, the wavelength is determined to be 16.46 cm.
To calculate the wavelength of the waves on the string, we can use the wave speed and the frequency. The speed of a wave ( is given by the formula speed = distance / time. Here, the distance is the length of the string, and the time is how long it takes for waves to reach the other end.
The speed of the wave is therefore 5.02 m / 0.5 s = 10.04 m/s. The frequency of the wave is given as 61 Hz. The wavelength ( can be found using the equation v = f, where v is the wave speed, f is the frequency, and is the wavelength.
By rearranging the equation to solve for the wavelength, we get
= v / f. Substituting the given values, we find
= 10.04 m/s / 61 Hz
= 0.1646 m. To convert the wavelength into centimeters, we multiply by 100, which gives us a wavelength of 16.46 cm.
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive the frequency as 1370 Hz. You are relieved that he is in pursuit of a different speeder when he continues past you, but now you perceive the frequency as 1330 Hz. What is the speed of the police car? The speed of sound in air is 343 m/s.
Answer:
Explanation:
Given
Apparent frequency [tex]f'=1370\ Hz[/tex]
Velocity of sound [tex]v=343\ m/s[/tex]
speed of observer [tex]v_o=35\ m/s[/tex]
Using Doppler effect Apparent frequency when source is approaching is given by
[tex]f'=f(\frac{v-v_0}{v-v_s})[/tex]
[tex]1370=f(\frac{343-35}{343-v_s})---1[/tex]
Apparent frequency when source moves away from observer
[tex]f''=f(\frac{v+v_0}{v+v_s})[/tex]
[tex]1330=f(\frac{343+35}{343+v_s})---2[/tex]
Divide 1 and 2 we get
[tex]\frac{f'}{f''}=\frac{\frac{343-35}{343-v_s}}{\frac{343+35}{343+v_s}}[/tex]
[tex]\frac{1370}{1330}=\frac{343-35}{343-v_s}\times \frac{343+v_s}{343+35}[/tex]
[tex]v_s=40\ m/s[/tex]
Thus speed of sound of police car is 40 m/s
The problem involves the Doppler effect and can be solved by applying the relevant formulas for when the source of sound is approaching and when it is receding. By setting up a system of two equations with the speed of the police car as the unknown, one can solve for it algebraically.
Explanation:This is a classic problem related to the Doppler effect, which describes the change in frequency of a wave in relation to an observer moving relative to the source of the wave. It requires us to know some principles about sound waves, including that their speed in the medium is constant, and their frequency and wavelength are inversely proportional.
To solve the problem, we need to apply the formula of the Doppler effect both when the police car is approaching and when it is moving away. The formula for the perceived frequency (f) when the source is moving towards the observer in a medium, like air, is given by f = f0*(v + vo) / (v - vs), where f0 is the emitted frequency, v is the speed of the medium (sound in this case), vo is the observer's speed, and vs is the source's speed.
When the source (police car) is moving away, the formula differs slightly and is given by f = f0 * (v - vo) / (v + vs). Here, using the frequencies when the car was approaching (1370 Hz) and moving away (1330 Hz), along with your speed (35.0 m/s) and the speed of sound (343 m/s), we have two equations with vs (the speed of the police car) as the unknown. This system of equations can be solved algebraically, yielding the speed of the police car.
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A mass m = 3.4 kg is at the end of a horizontal spring of spring constant k = 105 N/m on a frictionless horizontal surface. The block is pulled, stretching the spring a distance A = 6.5 cm from equilibrium, and released from rest.
(a) Write an equation for the angular frequency of the oscillation
(b) Calculate the angular frequency o of the oscillation in rad/seconds
Answer:
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. [tex]5.6rad/s[/tex]
Explanation:
a. from the spring-mass system which is explicitly describe by hooks law
from
F=-kx
which is in comparison to newtons general law of motion
F=ma
where the displacement x is expressed as
[tex]x=Asin(wt)\\[/tex]
and the acceleration is the second derivative of the displacement
[tex]a=-Aw^{2}sin(wt)\\[/tex]
hence final expression after substituting for the acceleration and the displacement is expressed as
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. for k=105N/m and m=3.4kg
we have the angular frequency to be
[tex]w=\sqrt{\frac{105}{3.4}}\\\\w=5.6rad/s[/tex]
Final answer:
The angular frequency (ω) can be found using the formula ω = sqrt(k/m). By plugging in the given values of mass (3.4 kg) and spring constant (105 N/m), the angular frequency is calculated to be approximately 5.56 rad/s.
Explanation:
The subject question pertains to simple harmonic motion (SHM) and specifically relates to the oscillatory motion of a mass attached to a horizontal spring. The question involves determining the angular frequency and calculating it in radians per second.
Part (a)
To write an equation for the angular frequency (ω) of the oscillation, we use the equation:
ω = sqrt(k/m)
where ω is the angular frequency, k is the spring constant, and m is the mass.
Part (b)
Plugging in the given values:
m = 3.4 kg
k = 105 N/m
We find ω using the formula:
ω = sqrt(105 N/m / 3.4 kg) = sqrt(30.8824 s-2) ≈ 5.56 rad/s
Therefore, the angular frequency of the oscillation is approximately 5.56 rad/seconds.
A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg point mass placed at the following distances from the center of the sphere: (a) 5.04 mm , and (b) 2.70 mm .
Answer:
[tex]0.0110284391534\ N[/tex]
[tex]0.0653784219002\ N[/tex]
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
[tex]m_1[/tex] = Mass of sphere = 2000 kg
[tex]m_2[/tex] = Mass of other sphere = 2.1 kg
r = Distance between spheres
Force of gravity is given by
[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N[/tex]
The gravitational force is [tex]0.0110284391534\ N[/tex]
[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N[/tex]
The gravitational force is [tex]0.0653784219002\ N[/tex]
A person is standing in an elevator that is moving downward and slowing down. Is the magnitude of the normal force on the person greater than, less than, or equal to the magnitude of the weight force on the person?
the magnitude of the normal force is greater than the magnitude of the weight of the person
Explanation:
As the elevator moves down, there are two forces acting on the person:
- The weight of the person, [tex]W=mg[/tex], where m is the mass of the person and g the acceleration due to gravity, acting downward
- The normal reaction exerted by the floor of the lift on the person, N, acting upward
This means that Newton's second law can be written as
[tex]\sum F = N-W= ma[/tex] (1)
where we chose upward as positive direction, and a is the acceleration of the elevator.
Here we know that the elevator is moving downward and it is slowing down: this means that the velocity is negative (upward), and the acceleration is in the opposite direction (upward), so the acceleration is positive.
Eq.(1) can be rewritten as
[tex]N=W+ma[/tex]
and as we said that [tex]a>0[/tex], this means that
[tex]N>W[/tex]
So the magnitude of the normal force is greater than the magnitude of the weight of the person.
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The magnitude of the normal force on a person in a downward-slowing elevator is greater than the magnitude of the weight force.
Explanation:When a person is standing in an elevator that is moving downward and slowing down, the magnitude of the normal force on the person is greater than the magnitude of the weight force on the person.
According to Newton's third law, the normal force and the weight force are equal in magnitude and opposite in direction. In this case, since the elevator is moving downward and slowing down, the acceleration of the person is in the upward direction. Therefore, the normal force exerted by the floor of the elevator on the person must be greater than the weight force in order to cause the upward acceleration.
It is important to note that if the elevator is in free-fall and accelerating downward at the acceleration due to gravity, the normal force becomes zero and the person appears to be weightless.
If the specimen is loaded until it is stressed to 65 ksi, determine the approximate amount of elastic recovery after it is unloaded. Express your answer as a length. Express your answer to three significant figures and include the appropriate units.
Answer:
ER = 0.008273 in
Explanation:
Given:
- Length of the specimen L = 2 in
- The diameter of specimen D = 0.5 in
- Specimen is loaded until it is stressed = 65 ksi
Find:
- Determine the approximate amount of elastic recovery after it is unloaded.
Solution:
- From diagram we can see the linear part of the curve we can determine the Elastic Modulus E as follows:
E = stress / strain
E = 44 / 0.0028
E = 15714.28 ksi
- Compute the Elastic strain for the loading condition:
strain = loaded stress / E
strain = 65 / 15714.28
strain = 0.0041364
- Compute elastic recovery:
ER = strain*L
ER = 0.0041364*2
ER = 0.008273 in
The approximate amount of elastic recovery after unloading a stressed specimen is zero.
Explanation:To determine the approximate amount of elastic recovery after unloading a stressed specimen, we need to consider the concept of elastic deformation. Elastic deformation refers to the temporary elongation or compression of a material when a stress is applied to it, and it returns to its original shape once the stress is removed.
Since the question does not provide specific information about the material or its elastic modulus, we cannot determine the exact amount of elastic recovery. However, we can generally say that the elastic recovery would be close to the original length of the specimen before it was loaded.
Therefore, we can assume that the approximate amount of elastic recovery would be zero, as the specimen would return to its original length.
A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.
Explanation:
It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.
Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17
(a) Therefore, we will calculate the mass as follows.
[tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]
= 8500 g
Now, formula to calculate weight of the system in N is as follows.
Weight = mass × g
= [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]
= 83.3 N (1 [tex]kg m/s^{2}[/tex] = 1 N)
Hence, the weight of the system is 83.3 N.
(b) Relation between specific volume and number of moles is as follows.
[tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]
Therefore, calculate the specific volume as follows.
[tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]
= 12 [tex]m^{3}/k mol[/tex]
Also,
[tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]
v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]
= 0.705882 [tex]m^{3}/kg[/tex]
Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex] and 0.705882 [tex]m^{3}/kg[/tex].
Answer:
a) [tex]w=83.385\ N[/tex]
b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Explanation:
Given:
no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]
volume of ammonia, [tex]V=6\ m^3[/tex]
We know the molecular formula of ammonia: [tex]NH_3[/tex]
The molecular mass of ammonia:
[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]
Now the mass of given ammonia:
[tex]m=n.M[/tex]
[tex]m=500\times 17[/tex]
[tex]m=8500\ g=8.5\ kg[/tex]
a)
Now weight:
[tex]w=m.g[/tex]
[tex]w=8.5\times 9.81[/tex]
[tex]w=83.385\ N[/tex]
b)
Specific volume:
[tex]\bar V=\frac{6}{0.5}[/tex]
[tex]\bar V=12\ m^3.kmol^{-1}[/tex]
also
[tex]\b V=\frac{V}{m}[/tex]
[tex]\b V=\frac{6}{8.5}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Calculate the energy of the quantum involved in the excitation of (i) an electronic oscillation of period 1.0 fs, (ii) a molecular vibration of period 10 fs, (iii) a pendulum of period 1.0 s. Express the results in joules and kilojoules per mole.
Answer:
a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Explanation:
E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period
a) Period = 1 fs = 1 × 10⁻¹⁵ s
f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s
f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) Period = 1s
f = 1/period = 1.0 Hz
E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Answer:
(i) E = 6.626 X 10⁻¹⁹ J = 400 KJ/mol
(ii) E = 6.626 X 10⁻²⁰ J = 40 KJ/mol
(iii) E = 6.626 X 10⁻³⁴ J = 4 X 10⁻¹³ KJ/mol
Explanation:
Energy associated with excitation of a quantum is given as;
E = hf
where;
E is the energy of excitation
h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹
f is the threshold frequency in s⁻¹
Thus, E = h/t
Part (i)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)
E = 6.626 X 10⁻¹⁹ J
In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol
Part (ii)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)
E = 6.626 X 10⁻²⁰ J
In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol
Part (iii)
E = (6.626 X 10⁻³⁴)/(1)
E = 6.626 X 10⁻³⁴ J
In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol
A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied to keep the block from sliding down the incline?
Answer:
Explanation:
Reaction force of inclined surface = mg cosθ
= Friction force acting in upward direction = μ x mg cosθ
If F be force required in upward direction to keep the block at rest on the plane
F + μ x mg cosθ = mg sinθ
F = mg sinθ - μ x mg cosθ
F = mg( sinθ - μ cosθ)
= 2 x 9.8 ( sin34 - 0.2 cos34 )
= 19.6 ( .559 - .1658)
= 7.7 N
This is the minimum force required .
The additional force that needs to be applied to keep a 2 kg block from sliding down a 34 degree incline, given a coefficient of static friction of 0.2, is 7.89 N.
Explanation:To find out how much additional force, F, needs to be applied to keep the 2 kg block from sliding down the 34° incline, we first need to calculate the gravitational force pulling the block down the incline. This force is given by F_gravity = m * g * sin(theta), where m = 2 kg, g = 9.8 m/s² (roughly the acceleration due to gravity on Earth), and theta = 34 degrees. Thus, F_gravity = 2 kg * 9.8 m/s² * sin(34°) = 11.10 N.
Next, we calculate the static frictional force that is preventing the block from sliding down the incline. This force is given by F_friction = m * g * cos(theta) * mu_s, where mu_s = 0.2 is the coefficient of static friction. Thus, F_friction = 2 kg * 9.8 m/s² * cos(34°) * 0.2 = 3.21 N.
The additional force, F, that must be applied to keep the block from sliding down the incline is the difference between the gravitational force and the frictional force. That's F = F_gravity - F_friction = 11.10 N - 3.21 N = 7.89 N.
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Reflected light from a thin film of oil gives constructive interference for light with a wavelength inside the film of λfilm. By how much would the film thickness need to be increased to give destructive interference?
A. 2λfilm
B. λfilm
C. λfilm/2
D. λfilm/4
The film thickness needs to be increased by C. λfilm/2 to achieve destructive interference.
Explanation:When light reflects off a thin film, it can undergo constructive or destructive interference depending on the thickness of the film and the wavelength of the light. For constructive interference, the path difference between the reflected rays should be an integer multiple of the wavelength, while for destructive interference, the path difference should be an odd multiple of half the wavelength.
Interference in thin films is a phenomenon where light waves reflect off both the top and bottom surfaces of a thin film, leading to constructive or destructive interference patterns, creating colors. Therefore, to achieve destructive interference, the film thickness needs to be increased by λfilm/2. Thus, the correct answer is option C.
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To switch from constructive to destructive interference, the film thickness must be increased by λ[tex]_f_i_l_m[/tex]/2.
In thin film interference, the thickness of the film and the wavelength of light are crucial in determining whether the interference is constructive or destructive. Constructive interference happens when the path length difference for the two reflected rays is an integral multiple of the wavelength inside the film (i.e., 2t = nλ[tex]_f_i_l_m[/tex], where n is an integer). Conversely, destructive interference occurs when this path length difference is a half-integral multiple of the wavelength (i.e., 2t = (2n + 1)λ[tex]_f_i_l_m[/tex] / 2).
Given that the film currently gives constructive interference for a wavelength of λ[tex]_f_i_l_m[/tex], the path length difference is an integral multiple of λ[tex]_f_i_l_m[/tex]. To switch to destructive interference, the thickness must be increased so the total path length difference becomes a half-integral multiple of λ[tex]_f_i_l_m[/tex]. This increase should be λ[tex]_f_i_l_m[/tex]/2.
The film thickness would need to be increased by λ[tex]_f_i_l_m[/tex]/2 to give destructive interference.
(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?
Answer
given,
height of the jump = 0.44 m
acceleration due to gravity, g = 9.8 m/s²
velocity at the height point = 0 m/s
initial speed = ?
Using equation of motion for speed calculation
v² = u² + 2 g h
0 = u² - 2 x 9.8 x 0.44
u = √8.624
u = 2.94 m/s
time taken to reach the highest point
v = u + g t
0 = 2.94 - 9.8 x t
t = 0.3 s
total time of flight will be equal to double of the time taken to reach the maximum height.
Total time = 2 x 0.3 = 0.6 s
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
A flea jumps straight up to a height (h) of 0.440 m. At the maximum height, the final speed (v) is zero. Given that the gravity (g) is 9.81 m/s², we can calculate the initial speed (u) using the following kinematic expression (We will assume y+ as the positive direction)
[tex]v^{2} = u^{2} + 2 \times g \times h \\\\(0m/s)^{2} = u^{2} + 2 \times (-9.81m/s^{2} ) \times 0.440m\\\\u = 2.94 m/s[/tex]
We can calculate the time (t) required to reach the maximum height using the following kinematic expression.
[tex]v = u + g \times t\\\\0m/s = 2.94m/s - 9.81m/s^{2} \times t\\\\t = 0.300 s[/tex]
The time in the air will be double the time to reach the maximum height.
[tex]t_{total} = 2 \times 0.300 s = 0.600 s[/tex]
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
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A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position. Refer to the figure. show answer No Attempt 25%
Part (a) Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive. Enter an expression for the normal force, FN, that the plane exerts on the block (in the y-direction) in terms of defined quantities and g. 25%
Part (b) Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
Part (c) Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m? 25%
Part (d) Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?
Answer:
Explanation:
given
spring constant k = 880 N/m
mass m = 150 kg
Normal force will be equal to the component of weight of mass m which is perpendicular to the inclined surface
= mgcosθ
So normal force
FN = mgcosθ j , as it acts in out of plane direction .
b )
Fricrion force acting in upward direction = μs mgcosθ
component of weight acting in downward direction = mgsinθ
restoring force by spring on block in downward direction
= kd
= 880d
F( total ) = (μs mgcosθ - mgsinθ - 880d )i
c )
for balance
mgsinθ = kd
sinθ = kd / mg
= 880 x .1 / 150 x 9.8
= 88 / 1470
.0598
θ = 3.4 degree
d )
d = mgsinθ / k
150 x 9.8 sin45 / 880
= 1.18 m
The problem involves the equations of force related to a block on an inclined plane with a spring. The solution involves using concepts of statics, the restoring force of a spring, the force of gravity on an inclined plane and trigonometric functions to derive the formulas and find the answers.
Explanation:Part (a) The normal force, FN, is the product of the gravitational motion and the cosine of the angle. Therefore, FN=mgcosθ.
Part (b) Just before the block begins to slide up the plane, the sum of the forces in the x-direction is equal to the difference between the restoring force of the spring and the force due to gravity along the plane. Therefore, ΣFx = k*d - mgsinθ.
Part (c) For a frictionless plane, the angle of the plane is found by observing that the force due to gravity must be equal to the restoring force of the spring, which gives θ = arcsin(k*d/(mg)). To get the angle in degrees for d = 0.1 m, you would plug in the values: θ = arcsin(880*0.1/(150*9.81)) in radians, and convert to degrees.
Part (d) If θ = 45 degrees and the surface is frictionless, the distance the spring will be compressed, d, can be found from the equation mg*sinθ = k*d, which gives d = m*g*sinθ/k. Substituting in these values, d = 150*9.81*sin(45)/880.
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A motorcycle has a velocity of 15 m/s, due south as it passes a carwith a velocity of 24 m/s, due north. What is the magnitude anddirection of the velocity of the motorcycle as seen by the driverof the car?
a. 9 m/s, north
b. 9 m/s, south
c. 15 m/s, north
d. 39 m/s, north
e. 39 m/s, south
Answer:
(b) 9m/s south
Explanation:
Case 1: A motorcycle has a velocity of 15 m/s, due south
Case 2: A car with a velocity of 24 m/s, due north.
Let the velocity of the car due south = Vs↓
Let the velocity of the car due north = Vn↑
Magnitude of the velocity of the motorcycle as seen by the driver of the car = V
V = Vn - Vs
= 24m/s - 15m/s = 9m/s↓
The magnitude and velocity of of the motorcycle as seen by the driver of the car = 9m/s south
The correct option is B
e. 39 m/s, south
Explanation:Let the velocity of the motorcycle be [tex]V_{M}[/tex]
Let the velocity of the car be [tex]V_{C}[/tex]
Let the velocity of the motorcycle relative to the car be = [tex]V_{MC}[/tex]
According to relativity of velocities in one dimension;
[tex]V_{MC}[/tex] = [tex]V_{M}[/tex] - [tex]V_{C}[/tex] --------------------------(i)
Now, take;
south to be negative (-ve)
north to be positive (+ve)
Therefore, we can say that;
[tex]V_{M}[/tex] = -15m/s [since the velocity is due south]
[tex]V_{C}[/tex] = +24m/s [since the velocity is due north]
Now, substitute the values of [tex]V_{M}[/tex] and [tex]V_{C}[/tex] into equation (i) as follows;
[tex]V_{MC}[/tex] = -15 - (+24)
[tex]V_{MC}[/tex] = -15 -24
[tex]V_{MC}[/tex] = -39 m/s
Remember that we have taken;
south to be negative (-ve)
north to be positive (+ve)
Since the result we got is negative, it means the speed is due south.
Therefore, the speed of the motorcycle as seen by the driver of the car is 3.9m/s, due south.
A 0.800kg block is attached to a spring with spring constant 16.0N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 34.0cm/s . What areA)The amplitude of the subsequent oscillations?B)The block's speed at the point where x= 0.250 A?
Answer:
(a) Amplitude=0.0760 m
(b) Speed=0.337 m/s
Explanation:
(a) For amplitude
We can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)mv^{2}+0=0+(1/2)kA^{2}\\ A^{2}=\frac{mv^{2}}{k} \\A=\sqrt{\frac{mv^{2}}{k}}\\ A=\sqrt{\frac{m}{k} }v\\ A=\sqrt{\frac{(0.800kg)}{16N/m} }(0.34m/s)\\A=0.0760m[/tex]
(b) For Speed
Again we can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)m(v_{i})^{2}+0=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\ (1/2)m(v_{i})^{2}=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\(1/2)m(v_{i})^{2}-(1/2)k(A/2)^{2}=(1/2)m(v_{f} )^{2}\\(1/2)[m(v_{i})^{2}-k(A/2)^{2}]=(1/2)m(v_{f} )^{2}\\(v_{f} )^{2}=1/m[m(v_{i})^{2}-k(A/2)^{2}]\\As\\x=0.250A\\(v_{f} )^{2}=(1/0.800kg)[0.800kg(0.34m/s)^{2}-(16N/m)(0.250(0.07602m)/2)^{2}\\(v_{f} )^{2}=0.1138\\ v_{f}=\sqrt{0.1138}\\ v_{f}=0.337m/s[/tex]
Final answer:
The amplitude and speed of the block at a specified position in SHM can be determined by using conservation of energy, equating the initial kinetic energy to the maximum potential energy at the amplitude, and calculating the speed via energy values at a given displacement from equilibrium.
Explanation:
Let's break down the problem step by step:
1. Amplitude of Subsequent Oscillations (A):
- When the block is hit with the hammer, it acquires an initial velocity of [tex]\(34.0 \, \text{cm/s}\)[/tex], which we'll convert to meters per second: [tex]\(v = 34.0 \, \text{cm/s} = 0.34 \, \text{m/s}\)[/tex].
- The mechanical energy of the system (block + spring) is conserved. At the maximum extension (amplitude) of the oscillation, the kinetic energy is zero.
- Therefore, the total mechanical energy at the maximum extension is equal to the potential energy stored in the spring:
[tex]\[ E = U = \frac{1}{2} k A^2 \][/tex]
- We can express the kinetic energy at the initial point as:
[tex]\[ K = \frac{1}{2} m v^2 \][/tex]
- Since the total mechanical energy is conserved, we have:
\[ E = K + U \]
[tex]\[ \frac{1}{2} k A^2 = \frac{1}{2} m v^2 \][/tex]
- Solving for the amplitude \(A\):
[tex]\[ A = \sqrt{\frac{m v^2}{k}} \][/tex]
Substituting the given values:
[tex]\[ A = \sqrt{\frac{0.800 \, \text{kg} \cdot (0.34 \, \text{m/s})^2}{16.0 \, \text{N/m}}} \][/tex]
Calculating:
[tex]\[ A \approx 0.34 \, \text{m} \][/tex]
Therefore, the amplitude of the subsequent oscillations is approximately 0.34 meters.
2. Block's Speed at [tex]\(x = 0.250A\)[/tex]:
- At any position \(x\), the mechanical energy \(E\) of the system is given by:
[tex]\[ E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \][/tex]
- At the maximum extension (amplitude), the kinetic energy is zero, so:
[tex]\[ E = U(x = A) = \frac{1}{2} k A^2 \][/tex]
- We can find the speed of the block at any position \(x\) using the amplitude \(A\):
[tex]\[ v = \sqrt{\frac{k}{m} (A^2 - x^2)} \][/tex]
Substituting the given value [tex]\(x = 0.250A\)[/tex]:
[tex]\[ v = \sqrt{\frac{16.0 \, \text{N/m}}{0.800 \, \text{kg}} \left(0.34^2 - (0.250 \cdot 0.34)^2\right)} \][/tex]
Calculating:
[tex]\[ v \approx 0.24 \, \text{m/s} \][/tex]
Therefore, the block's speed at the point where [tex]\(x = 0.250A\)[/tex] is approximately 0.24 meters per second
From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering 30 600 km. What was the magnitude of his average speed during that time period
Answer:
average speed = 0.146 m/s
Explanation:
given data
distance = 30600 km
time = January 26, 1977 to September 18, 1983
solution
we get here time that is January 26, 1977 to September 18, 1983
so it is = 6 year and 7 months and 22 days
and that is = 2422 days = 2.09 × [tex]10^{8}[/tex] seconds
and distance is = 30600 km = 30600 × 10³ m
so here average speed will be as
average speed = [tex]\frac{distance}{time}[/tex] ...........1
average speed = [tex]\frac{30600*10^3}{2.09*10^8}[/tex]
average speed = 0.146 m/s
A particle position in meters is given by the function x(t) = ct^4 + dt^2 + f. where c = 6m/s^4, d=8m/s^2, and f=-6m and t is in seconds.
Answer:
See explanation below.
Explanation:
For this case we have the following function:
[tex] x(t)= ct^4 +dt^2 +f[/tex]
Where [tex] c= 6 m/s^4 , d = 8m/s^2 , f=-6m [/tex]
If we replace those values we got:
[tex] x(t) = 6t^4 + 8t^2 -6[/tex]
If we want to find the position after t = 2.358 seconds for example we ust need to replace in the position function t = 2.358 and we got:
[tex] x(t=2.358) = 6(2.358)^4 + 8(2.358)^2 -6 \approx 224 m[/tex]
If we want to find the velocity we need to take the derivate of the position function and we got:
[tex] \frac{dx}{dt}=v(t) = 4ct^3 + 2dt[/tex]
[tex] v(t)= 24 t^3 + 16 t[/tex]
If we want to find the instantaneous velocity we just need to replace on v(t) a value for t
And the accelaration would be given by the second derivate of the position:
[tex] \frac{dv}{dt}= 72 t^2[/tex]
If we want to find the instantaneous acceleration we just need to replace on v(t) a value for t
Final answer:
Explaining the position function of a particle in meters at a given time using a specific function with provided values for constants.
Explanation:
Position Function: The position of a particle in meters at time 't' is described by the function x(t) = ct² + dt² + f, with given values for c, d, and f.
Explanation: To find the position of the particle using this function, substitute the values of c, d, f, and the specific time 't'. This will give you the position of the particle at that time.
Example: If c = 6m/s², d = 8m/s², f = -6m, and t = 2s, then you can plug in these values to find the position of the particle at t = 2s.