Answer:
a)(2×40+2×80)/(2+2)=240/4=60
b)(240)/(120/40+120/80)=240/(3+1.5)
240/4.5=53.33333
no
A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat was later found is B = 30.0 km, 15.0° north of east.
How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector C?
A)65.3 km
B)39.7 km
C)26.5 km
D)54.0 km
E)42.5 km
Answer:
d = 39.7 km
Explanation:
initial position of the boat is 45 km away at an angle of 15 degree East of North
so we will have
[tex]r_1 = 45 sin15 \hat i + 45 cos15 \hat j[/tex]
[tex]r_1 = 11.64 \hat i + 43.46\hat j[/tex]
after some time the final position of the boat is found at 30 km at 15 Degree North of East
so we have
[tex]r_2 = 30 cos15\hat i + 30 sin15 \hat j[/tex]
[tex]r_2 = 28.98\hat i + 7.76 \hat j[/tex]
now the displacement of the boat is given as
[tex]d = r_2 - r_1[/tex]
[tex]d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)[/tex]
[tex]d = 17.34 \hat i - 35.7 \hat j[/tex]
so the magnitude is given as
[tex]d = \sqrt{17.34^2 + 35.7^2}[/tex]
[tex]d = 39.7 km[/tex]
An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:
2SO2(g) + O2(g) ⇌ 2SO3(g)
(a) At equilibrium, PSO3 = 300. atm and PO2 = 100. atm. Calculate PSO2.
Answer: [tex]p_{SO_2}=0.017atm[/tex]
Explanation:
We are given:
[tex]K_c=1.7\times 10^8[/tex]
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
Where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?
[tex]K_c[/tex] = equilibrium constant in terms of concentration
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]600K[/tex]
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-3=-1[/tex]
Putting values in above equation, we get:
[tex]K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6[/tex]
The chemical reaction follows the equation:
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
The expression for [tex]K_p[/tex] for the given reaction follows:
[tex]K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}[/tex]
We are given:
[tex]p_{SO_3}=300atm[/tex] [tex]p_{O_2}=100atm[/tex]
Putting values in above equation, we get:
[tex]3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}[/tex]
[tex]p_{SO_2}=0.017atm[/tex]
Hence, the partial pressure of the [tex]SO_2[/tex] at equilibrium is 0.017 atm.
Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a velocity of 40 km/h in a direction 38° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B?
Answer:
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr
Explanation:
Given data:
A position is 4km north and 2.5 km east to B
Ship A velocity = 22 km/hr
ship B velocity = 40 km/hr
A velocity wrt to velocity of B
[tex]\vec{V_{AB}} =\vec{V_A} - \vec{V_B}[/tex]
[tex]\vec{V_A} = 22 km/hr[/tex]
[tex]\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}[/tex]
[tex]= 31.52\hat{i} + 24.62 \hat{j}[/tex]
putting respective value to get velocity of A with respect to B
[tex]\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})[/tex]
[tex]\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}[/tex]
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr
A wave on the ocean surface with wavelength 44 m travels east at a speed of 18 m/s relative to the ocean floor. If, on this stretch of ocean, a powerboat is moving at 14 m/s (relative to the ocean floor), how often does the boat encounter a wave crest, if the boat is traveling (a) west, and (b) east?
Answer:
A)t=1.375s
B)t=11s
Explanation:
for this problem we will assume that the east is positive while the west is negative, what we must do is find the relative speed between the wave and the powerboat, and then with the distance find the time for each case
ecuations
V=Vw-Vp (1)
V= relative speed
Vw= speed of wave
Vp=Speesd
t=X/V(2)
t=time
x=distance=44m
A) the powerboat moves to west
V=18-(-14)=32m/s
t=44/32=1.375s
B)the powerboat moves to east
V=18-14=4
t=44/4=11s
Final answer:
The frequency at which the boat encounters a wave crest is 0.09 Hz regardless of the direction the boat is traveling.
Explanation:
To find the frequency at which the boat encounters a wave crest, we need to determine the time it takes for one wave crest to pass by the boat. The speed of the boat relative to the ocean floor doesn't affect the frequency, only the speed of the wave. The formula to calculate the frequency of a wave is:
Frequency = Wave Speed / Wavelength
When the boat is traveling west:Wave speed = Speed of the wave + Speed of the boat (since the boat is traveling in the opposite direction)
Frequency = (Speed of the wave + Speed of the boat) / Wavelength
Frequency = (18 m/s + (-14 m/s)) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz
Wave speed = Speed of the wave - Speed of the boat (since the boat is traveling in the same direction)
Frequency = (Speed of the wave - Speed of the boat) / Wavelength
Frequency = (18 m/s - 14 m/s) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz
The decibel is the unit of Sound Level, which is a convenient way of representing the large range of sound intensities that the ear can detect. From the definition of the decibel we can see that increasing the sound intensity by a factor of ten results in:
Answer:
Increase of ten yields 10 dB increase
Explanation:
The decibel is a logarithmic scale that shortens a large range of numbers into smaller and more manageable numbers. The decibel is found using:
[tex]dB = 10log(r )[/tex]
Where r represents a ratio of a measured magnitude (pressure or intensity level in sound) to a reference pressure or intensity value. However, in our problem we only need to represent an increase factor of ten:
[tex]dB = 10log(\frac{10}{1} )\\\\dB = 10*1=10 dB[/tex]
A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hours. What is the plane's speed in still air, and how fast is the wind blowing?
Answer:
airplane speed 135mph windspeed 45 mph
Explanation:
This information helps us to write down a system of linear equations
When going head wind, the speed of the wind is substracted from that of the airplane and on the return trip it is added, then:
A:=Airlplane speed
W:= Wind speed
(A+W)*1h=180mi (1)
(A-W)*2h=180mi (2)
then from (1) A=180-W (3), replacing this in (2) we get (180-W-W)*2h =180mi, then
360-4W=180, or 180=4W, then W=45 mph. Replacing this in (3) we have that A=180-45=135 mph.
The problem was solved by setting up equations based on the given distances and times. By solving these equations, it was determined that the plane's speed in still air is 105 miles per hour, and the wind's speed is 15 miles per hour.
To solve this problem, we can set up two equations using the relationship between distance, speed, and time. We will denote the plane's speed in still air as P and the wind's speed as W.
When flying against the wind, the plane's effective speed is P - W, and the time taken to cover the 180 miles is 2 hours. We can represent this with the equation:
180 = 2(P - W) ... (1)
On the return trip, with a tailwind, the plane's effective speed is P + W, and the time taken is 1.5 hours. This gives us a
second equation:
180 = 1.5(P + W) ... (2)
By solving these two equations simultaneously, we can find the values of P and W. Multiplying equation (2) by 2 to eliminate the fractions, we get:
360 = 3(P + W) ... (2')
Now, we subtract equation (1) from equation (2'):
360 - 180 = 3(P + W) - 2(P - W)
180 = 3P + 3W - 2P + 2W
180 = P + 5W
Using equation (1), we express P in terms of W:
180 = 2P - 2W
P = 90 + W ... (3)
Substituting (3) into the equation we got after subtracting:
180 = (90 + W) + 5W
180 = 90 + 6W
90 = 6W
W = 15 miles per hour
Now, we substitute the value of W back into equation (3) to find P:
P = 90 + 15
P = 105 miles per hour
The plane's speed in still air is 105 miles per hour and the wind's speed is 15 miles per hour.
A pinball bounces around its machine before resting between two bumpers. Before the ball came to rest, its displacement (all angle measuresup from x axis) was recorded by a series of vectors.83 cm at 90 degrees59cm at 147 degrees69cm at 221 degrees45cm at 283 degrees69cm at 27 degreesWhat is the magnitude and the direction?
Answer:
You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.
Mathematically:
x = [83cos90+55cos141+69cos229+41cos281+61co... cm = -27 cm
y = [83sin90+55sin141+69sin229+41sin281+61si... cm = 55 cm
so d = √(x² + y²) = 61 cm
and Θ = arctan(55/-27) = -64º +180º (to get into QII) = 116º
Explanation:
To calculate the pinball's total displacement, you sum up the x and y components of each vector separately and find the resultant vector's magnitude and direction, which is approximately 96.60 cm at an angle of 88.7 degrees from the horizontal.
Explanation:To find the total displacement of the pinball, we need to add the vectors given by their magnitude and direction. We first break down each vector into its horizontal (x) and vertical (y) components using trigonometric functions:
The first vector is 83 cm at 90 degrees, which gives us 0 cm in the x-direction and 83 cm in the y-direction.The second vector is 59 cm at 147 degrees, resulting in -50.47 cm x and 44.32 cm y.The third vector is 69 cm at 221 degrees, resulting in -51.55 cm x and -51.55 cm y.The fourth vector is 45 cm at 283 degrees, resulting in 42.43 cm x and -10.79 cm y.The final vector is 69 cm at 27 degrees, resulting in 61.56 cm x and 31.15 cm y.To find the resultant vector, sum the x and y components separately:
Sum of x-components: 0 - 50.47 - 51.55 + 42.43 + 61.56 = 1.97 cmSum of y-components: 83 + 44.32 - 51.55 - 10.79 + 31.15 = 96.13 cmThe magnitude of the resultant displacement vector (S) is calculated using the Pythagorean theorem:
S = √(x² + y²) = √(1.97² + 96.13²) = √(9330.56) = 96.60 cm approximately
The angle made with the horizontal (θ) is found using the tangent function:
θ = arctan(y/x) = arctan(96.13/1.97) ≈ 88.7 degrees
The distance between two adjacent peaks on a wave is called the wavelength. The wavelength of a beam of ultraviolet light is 113 nanometers (nm). What is its wavelength in meters?
Answer:
0.000000113 or 1.13*[tex]10^{-7}[/tex] meters
Explanation:
One nanometer is [tex]10^{-9}[/tex] meters. So 113 nanometers would be 113*[tex]10^{-9}[/tex], or 1.13*[tex]10^{-7}[/tex] meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.
a vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.
Answer:
209 m
Explanation:
The y-component of a vector is the magnitude times the sine of the angle.
y = 253 sin 55.8°
y = 209
Answer:
The answer to your question is: 209 m
Explanation:
Data
length = 253 m
d = 55.8°
y - component = ?
Formula
To solve this problem we need to use a right triangle a the trigonometric functions (sine, cosine, tangent, etc)
Now we have to choose among the trigonometric functions which one to use that relates the opposite side and the hypotenuse.
And that one is sineФ = os/h
we clear os = sineФ x h
os = sine 55.8 x253
os = 209 m
soh, cah, toa
The densities of cardboard, aluminum, and lead are 0.6 g/cm^3, 2.7 g/cm^3, and 11.4 g/cm^3, respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 49 cm of cardboard, or 42 cm of aluminum, or 17 cm of lead to stop half of the Heibrinkions emitted from a source. Calculate the absorber thickness for each material.
__________ g/cm^2 for cardboard
__________ g/cm^2 for aluminum
__________ g/cm^2 for lead
Answer:
For cardboard = 29.4 g/cm²
For aluminium = 113.4 g/cm²
For lead = 193.8 g/cm²
Explanation:
Given:
Density of the cardboard, d₁ = 0.6 g/cm³
Density of the aluminium, d₂ = 2.7 g/cm³
Density of the lead, d₃ = 11.4 g/cm³
Length of the cardboard, L₁ = 49 cm
Length of the aluminium, L₂ = 42 cm
Length of the lead, L₃ = 17 cm
Now,
The absorber thickness is calculated as:
= Density × Length
therefore,
For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²
For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²
For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²
The weight of a body above sea level varies inversely with the square of the distance from the center of Earth. If a woman weighs 131 pounds when she is at sea level, 3960 miles from the center of Earth, how much will she weigh when she is at the top of a mountain, 4.8 miles above sea level?
Answer:
89.16pounds
Explanation:
The equation that defines this problem is as follows
W=k/X^2
where
W=Weight
K= proportionality constant
X=distance from the center of Earth
first we must find the constant of proportionality, with the first part of the problem
k=WX^2=131x3960^2=2054289600pounds x miles^2
then we use the equation to calculate the woman's weight with the new distance
W=2054289600/(4800)^2=89.16pounds
The height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula h = −16t2 + 136t, where h is the height in feet and t is the time in seconds. Find the time required for the object to return to its point of departure.
Explanation:
Given that, the height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula :
[tex]h=-16t^2+136t[/tex]
Where
h is the height in feet
t is the time in seconds
Let t is the time required for the object to return to its point of departure. At this point, h = 0
[tex]-16t^2+136t=0[/tex]
t = 8.5 seconds
So, the time required for the object to return to its point of departure is 8.5 seconds.
A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the instant it touches the runway, what is the minimum time needed before it can come to rest?
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a
An air bubble has a volume of 1.1 cm3 when it is released by a submarine 110 m below the surface of a freshwater lake. What is the volume of the bubble (in cm3) when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent. (The density of water is 1,000 kg/m3.)
Answer:
[tex]V = 12.85 cm^3[/tex]
Explanation:
As we know that initially the air bubble is at depth 110 m
so the pressure of the air bubble is given as
[tex]P = P_o + \rho gh[/tex]
[tex]P = 1.01 \times 10^5 + (1000)(9.81)(110)[/tex]
[tex]P = 1.18 \times 10^6 Pa[/tex]
initial volume of the bubble is given as
[tex]V = 1.1 cm^3[/tex]
now we know that here temperature of air bubble is constant
so we have
[tex]P_1 V_1 = P_2 V_2[/tex]
[tex](1.18 \times 10^6)(1.1 cm^3) = (1.01 \times 10^5) V[/tex]
[tex]V = 12.85 cm^3[/tex]
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.40 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.80 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance. What is the maximum height above ground reached by the helicopter?
Answer:
314.92 m
Explanation:
Acceleration of the helicopter = 5.4 m/s² = a
Time taken by the helicopter to reach maximum height = 10.8 s = t
Initial velocity = 0 = u
Final velocity = v
Displacement = s
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 10.8+\frac{1}{2}5.4\times 10.8^2\\\Rightarrow s=314.92\ m[/tex]
Maximum height above ground reached by the helicopter is 314.92 m
A swan on a lake becomes airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.50 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2 , what distance Δx does it travel before becoming airborne?
Answer:
The distance traveled by the swan is 60.35 meters.
Explanation:
Given that,
A swan accelerate from rest (u = 0) to 6.5 m/s to take off.
Acceleration of the swan, [tex]a=0.35\ m/s^2[/tex]
We need to find the distance Δx it travel before becoming airborne. From the third equation of motion as :
[tex]\Delta x=\dfrac{v^2-u^2}{2a}[/tex]
[tex]\Delta x=\dfrac{(6.5)^2}{2\times 0.35}[/tex]
[tex]\Delta x=60.35\ m[/tex]
So, the distance traveled by the swan is 60.35 meters. Hence, this is the required solution.
Every 4.5 Billion years, half of the atoms in a sample of uranium-238 will undergo radioactive decay and become atoms of lead-206. Suppose you lived in another planetary system around a faraway star and found meteorite that was originally made of uranium-238, but is now one-quarter uranium-238 and three quarters lead-206. what would be your best estimate of the planetary system?
Answer:
The planetary system is 9 billion years old
Explanation:
In 4.5 billion years, the meteorite was half uranium and half lead. For this amount of uranium to be reduced by half again (reaching 1/4 uranium and 3/4 lead), another 4.5 billion years must have passed. Then, the planetary system should be 9 (2 x 4.5) billion years old.
A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters?
Answer:
[tex]v_{f} =4.9\frac{m}{s}[/tex] : velocity of the crate after the person has pushed the crate a distance of 6 meters
Explanation:
Crate kinetics
Crate moves with uniformly accelerated movement
v f²=v₀²+2a*d (formula 1)
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Known data
v₀=0 The speed of the crate is equal to zero because part of the rest.
a= 2m/s²
d= 6m
Distance calculating
We replace data in the Formula (1)
v f²=0+2*2*d
v f²=2*2*6
v f²=24
[tex]v_{f} =\sqrt{24}[/tex]
[tex]v_{f} =4.9\frac{m}{s}[/tex]
To find the final velocity of a 50-kg crate after being pushed for 6 meters with an acceleration of 2.0 m/s², the kinematic equation v² = u² + 2as is used, resulting in a final velocity of 4.9 m/s.
Explanation:To calculate the velocity of a 50-kg crate after being pushed a distance of 6 meters with an acceleration of 2.0 m/s2 from rest, we use the kinematic equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance moved. Since the crate starts from rest, the initial velocity, u, is 0 m/s. Substituting the given values we get:
v2 = 0 m/s + 2(2.0 m/s2)(6 m),
v2 = 24 m2/s2,
v = √(24 m2/s2),
v = 4.9 m/s.
Therefore, the velocity of the crate after being pushed 6 meters will be 4.9 m/s.
A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?
Answer:
75 N
Explanation:
t = Time taken = 2 seconds
u = Initial velocity
v = Final velocity
s = Displacement = 30 m
a = Acceleration
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2[/tex]
The acceleration due to gravity on the planet is 15 m/s²
Force
F = ma
[tex]F=5\times 15\\\Rightarrow F=75\ N[/tex]
The gravitational force exerted on the object near the planet’s surface is 75 N
We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface
F=75N
From the question we are told
A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m.
Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?
Generally the equation for the Motion is mathematically given as
[tex]s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2[/tex]
Therefore
F=ma
F=5*15
F=75N
Hence, the gravitational force exerted on the 5kg object near the planet’s surface
F=75N
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A 3 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is < 0, -mg, 0 >, where g is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the +x direction towards where the ball lands, and the +y direction taken to be upwards.)
Answer:
F=<0,-29.4,0>N
D = <0,-10,0>m
Work=F·D=294 Joules
Explanation:
Force:
The gravity has a negative direction in the axis Y in our coordinate system:
F=<0,-mg,0>=<0,-3*9.8,0>=<0,-29.4,0>N
Displacement:
The initial position A is <0, 33, 0>m
The final position B is <0, 23,0>m
The displacement vector is D=B-A = <0,-10,0>m
Gravitational Work:
The work is the scalar product between the force and the displacement
Work=F·D=(-29.4)*(-10)=294 Joules
For a standard production car, the highest roadtested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mph) in 3.275 s. Find the magnitude of the car's acceleration.
Answer:
a=8.1832m/s^2
Explanation:
Vo=initial speed=0m/s
Vf=final speed=26.8m/s
t=time=3.275s
the vehicle moves with a constant acceleration therefore we can use the following equation
A=aceleration=(Vf-Vi)/t
A=(26.8m/s-0m/s)/3.275s=8.1832m/s^2
the magnitude of the car´s aceleration is 8.1832m/s^2
A flagpole consists of a flexible, 7.14 m tall fiberglass pole planted in concrete. The bottom end of the flagpole is fixed in position, but the top end of the flagpole is free to move. What is the lowest frequency standing wave that can be formed on the flagpole if the wave propagation speed in the fiberglass is 2730 m/s?
Answer:
The lowest frequency is 95.6 Hz
Explanation:
The standing waves that can be formed in this system must meet some conditions, such as until this is fixed at the bottom here there must be a node (point without oscillation) and being free at its top at this point there should be maximum elongation (antinode)
For the lowest frequency we have a node at the bottom point and a maximum at the top point, this corresponds to ¼ of the wavelength, so the full wave has
λ = 4L
As the speed any wave is equal to the product of its frequency by the wavelength
v = f λ
f = v / λ
f = v / 4L
f = 2730 / (4 7.14)
f= 95.6 1 / s = 95.6 Hz
The lowest frequency of the standing wave that can be formed on the flagpole is 95.59 Hz.
Data obtained from the question Length (L) = 7.14 mWavelength (λ) = 4L = 4 × 7.14 = 28.56 mVelocity (v) = 2730 m/sFrequency (f) =? How to determine the frequencyThe velocity, frequency and wavelength of a wave are related according to the following equation:
Velocity (v) = wavelength (λ) × frequency (f)
v = λf
With the above formula, we can obtain the frequency as follow:
v = λf
2730 = 28.56 × f
Divide both side by 28..56
f = 2730 / 28.56
f = 95.59 Hz
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Is it possible for two pieces of the same metal to have different recrystallization temperatures? Is it possible for recrystallization to take place in some regions of a part before it does in other regions of the same part? Explain
Recrystallization is a temperature-induced process that can vary among materials. Metal parts may recrystallize at different temperatures, and recrystallization can happen unevenly within a single part.
Recrystallization is a process where a material undergoes structural changes due to increased temperature. Metallic glasses can have different recrystallization temperatures based on the composition of the metals. Recrystallization can occur non-uniformly in a part, with some regions undergoing the process before others due to variations in temperature or cooling rates.
True or False: When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly.
Final answer:
The statement regarding receiving credit for correctly positioned endpoints using an endpoint mover is typically true. Credit is awarded for each endpoint positioned correctly in a digital math exercise or tool, which may involve placing endpoints according to specific criteria.
Explanation:
The statement "When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly" is generally True. In the context of mathematical tools or interactive digital platforms, an endpoint mover likely refers to a feature that allows users to manipulate the endpoints of a line segment or other geometric object. When students are tasked with positioning these endpoints correctly according to given criteria, they would typically receive credit or points for each endpoint that is placed accurately.
For example, in a digital math exercise, if you're asked to position the endpoints of a line segment so that one end is at the point (3,2) and the other end is at the point (-1,5), and you succeed in placing them correctly, you would receive credit for both endpoints. However, details may vary based on the specific platform or educational program, so it's essential to follow the instructions provided.
Solve the following equation.
LaTeX: \left|\frac{x}{7}\right|=1| x 7 | = 1
Group of answer choices
LaTeX: x=7 x = 7
LaTeX: x=-4,\:x=-16 x = − 4 , x = − 16
LaTeX: x=7,\:x=-7 x = 7 , x = − 7
LaTeX: x=-5,\:x=-13
Answer:
x = +7, x = -7
Explanation:
The equation to solve is:
[tex]\left|\frac{x}{7}\right|=1[/tex]
Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:
1) [tex]\frac{x}{7}=+1[/tex]
Solving this one, we find
[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]
2) [tex]\frac{x}{7}=-1[/tex]
Solving this one, we find
[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]
So the two solutions are +7 and -7.
A ball is dropped from a tower that is 512 feet high. Use the formula below to find height of the ball 5 seconds after it was dropped? h=512−14t2where h represents the height, in feet, and t represents the time, in seconds, after it was dropped.
Answer:
The height of the ball after 5 seconds is 162 ft.
Explanation:
First, replace the variable t with how many seconds the ball has dropped, which in this case is 5.
[tex]h = 512 - 14 {t}^{2} \\ h = 512 - 14 \times {5}^{2} [/tex]
Solve.
[tex]h = 512 - 14 \times {5}^{2} \\ h = 512 - 14 \times 25 \\ h = 512 - 350 \\ h = 162[/tex]
For a duration of 5 seconds, the ball had managed to drop 350 feet, with 162 feet left to go to touch ground level.
A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 4.50 M and 1.11 ⋅ 10−2 M, respectively. The cell emf is ________ V.
Answer : The cell emf for this cell is 0.077 V
Solution :
The balanced cell reaction will be,
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = ?
[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M
[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]
[tex]E_{cell}=0.077V[/tex]
Therefore, the cell emf for this cell is 0.077 V
The cell emf, or electromotive force, for a voltaic cell, like the Zn2+-Zn kind described, can be calculated with the Nernst equation. The emf is determined by the standard electrode potential as well as the concentrations of the redox species in each half-cell.
Explanation:A voltaic cell runs on a spontaneous redox reaction occurring indirectly, with the oxidant and reductant redox couples contained in separate half-cells. In your example, both half-cells are Zn2+-Zn electrodes where the reaction is Zn2+ + 2e− → Zn (s) with a standard electrode potential, E°, of -0.763 V. The cell emf, or electromotive force, is calculated through the Nernst Equation:
Ecell = E° - (RT/nF) * ln(Q)
where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, and F is Faraday's constant. Since we're dealing with a concentration cell, Q is given by the ratio of the concentrations of the reduced and oxidized species. Depending on the temperature and assuming that Zn2+ is reduced at the cathode and oxidized at the anode, the cell emf can be calculated accordingly.
Learn more about Cell EMF here:https://brainly.com/question/34647069
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Alice and Bob are each riding horses on a carousel. Alice's horse is twice as far from the axis of spin of the carousel as Bob's horse. Let ω A be the angular velocity of Alice's horse, and let ω B be the angular velocity of Bob's horse. Which of the following is true?
a. ω A = ω B
b. ω A > ω B
c. ω A < ω B
Answer:
option (a)
Explanation:
the angular velocity of the carousel is same througout the motion, so the angular velocity of all the horses is same, but the linear velocity is different for different horses.
As the angular displacement of all the horses are same in the same time so the angular velocity is same.
The relation between the linear velocity and the angular velocity is given by
v = r ω
where, v is linear velocity and r be the distance between the horse and axis of rotation and ω be the angular velocity.
So, the angular velocity of Alice horse is same as the angular velocity of Bob horse.
ωA = ωB
Thus, option (a) is true.
The correct option is Option A ( ω A = ω B). Both Alice and Bob's horses on the carousel have the same angular velocity, regardless of their distance from the axis. Hence, ω A == ω B.
The key to answering the question lies in understanding the concept of angular velocity in rotational motion.Angular velocity (">ω") is the rate at which an object rotates around an axis and is usually measured in radians per second.In a carousel, if two points rotate with the same angular velocity but are at different distances from the axis of rotation, they still maintain the same angular velocity. Therefore, the fact that Alice's horse is twice as far from the axis of spin as Bob's horse doesn't affect the angular velocities.Thus, the correct answer is:
a. ω A = ω B
Consider the three displacement vectors
A=(3i+3j)meters,
B-(i-4j) m
C=(-2i+5j) m
Use the Component method to determine
a) the magnitude and direction of the vector D= A+B+C
b) the magnitude And direction of E=-A-B+C
Answer:
Explanation:
[tex]\overrightarrow{A} = 3\widehat{i}+3\widehat{j}[/tex]
[tex]\overrightarrow{B} = \widehat{i}-4\widehat{j}[/tex]
[tex]\overrightarrow{C} = -2\widehat{i}+5\widehat{j}[/tex]
(a)
[tex]\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}[/tex]
[tex]\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}[/tex]
[tex]\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}[/tex]
Magnitude of [tex]\overrightarrow{D}[/tex] = [tex]\sqrt{2^{2}+4^{2}}[/tex]
= 4.47 m
Let θ be the direction of vector D
[tex]tan\theta =\frac{4}{2}[/tex]
θ = 63.44°
(b)
[tex]\overrightarrow{E} =
- \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}[/tex]
[tex]\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}[/tex]
[tex]\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}[/tex]
Magnitude of [tex]\overrightarrow{E}[/tex] = [tex]\sqrt{6^{2}+6^{2}}[/tex]
= 8.485 m
Let θ be the direction of vector D
[tex]tan\theta =\frac{6}{-6}[/tex]
θ = 135°
a) For vector [tex]\( \mathbf{D} \)[/tex]: Magnitude: [tex]\( 4 \, \text{meters} \)[/tex] ,Direction: [tex]\( 90^\circ \)[/tex]
b) For vector [tex]\( \mathbf{E} \)[/tex]: Magnitude: [tex]\( 7.21 \, \text{meters} \)[/tex] ,Direction: [tex]\( 123.69^\circ \)[/tex]
To find the resultant vectors [tex]\( \mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C} \) and \( \mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C} \)[/tex] using the component method, we need to break each vector into its [tex]\( i \)[/tex] and [tex]\( j \)[/tex] components, sum these components, and then find the magnitude and direction of the resulting vectors.
Given vectors:
[tex]\[ \mathbf{A} = 3\mathbf{i} + 3\mathbf{j} \][/tex]
[tex]\[ \mathbf{B} = -\mathbf{i} - 4\mathbf{j} \][/tex]
[tex]\[ \mathbf{C} = -2\mathbf{i} + 5\mathbf{j} \][/tex]
a) Vector [tex]\(\mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C}\)[/tex]
1. Sum the components:
[tex]\[ \mathbf{D} = (3\mathbf{i} + 3\mathbf{j}) + (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Combine like terms:
[tex]\[ D_i = 3 - 1 - 2 = 0 \][/tex]
[tex]\[ D_j = 3 - 4 + 5 = 4 \][/tex]
So, the components of [tex]\( \mathbf{D} \)[/tex] are:
[tex]\[ \mathbf{D} = 0\mathbf{i} + 4\mathbf{j} \][/tex]
2. Magnitude of [tex]\( \mathbf{D} \)[/tex]:
[tex]\[ |\mathbf{D}| = \sqrt{D_i^2 + D_j^2} = \sqrt{0^2 + 4^2} = 4 \, \text{meters} \][/tex]
3. Direction of [tex]\( \mathbf{D} \)[/tex]:
The angle [tex]\( \theta_D \)[/tex] from the positive x-axis is:
[tex]\[ \theta_D = \tan^{-1}\left(\frac{D_j}{D_i}\right) = \tan^{-1}\left(\frac{4}{0}\right) = 90^\circ \][/tex]
b) Vector [tex]\(\mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C}\)[/tex]
1. Sum the components:
[tex]\[ \mathbf{E} = - (3\mathbf{i} + 3\mathbf{j}) - (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Simplify the negative signs:
[tex]\[ \mathbf{E} = (-3\mathbf{i} - 3\mathbf{j}) + (\mathbf{i} + 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Combine like terms:
[tex]\[ E_i = -3 + 1 - 2 = -4 \][/tex]
[tex]\[ E_j = -3 + 4 + 5 = 6 \][/tex]
So, the components of [tex]\( \mathbf{E} \)[/tex] are:
[tex]\[ \mathbf{E} = -4\mathbf{i} + 6\mathbf{j} \][/tex]
2. Magnitude of [tex]\( \mathbf{E} \)[/tex]:
[tex]\[ |\mathbf{E}| = \sqrt{E_i^2 + E_j^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \, \text{meters} \][/tex]
3. Direction of [tex]\( \mathbf{E} \)[/tex]:
The angle [tex]\( \theta_E \)[/tex] from the positive x-axis is:
[tex]\[ \theta_E = \tan^{-1}\left(\frac{E_j}{E_i}\right) = \tan^{-1}\left(\frac{6}{-4}\right) = \tan^{-1}\left(-1.5\right) \][/tex]
Since [tex]\( E_i \)[/tex] is negative and [tex]\( E_j \)[/tex] is positive, [tex]\( \theta_E \)[/tex] is in the second quadrant:
[tex]\[ \theta_E = 180^\circ - \tan^{-1}(1.5) \approx 180^\circ - 56.31^\circ = 123.69^\circ \][/tex]
a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Answer:
magnitude=34.45 m
direction=[tex]55.52\°[/tex]
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
For the magnitude we will use the formula to calculate the distance [tex]d[/tex] between two points:
[tex]d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}[/tex] (1)
[tex]d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}[/tex] (2)
[tex]d=\sqrt{1186.81 m^{2}}[/tex] (3)
[tex]d=34.45 m[/tex] (4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
[tex]tan \theta=\frac{Y2-Y1}{X2-X1}[/tex] (5)
[tex]tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}[/tex] (6)
[tex]tan \theta=\frac{24.8}{19.5}[/tex] (7)
Finding [tex]\theta[/tex]:
[tex]\theta= tan^{-1}(\frac{24.8}{19.5})[/tex] (8)
[tex]\theta= 55.52\°[/tex] (9) This is the direction of the vector