Complete Question
(10 points) Consider a single crystal of some hypothetical metal that has the FCC crystal structure, and its critical resolved shear stress is 3.42 MPa. The crystal is oriented such that a tensile stress is applied at an angle of 39.2 degrees to the slip plane. Slip can occur in two directions( 18.4 and 74.2 degrees to the tensile force).
(a) (5 points) Which direction, is slip favored?
b) (5 points) What is the yield strength of this crystal during this tensile test ?
Answer:
a
the slip would occur 18.4° to the tensile force
b
The yield strength is [tex]\sigma_y = 4.65\ MPa[/tex]
Explanation:
From the question we are told that
The Resolved shear force is [tex]\sigma = 3.2MPa[/tex]
The angle in which the tensile stress is applied is [tex]\O = 39.2^o[/tex]
The first direction of slip is [tex]\theta _ 1 =18.4^o[/tex]
The second direction of the slip is [tex]\theta_2 = 74.2^o[/tex]
Generally the condition for direction in which slip is likely to occur(
direction in which slip is favored ) is that [tex](cos( \O) cos (\theta) )[/tex]
Must be Maximum for that direction
Since [tex]Cos (\O)[/tex] is constant for both direction we would look at the the cos of the the angle for both direction
[tex]Cos (\theta_ 1) = Cos(18.4^o) =0.9488[/tex]
[tex]Cos (\theta_ 2) = Cos(74.2^o) =0.2722[/tex]
From this calculation we can see that the slip would occur 18.4° to the tensile force
Generally critical resolved shear stress is mathematically represented as
[tex]\sigma = \sigma_y * (cos(\O) cos(\theta_1))[/tex]
Where [tex]\sigma_y[/tex] is the yield strength
Making [tex]\sigma_y[/tex] the subject
[tex]\sigma_y = \frac{\sigma }{[cos (\O) cos(\theta_1)]}[/tex]
Substituting value
[tex]\sigma_y = \frac{3.4*10^{6}}{cos (39.2) (cos 18.4)}[/tex]
[tex]\sigma_y = 4.65\ MPa[/tex]
The two containers of water below are completely insulated so that no heat can be transferred in or out. The water in both containers started at room temperature (20 0C), and heat was transferred to both using heating coils until they reached the indicated final temperatures.
Which container had more heat transferred on it.?
Complete Question
The complete question is shown on the first uploaded image s
Answer:
The container that had more heat transfer is container B
Explanation:
From the question we are told that
The initial temperature of water i both container is [tex]\theta_i = 20^oC[/tex]
generally the rate of heat transfer is mathematically represented as
[tex]Q = mc\Delta T[/tex]
in the question we can see that the temperature is same for the both water in the container but their masses are the same and from the equation we can see that each parameter varies directly with the rate of heat transfer so it then mean that the water in container B would have more heat transfer because its mass is greater.
The container with more water had more heat transferred to it, as the heat required to change the temperature of a substance is proportional to its mass and specific heat capacity.
Explanation:The question addresses a scenario involving two water containers with different volumes heated to certain temperatures and asks which container had more heat transferred to it. This is a classic example in the study of calorimetry, where heat transfer results in temperature changes. To determine which container received more heat, the concept of heat capacity and the formula Q = mcΔT (where Q is heat energy, m is mass, c is specific heat capacity, and ΔT is the change in temperature) can be applied.
In this case, since both containers started at room temperature and reached their respective final temperatures, the amount of heat absorbed or lost is directly proportional to the mass (volume of water) and their specific heat capacity. Given that water's specific heat capacity is constant, the container with the larger volume of water would require a larger amount of heat to reach its final temperature.
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Two rather unequal cars are competing in a drag race. For simplicity, let's assume that the accelerations of the cars are constant. Car 1 has an acceleration of 3.3 m/s2, while Car 2 has an acceleration of 5.9 m/s2. Car 2 gets off to a late start, 0.5 seconds after Car 1.
1. At what time after Car 1 took off will it be overtaken by Car 2?
2. How far down the track will this happen?
3. How much faster will Car 2 be than Car 1 by that time?
Answer:
car1: a=3.1m/s^2 , car2: a=6.1m/s^2
(a) 1/2*3.1*t^2= 1/2*6.1*(t-0.9)^2
1.55t^2= 3.05(t^2-1.8t+0.81)= 3.05t^2-5.49t+2.4705
1.5t^2-5.49t+2.4705= 0
t= 3.13457 = 3.14[s] after.
(b) d= 1/2*3.1*3.13457^2= 15.23[m] approx.
(c) car1: v=at = 3.1*3.13457= 9.717m/s
car2: v=at = 6.1*(3.13457-0.9)= 13.631m/s
13.631-9.717= 3.914 = 3.91[m/s] faster than car1.
At the time after 3.14 seconds, car1 is overtaken by car2, this will happen after 15.23 meters and car2 is 3.914 m/s faster than car1.
What is Acceleration?In mechanics, acceleration is the measure of how quickly an object's velocity changes in relation to time. The quantity is a vector. An object's acceleration is determined by the direction of the net force exerted on it.
Acceleration is a vector quantity since it has a magnitude and a direction. A vector quantity is also velocity. Acceleration is defined as the change in velocity vector over a time interval divided by the time interval.
There are several types of acceleration :
Uniform AccelerationNon-Uniform AccelerationAverage Acceleration(a)
1/2 × 3.1× t²
⇒ 1/2 × 6.1 × (t-0.9)²
⇒1.55t²= 3.05 (t² - 1.8t + 0.81)
⇒ 3.05t² -5.49t + 2.4705
=1.5t² -5.49t +2.4705= 0
t= 3.14 seconds.
(b)
d= 1/2 × 3.1 × 3.14²
= 15.23 meters
(c)
car1: v=at
⇒3.1 × 3.14
= 9.717 m/s
car2:
v=at
⇒ 6.1 × (3.14 - 0.9)
= 13.631 m/s
13.631-9.717= 3.914 m/s
Hence, car2 is 3.914 m/s faster than car1.
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A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 1.5 T to 5.1 T in 2.0 s. If the coil has a total resistance of 0.20 Ω, what is the magnitude of the induced current?
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
When light of wavelength 236 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.99 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface?
(Use 1 eV = 1.602 ✕ 10−19 J, e = 1.602 ✕ 10−19 C, c = 2.998 ✕ 108 m/s, and h = 6.626 ✕ 10−34 J · s = 4.136 ✕ 10−15 eV · s as necessary.)
Answer:
Explanation:
Energy of photon = h c / λ , h is planks constant , c is velocity of light and λ is wave length
= 6.626 x 10⁻³⁴ x 2.998 x 10⁸ / 236 x 10⁻⁹
= .08417 x 10⁻¹⁷ J
Kinetic energy of photoelectron = 1.99 e V
= 1.99 x 1.602 x 10⁻¹⁹ J
= 3.18798 x 10⁻¹⁹
= .03188 x 10⁻¹⁷ J
Diff of energy = .08417 - .03188 x 10⁻¹⁷
= .05229 x 10⁻¹⁷ J
This will be the work function of the metal
If λ be the maximum wave-length required
h c / λ = .05229 x 10⁻¹⁷
6.626 x 10⁻³⁴ x 2.998 x 10⁸ / λ = .05229 x 10⁻¹⁷
λ = 6.626 x 10⁻³⁴ x 2.998 x 10⁸ / .05229 x 10⁻¹⁷
= 379.89 x 10⁻⁹ m
= 379.89 nm .
A person is pushing a fully loaded 21.60 kg wheelbarrow at constant velocity along a plank that makes an angle ????=20.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length L. What is the magnitude of the total force Fx the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is g=9.81 m/s2.
Answer:
Explanation:
The weight of the wheelbarrow will act downwards . Its component parallel to the plank will act downwards along the plank .
The value of component = mgsinθ
21.6 x 9.8 x sin20
= 72.4 N
The person shall have to apply force equal to this component so that the barrow moves with uniform velocity.
Force required = 72.4 N .
A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.20 kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down.
how high above the valley floor will the combined chunks go?
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
Overload refers to: A. Performing a weight-lifting exercise with the resistance (load) held overhead B. Using a demand (load) above the normal demand faced by a muscle C. The principle that strength will be best developed when the resistance (load) exceeds the individual's physical abilities (e.g., 8 reps of the 5-RM) D. The principle that a resistance (load) must be presented repeatedly in order to elicit any adaptations E. All of the answers are correct
Answer:
E. All of the answers are correct
Explanation:
Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.
In order to do this, the athlete's limit must be surpassed albeit gradually at first then picked up later over time.
Final answer:
Overload refers to using a demand (load) above the normal demand faced by a muscle, fundamental for muscle growth and strength increase. It involves progressively increasing the workload to stimulate muscle adaptation and improvements. Overload, progression, and specificity are key principles in effective training programs.
Explanation:
Overload refers to B. Using a demand (load) above the normal demand faced by a muscle. This is a fundamental concept in strength training and physical conditioning. Overload is the principle that for muscles to grow stronger or for endurance to increase, they must be challenged with a demand that is greater than what they are accustomed to. This concept involves progressively increasing the workload on the muscle to stimulate adaptation and improvements in muscle strength, endurance, and size.
The principle of overload is based on the body's ability to adapt to increased demands. When you perform exercises that are more challenging than your muscles are used to, your body adapts by making those muscles stronger, which in turn makes them capable of handling the increased load. This can be achieved through various methods such as increasing the weight lifted, the number of repetitions performed, or the intensity of the exercise.
It is also important to note that overload, along with progression and specificity, form the three foundational principles of training that guide the development of effective strength and conditioning programs. These principles ensure that exercises not only challenge the body but do so in a way that promotes optimal growth, strength, and performance enhancements over time.
Is it possible for an inverted pyramid to exist, like the one show here? Explain your answer.
The inverted pyramid of energy contradicts the unidirectional flow of energy in ecosystems, violating thermodynamic principles. Natural energy pyramids depict decreasing energy availability at higher trophic levels, reflecting ecological dynamics and energy conservation.
The concept of an inverted pyramid of energy contradicts the fundamental principles of energy flow in ecosystems. In ecological systems, energy is transferred through trophic levels in a unidirectional manner, following the second law of thermodynamics. Primary producers, such as plants, harness solar energy and convert it into chemical energy through photosynthesis. Herbivores then consume these plants, transferring some of the energy to the next trophic level. Carnivores, in turn, consume herbivores, and so on.
An inverted pyramid of energy would imply a scenario where the energy at higher trophic levels exceeds that at lower levels, contrary to the established flow. This goes against the principles of energy conservation and the inefficiencies inherent in energy transfer between trophic levels, where energy is lost as heat in each transfer.
The inverted pyramid concept is not observed in natural ecosystems due to the energy losses associated with metabolism, heat production, and other inefficiencies. The pyramid shape represents the decreasing energy availability at higher trophic levels, a pattern consistent with the laws of thermodynamics and ecological dynamics.
In summary, an inverted pyramid of energy is not feasible within ecological systems, as it contradicts established principles of energy transfer and conservation in ecosystems.
The question probable may be:
Is it possible to have an inverted pyramid of energy?? why or why not??
(1 point) Suppose a spring with spring constant 1 N/m is horizontal and has one end attached to a wall and the other end attached to a mass. You want to use the spring to weigh items. You put the spring into motion and find the frequency to be 1.4 Hz (cycles per second). What is the mass? Assume there is no friction.
Answer:
Explanation:
Given that,
The spring constant
K = 1N/m
Frequency of motion
f = 14Hz
We want calculate the mass m?
The frequency of spring system is related to the mass by
From w = √k/m
Where w = 2πf
f = 1/2π √k/m
Where,
w is angular frequency in rad/s
m is mass of object attached in kg
k is the spring constant in N/m
f Is the frequency in Hz
Then, make m subject of formula
Multiply both sides by 2π
2πf = √k/m
Square both sides
4π²f² = k/m
Then, k= 4π²f² × m
m = k / 4π²f²
m = 1 / (4π² × 1.4)
m = 1 / 55.27
m = 0.0181 kg
m= 18.1 g
The mass of the object attached in 18.1 g or 0.0181 kg
Final answer:
To find the mass attached to a spring with a known frequency of 1.4 Hz and a spring constant of 1 N/m, we use the formula for the frequency of a spring-mass system, resulting in a mass of approximately 0.081 kg.
Explanation:
The question asks about finding the mass attached to a spring when the frequency of motion is known, assuming no friction. The formula to determine the frequency of a spring-mass system is given by f = (1/2π)∙(k/m)^½, where f is the frequency in hertz, k is the spring constant in Newtons per meter, and m is the mass in kilograms. Given the frequency of 1.4 Hz and a spring constant of 1 N/m, we rearrange the formula to solve for m: m = k / (2πf)^². Substituting the given values, we calculate the mass as follows: m = 1 / (2π(1.4))^² ≈ 0.081 kg. Therefore, the mass attached to the spring is approximately 0.081 kilograms.
Sound travels through water at a speed of 1500 m/s. If the frequency of a sound is 1000 Hz, what is the wavelength?
Answer:
1.5m
Explanation:
Velocity=1500m/s
Frequency=1000hz
Wavelength =velocity ➗ frequency
wavelength =1500 ➗ 1000
Wavelength=1.5m
The wavelength of a sound, given a frequency of 1000 Hz and a speed of 1500 m/s, is 1.5 meters as determined by using the formula Wavelength = Speed / Frequency.
Explanation:In the field of Physics, the wavelength of a sound is calculated using the speed of sound and its frequency. The speed at which sound travels is given as 1500 m/s, and the frequency of sound as 1000 Hz. The formula used to calculate wavelength is Speed = Frequency x Wavelength. In this case, rearranging to find wavelength gives us, Wavelength = Speed Frequency. Therefore, the wavelength of the sound is 1500m/s / 1000Hz, which equals 1.5 meters.
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A thin block of soft wood with a mass of 0.0720 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 556 m/s at a block of wood and passes completely through it. The speed of the block is 18.0 m/s immediately after the bullet exits the block. (a) Determine the speed of the bullet as it exits the block.
Answer:
Explanation:
Let the mass of the thin block be M = .072 kg .
mass of bullet m = 4.67 x 10⁻³ kg.
speed of bullet v = 556 m/s.
speed of block after billet exits the block = 18 m/s.
We shall apply law of conservation of momentum
momentum of bullet block system
.072 x 0 + 4.67 x 10⁻³ x 556 = .072 x 18 + 4.67 x 10⁻³ x V here V is velocity of bullet after exit
= 2.59652 = 1.296 + 4.67 x 10⁻³ x V
4.67 x 10⁻³ x V = 1.30052
V = .278 x 10³ m /s
= 278 m /s
A capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 6.0 kΩ resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.46 mA . Part A What is the battery's emf?
Answer:
Battery voltage will be equal to 9.65 volt
Explanation:
We have given capacitance [tex]C=20\mu F=20\times 10^{-6}F[/tex]
Resistance [tex]R=6kohm=6000ohm[/tex]
Time constant of RC circuit is
[tex]\tau =RC=20\times 10^{-6}\times 6000=0.12sec[/tex]
Time is given t = 0.15 sec
Current i = 0.46 mA
Current in RC circuit is given by
[tex]i=\frac{V}{R}e^{\frac{-t}{\tau }}[/tex]
[tex]0.00046=\frac{V}{6000}e^{\frac{-0.15}{0.12 }}[/tex]
[tex]0.00046=\frac{V}{6000}\times 0.286[/tex]
V = 9.65 volt
So battery emf will be equal to 9.65 volt
To find the emf, use Ohm's Law and the charging formula for a capacitor. With the current, resistance, and capacitor values, calculate the voltage across the resistor. Then, use that to solve for emf with the formula V = emf(1 - e^-t/RC).
Explanation:To determine the electromotive force (emf) of the battery, we can use the relationship between current, resistance, and emf in the circuit containing a resistor and a capacitor. The relevant formula derived from the natural charging process of a capacitor through a resistor is:
V = emf(1 - e-t/RC) (charging)
Since the question provides the current (I) 0.15 seconds after the circuit is closed, we know that:
I = V/R, where V is the voltage across the resistor at time t, R is the resistance, and I is the current.
Let's solve for emf using the given information:
Identify the given values: R = 6.0 kΩ and C = 20 μF.Adjust units for computations: R = 6000 Ω and C = 20 x 10-6 F.Calculate the voltage (V) across the resistor using Ohm's law: V = I * R = 0.46 mA * 6000 Ω.Convert the current to amperes: I = 0.46 x 10-3 A.Substitute V and I into the equation to find emf: 0.46 x 10-3 A * 6000 Ω = emf(1 - e-0.15/RC).Calculate RC: RC = R * C = 6000 Ω * 20 x 10-6 F = 0.12 s.Substitute RC into the exponent: emf = V/(1 - e-0.15/0.12).After calculations, the emf can be found to be approximately the voltage drop across the resistor divided by the remaining fraction given by the exponential term. This result is the battery's emf.Learn more about Capacitor Charging Circuit here:https://brainly.com/question/31415514
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Which statement best explains why objects are pulled towards earth's center
Answer:
I didn't see any statement. But here is the answer:--
Explanation:
But objects are pulled towards earth's center by gravity. classical mechanics can not define the gravity but general relativity describe it as a curve of space-time by a concentrated energy or mass. As the mass of an object increases the more space time will be curved in the direction of positive energy so the center of the earth is the minimum point of the curved space-time by earth.That's why smaller objects than earth will fall to the center of the earth.
The statement best explains why objects are pulled toward Earth’s center is : Earth has a much greater mass than objects on its surface .Gravity is the attractive force acting between masses. This force is why objects fall to the ground when dropped.
This phenomenon is explained by gravity. As Newton's Law of Universal Gravitation says, every object with mass attracts every other object with mass. The Earth, having a much greater mass compared to objects on its surface, exerts a significant gravitational force pulling objects towards its center. This is why when you drop an object, it falls to the ground. The gravitational force of Earth keeps us grounded and is responsible for attracting objects towards its center.
Complete Question
Which statement best explains why objects are pulled toward Earth’s center?
Earth has a magnetic force that is strongest at its core.
Earth has a much greater mass than objects on its surface.
The weight of the atmosphere pushes objects toward Earth's surface.
The strength of the Sun’s gravity pushes down on objects at Earth's surface.
A rogue band of colonists on the moon declares war and prepares to use a catapult to launch large boulders at the earth. Assume that the boulders are launched from the point on the moon nearest the earth. For this problem you can ignore the rotation of the two bodies and the orbiting of the moon.1) What is the minimum speed with which a boulder must be launched to reach the earth? Hint: The minimum speed is not the escape speed. You need to analyze a three-body system.2) Ignoring air resistance, what is the impact speed on earth of a boulder launched at this minimum speed?
Answer:
2.3 km/s11 km/sExplanation:
The system is made up of three body system which are the boulder, earth and the moon also the sum of potential ad kinetic energies are assumed to be the same also note that the boulder was launched from the moon at an initial velocity
A ) Minimum speed of the boulder for it to reach the earth = 2.3 km/s
B) ignoring air resistance the impact speed on earth of a boulder launched at this minimum speed = 11 km/s
find attached the solution in details
A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a distance d1= 0.850 m to the left of sawhorse B. When the cat is a distance d2= 1.11 m to the right of sawhorse B, the plank just begins to tip.
If the cat has a mass of 2.9 kg, how far to the right of sawhorse B can it walk before the plank begins to tip?
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
You want to manufacture a guitar such that the instrument will be in tune when each of the strings are tightened to the same tension. The middle (D) string on the guitar should have fundamental frequency 146.83 Hz. The highest (E) string should have fundamental frequency 329.63 Hz. If the D string has linear mass density 0.00256kg/m, what should be the mass density of the E string
Answer:
0.000507 kg/m
Explanation:
L = Length of string
T = Tension
[tex]\mu[/tex] = Mass density of string
E denotes the E string
D denotes the D String
Frequency is given by
[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}[/tex]
So
[tex]f\propto \sqrt{\dfrac{1}{\mu}}[/tex]
[tex]\dfrac{f_D}{f_E}=\sqrt{\dfrac{\mu_E}{\mu_D}}\\\Rightarrow \mu_E=\dfrac{f_D^2}{f_E^2}\mu_D\\\Rightarrow \mu_E=\dfrac{146.83^2}{329.63^2}\times 0.00256\\\Rightarrow \mu_E=0.000507\ kg/m[/tex]
The mass density of the E string is 0.000507 kg/m
An airplane is ying in the direction 25 west of north at 800 km/h. (a) (4 points) Find the component form of the velocity of the airplane, assuming that the positive x-axis represents due east and the positive y-axis represents the north. (b) (8 points) What speed and direction should the wind be in order for the airplane to now y 500 km/h due east
Answer:
Explanation:
FIND THE SOLUTION BELOW
Answer:
a) v = (-338.1î + 725ĵ) km/h
b) Velocity of the wind = (-838.1î + 725ĵ) km/h
Magnitude of the wind = 1108.17 km/h
Direction = -40.9°; that is, 40.9° in the clockwise direction from the positive x-axis.
Explanation:
a) Airplane is flying 800 km/h in a direction 25° west of North.
Velocity of the plane = (vₓ, vᵧ)
vₓ = v cos θ
vᵧ = v sin θ
v = magnitude of the velocity = 800 km/h
θ = angle the velocity makes with the positive x-axis = 90° + 25° = 115°
vₓ = v cos θ
vₓ = 800 cos 115° = - 338.1 km/h
vᵧ = v sin θ
vᵧ = 800 sin 115° = 725 km/h
v = (-338.1î + 725ĵ) km/h
b) What speed and direction should the wind be in order for the airplane to now fly 500 km/h due east
Relative velocity of airplane with respect to the wind
= (velocity of the airplane) - (velocity of the wind)
Note that the velocities on the right hand side are with respect to earth's frame of reference.
Relative velocity of airplane with respect to the wind = 500 km/h east = (500î) km/h
velocity of the airplane = (-338.1î + 725ĵ) km/h
velocity of the wind = ?
500î = (-338.1î + 725ĵ) - (velocity of the wind)
(velocity of the wind) = (-338.1î + 725ĵ) - 500î
= (-838.1î + 725ĵ) km/h
Velocity of the wind = (-838.1î + 725ĵ) km/h
Magnitude = √[(-838.1)² + (725²)] = 1108.17 km/h
Direction = tan⁻¹ (725 ÷ -838.1) = -40.9°
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A chemist must dilute 58.00ml of 13.5 aqueous silver(II) oxide solution until the concentration falls to 5.00M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answer:
Explanation:
The given solution of silver oxide = 58 ml of 13.5 M of silver oxide
When we increase its volume , its molar concentration becomes less . To calculate the concentration of diluted solution , we can apply the following relation .
S₁ V₁ = S₂ V₂ .
S₁ is molar strength when volume is V₁ . S₂ is molar strength when volume is V₂
Puting the given values in the formula above ,
58 x 13.5 M = V₂ x 5
V₂ = 58 x 13.5 / 5
= 156.6 ml .
= 157 ml after rounding off.
If the entire apparatus were submerged in water, would the width of the central peak change? View Available Hint(s) If the entire apparatus were submerged in water, would the width of the central peak change? The width would increase. The width would decrease. The width would not change.
Answer:
The width would decrease
Explanation:
The width of central maximum of screen is proportional to wavelength of light and wavelength of light in water is less than that of air
If the apparatus were submerged in water, the width of the central peak as a part of the light diffraction pattern would decrease. This is because water slows down light more than air does - reducing the wavelength and therefore narrowing the central peak.
Explanation:If the entire apparatus were submerged in water, the width of the central peak would indeed change. This is due to the phenomenon of diffraction, which describes the way waves spread out when they pass through an opening. The central peak referred to in the question is a part of the diffraction pattern observed when light passes through a slit.
When diffraction occurs, a pattern of bright and dark spots, or 'fringes', is created. The width of these fringes is determined by the wavelength of the light and the size of the opening. If we were to submerge the entire apparatus in water, the diffraction pattern would change. This is because water has a higher refractive index than air, meaning it slows down the light more. As a result, the wavelength of the light in water becomes smaller compared to that in air. According to the formula for diffraction, the smaller the wavelength, the narrower the central peak. So in conclusion, if the apparatus were submerged in water, the width of the central peak would decrease.
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A proton is located at the origin and an electron is located at (1.0, 1.0) mm:
(a) Determine the electric dipole moment of these two particles in unit vector notation.
(b) If we put this electric dipole moment in an external electric field vector(E) = 300vector(i) (N/C), calculate the work done by the electric field to rotate the dipole so that it becomes in the same direction as the field.
Answer:
(a). [tex]{\vec{p} =(1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j})m \cdot C.}[/tex]
(b). [tex]U = 4.8*10^{-20}J.[/tex]
Explanation:
(a).
The electric dipole moment of the charges is
[tex]\vec{p} = q \vec{r}[/tex]
In our case
[tex]\vec{r} = (1.0*10^{-3}\bold{i}+1.0*10^{-3}\bold{j})m[/tex]
and
[tex]q =1.6*10^{-19}C[/tex];
therefore, the dipole moment is
[tex]\vec{p} =1.6*10^{-19}C *(1.0*10^{-3}\bold{i}+1.0*10^{-3}\bold{j})m[/tex]
[tex]\boxed{\vec{p} =(1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j})m \cdot C.}[/tex]
(b).
The work done [tex]U[/tex] by an external electric field [tex]\vec{E}[/tex] is
[tex]U = -\vec{p}\cdot \vec{E}[/tex]
[tex]U = [1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j}] \cdot[300\bold{i}][/tex]
[tex]\boxed{U = 4.8*10^{-20}J.}[/tex]
Formulate and write a computer program to determine the effects of pressure ratio, minimum/maximum temperature ratio, compressor efficiency, and turbine efficiency on overall cycle efficiency, net-work output per unit mass, and back-work ratio of an actual gas turbine (Brayton cycle).
Answer:
A gas turbine, also called a combustion turbine, is a type of continuous and internal combustion engine. The main elements common to all gas turbine engines are:
an upstream rotating gas compressor
a combustor
a downstream turbine on the same shaft as the compressor.
A fourth component is often used to increase efficiency (on turboprops and turbofans), to convert power into mechanical or electric form (on turboshafts and electric generators), or to achieve greater thrust-to-weight ratio (on afterburning engines).
The basic operation of the gas turbine is a Brayton cycle with air as the working fluid. Atmospheric air flows through the compressor that brings it to higher pressure. Energy is then added by spraying fuel into the air and igniting it so the combustion generates a high-temperature flow. This high-temperature high-pressure gas enters a turbine, where it expands down to the exhaust pressure, producing a shaft work output in the process. The turbine shaft work is used to drive the compressor; the energy that is not used for compressing the working fluid comes out in the exhaust gases that can be used to do external work, such as directly producing thrust in a turbojet engine, or rotating a second, independent turbine (known as a power turbine) which can be connected to a fan, propeller, or electrical generator. The purpose of the gas turbine determines the design so that the most desirable split of energy between the thrust and the shaft work is achieved. The fourth step of the Brayton cycle (cooling of the working fluid) is omitted, as gas turbines are open systems that do not use the same air again.
Gas turbines are used to power aircraft, trains, ships, electrical generators, pumps, gas compressors, and tanks.[1]
Explanation: As applied above.
Two plane mirrors are facing each other. They are parallel, 6 cm apart, and 24 cm in length, as the drawing indicates. A laser beam is directed at the top mirror from the left edge of the bottom mirror. What is the smallest angle of incidence with respect to the top mirror, such that the laser beam hits the mirrors in each way
Answer: 33.69 degrees
Explanation:
Given
Distance between the two mirrors, = 6 cm
Length of the two mirrors, = 24 cm
Assuming the laser is shone from just clear of the left edge of the bottom mirror,
a)
Tan L = (O/A)
Tan L = (6 / 12)
Tan L = 0.5
L = Tan^-1 (0.5)
L = 26.57 degrees
(90 - 26.57) = 63.43 deg. to the normal.
Tan L = (6 / [24/6])
Tan L = (6 / 4)
Tan L = 1.5
L = Tan^-1 (1.5)
L = 56.31 degrees
(90 - 56.31) = 33.69 deg. to the normal.
Thus, the angle at which, with respect to the mirror, the laser beam hits the mirrors in each way is 33.69 degrees
Answer:
86degrrss
Explanation:
tan L =BC/AB
= 3/12 = 1/4
L= tan^-1 0.25
=4.0 degrees
= 90-4= 86degrees
A block of mass m attached to the end of a spring of spring constant k undergoes simple harmonic motion with amplitude A and angular frequency ω. The position of the block is described by a cosine function with an initial phase angle ϕ = 0. Which of the variables m, k, A, or ϕ should you increase if you want to increase the frequency of oscillation?
Answer:
To increase the frequency the spring constant k should be increased.
Explanation:
In simple harmonic motion, the frequency is expressed as;
f = (1/2π)√(k/m)
where;
k is spring constant
m is mass of the object
From the frequency formula i wrote, It’s obvious that frequency depends on only the spring constant and mass while Amplitude(A) and phase angle(ϕ) are not related and would have no effect on the frequency. Thus, by inspection, an increase in spring constant increases frequency while an increase in mass decreases frequency
CI. Design challenge: Your goal is to build a compound microscope than can at least double the size of the object. We will constrain ourselves to values that will work later in a simulation, which has the unrealistic- for microscopes - focal lengths in meters. (Imagine instead that those values are in cm) You need to choose two converging lenses. You have available lenses with focal lengths of 2, 4, 8 and 12 m. First choose an objective lens focal length, and a distance to place the object from the lens. Then choose an eyepiece lens focal length and a distance to place the lens from the objective lens. C2. Draw your microscope design including a ray diagram. Your diagram should include: 1. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the objective. 2. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the eyepiece. C3. Calculate the magnification of your microscope
Answer:
2. 17.7cm
3. -7 that is magnification
Explanation:
See attached handwritten document
Two speakers are placed 500 meters away from each other. A person stands right in the middle of the two speakers. One of the speakers sends a sound wave traveling at 343m/s towards the person standing in the middle and the other speaker sends a sound wave traveling at 342m/s. How long after the person in the middle hears the first sound wave will he hear the second wave ( ) (please answer the question with unit of seconds)
Answer:
the second wave will be heard at [tex]2*10^{-3} \ sec[/tex] after the first sound wave.
Explanation:
Given that:
two speakers are placed 500 meters away from each other and a person is standing in the middle;
that implies that the distance of the speed of the wave travelling is divided into two equal which is 250 m
Now; from the first speaker, time to reach sound wave is;
[tex]t_1 = \frac{250}{v_1}[/tex]
[tex]t_1 = \frac{250}{343}[/tex]
[tex]t_1 = 0.728 \ sec[/tex]
Also from the second speaker; time to reach sound wave is;
[tex]t_2 = \frac{250}{v_2}[/tex]
[tex]t_2 = \frac{250}{342}[/tex]
[tex]t_2 =0.730 \ sec[/tex]
[tex]t_2 - t_1 = (0.730 - 0.728 )\ sec[/tex]
[tex]= 0 .002 \ sec\\\\= 2*10^{-3} \ sec[/tex]
Thus, the second wave will be heard at [tex]2*10^{-3} \ sec[/tex] after the first sound wave.
Consider a ball in projectile motion under gravitational force so that it is has motion in both the vertical and horizontal directions. The effects of air friction may be ignored. When the ball reaches the highest point in its trajectory, what can be said about the magnitudes of the ball's velocity and acceleration vectors? Select the correct answer
a. The horizontal components of both the velocity and acceleration are zero.
b. The magnitude of velocity is zero, but the acceleration is a constant
c. The magnitudes of both velocity and acceleration are zero
d. The magnitude of acceleration is zero but the ball has a positive velocity
e. The magnitude of velocity is at its minimum nonzero value, but the magnitude of acceleration is a constant
At the highest point in the ball's trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant.
Explanation:When the ball reaches the highest point in its trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant. Since the ball is at the highest point, it momentarily comes to rest in the vertical direction, resulting in its velocity magnitude being zero in that direction. However, the ball still has a horizontal velocity component, which is constant throughout its motion.
An Arrow (0.5 kg) travels with velocity 60 m/s to the right when it pierces an apple (1 kg) which is initially at rest. After the collision, the arrow and the apple are stuck together. Assume that no external forces are present and therefore the momentum for the system is conserved. What is the final velocity (in m/s) of apple and arrow after the collision
Answer:
Velocity after collision will be 20 m/sec
Explanation:
We have given mass of arrow [tex]m_1=0.5kg[/tex]
Mass of arrow [tex]v_1=60m/sec[/tex]
Mass of apple [tex]m_2=1kg[/tex]
Apple is at rest so velocity of apple [tex]v_2=0m/sec[/tex]
According to conservation of momentum
Momentum before collision is equal to momentum after collision
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.5\times 60+1\times 0=(0.5+1)v[/tex]
[tex]1.5v=30[/tex]
v = 20 m/sec
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The cylinder has a net linear charge density 2λ. Assume λ is positive. Part A Find expressions for the magnitude of the electric field strength inside the cylinder, r
Answer:
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
Explanation:
We are given that
Linear charge density of wire=[tex]\lambda[/tex]
Radius of hollow cylinder=R
Net linear charge density of cylinder=[tex]2\lambda[/tex]
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem
[tex]\oint E.dS=\frac{q}{\epsilon_0}[/tex]
[tex]q=\lambda L[/tex]
[tex]E(2\pi rL)=\frac{L\lambda}{\epsilon_0}[/tex]
Where surface area of cylinder=[tex]2\pi rL[/tex]
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
What happens to a path of a light ray parallel to the principal axis, after it passes through a converging
lens?
Answer: The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. ... All three rays should intersect at exactly the same point.
Explanation: Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.
When a light ray parallel to the principal axis passes through a converging lens, it is bent towards the principal axis and converges to a point called the focal point.
Explanation:When a light ray parallel to the principal axis passes through a converging lens, it is bent towards the principal axis and converges to a point. This point is known as the focal point of the lens. The path of the light ray after passing through the lens depends on the distance of the object from the lens and the focal length of the lens.
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A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to the distance between interference fringes? Assume the angles involved remain small. A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to the distance between interference fringes? Assume the angles involved remain small. The distance between interference fringes remains the same. The distance between interference fringes also decreases. The distance between interference fringes increases. The effect cannot be determined unless the distance between the slits and the screen is known.
Answer:
Option C. The distance between interference fringes increases.
Explanation:
The distance between interference fringes for small angles can be given by the formula:
[tex]y = \frac{\lambda D}{d}[/tex]..........(1)
Where D = Distance between the slits and the screen
[tex]\lambda =[/tex] the wavelength of light
d = separation between the two slits
from the formula given in equation (1)
[tex]y \alpha 1/d[/tex]
It is obvious from the relationship above that the distance between interference fringes is inversely proportional to the separation between the slits.
Therefore, if the separation between slits is increased, the distance between interference fringes is increased.