Answers:
a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]
b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]
c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
Explanation:
a) According to Archimedes’ Principle:
A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.
In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium. This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.
Hence:
[tex]W_(cylinder)=B[/tex] (1)
Where:
[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.
[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.
We can rewrite (1) as follows:
[tex]m_{cylinder}g=m_{water}g[/tex] (2)
On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:
[tex]\rho=\frac{m}{V}[/tex] (3)
isolating the mass:
[tex]m=\rho V[/tex] (4)
Now we can express (2) in terms of the density and the volume of cylinder and water:
[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)
In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).
At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:
[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)
Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.
[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)
Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.
Substituting (6) and (7) in (5):
[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)
Clearing [tex]\rho_{cylinder}[/tex]:
[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)
Simplifying;
[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)
Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:
[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)
[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder
b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.
Applying again the Archimedes principle:
[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)
Isolating [tex]\rho_{liq}[/tex]:
[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)
Where:
[tex]V_{cylinder}=\pi r^{2} h[/tex]
[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]
Then:
[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)
[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)
[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)
[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid
c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.
This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
A stone is thrown vertically upward from ground level at t = 0. At t=2.50 s, it passes the top of a tall building, and 1.50 s later, it reaches its maximum height. What is the height of the tall building? We assume an answer in meters.
Answer:67.45 m
Explanation:
Given
at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum height
let u is the initial velocity of stone
v=u+at
0=u-gt
[tex]u=9.81\times 4=39.24 m/s[/tex]
Let us take h be the height of building
[tex]h=ut+\frac{-1}{2}gt^2[/tex]
[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]
h=67.45 m
You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 19.0 m from yours, in the direction 19.0° north of east. Karl's tent is 45.0 m from yours, in the direction 39.0° south of east. What is the distance between Karl's tent and Joe's tent?
Answer:
Distance between Karl and Joe is 38.467 m
Solution:
Let us assume that you are at origin
Now, as per the question:
Joe's tent is 19 m away from yours in the direction [tex]19.0^{\circ}[/tex] north of east.
Now,
Using vector notation for Joe's location, we get:
[tex]\vec{r_{J}} = 19cos(19.0^{\circ})\hat{i} + 19sin(19.0^{\circ})\hat{j}[/tex]
[tex]\vec{r_{J}} = 17.96\hat{i} + 6.185\hat{j} m[/tex]
Now,
Karl's tent is 45 m away from yours and is in the direction [tex]39.0^{\circ}[/tex]south of east, i.e., [tex]- 39.0^{\circ}[/tex] from the positive x-axis:
Again, using vector notation for Karl's location, we get:
[tex]\vec{r_{K}} = 45cos(-319.0^{\circ})\hat{i} + 45sin(- 39.0^{\circ})\hat{j}[/tex]
[tex]\vec{r_{K}} = 34.97\hat{i} - 28.32\hat{j} m[/tex]
Now, obtain the vector difference between [tex]\vec{r_{J}}[/tex] and [tex]\vec{r_{K}}[/tex]:
[tex]\vec{r_{K}} - \vec{r_{J}} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m[/tex]
[tex]\vec{d} = \vec{r_{K}} - \vec{r_{J}} = 17.01\hat{i} - 34.51\hat{j} m[/tex]
Now, the distance between Karl and Joe, d:
|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|
[tex]d = \sqrt{(17.01)^{2} + (34.51)^{2}} m[/tex]
d = 38.469 m
The distance between Karl's and Joe's tent is:
The distance between Joe's tent and Karl's tent is approximately 36.84 m.
Explanation:To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:
Joe's tent: 19.0 m at 19.0° north of east Karl's tent: 45.0 m at 39.0° south of east
Next, we can use the components to find the displacement from Joe's tent to Karl's tent:
For Joe's tent: North component = 19.0 m * sin(19.0°) = 6.36 m, East component = 19.0 m * cos(19.0°) = 17.88 m For Karl's tent: North component = -45.0 m * sin(39.0°) = -27.10 m, East component = 45.0 m * cos(39.0°) = 34.37 m
Using the components, we can calculate the displacement from Joe's tent to Karl's tent:
North displacement = -27.10 m - 6.36 m = -33.46 m East displacement = 34.37 m - 17.88 m = 16.49 m
Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:
Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m
Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.
A super snail initially traveling at 2 m/s accelerates at 1 m/s^2 for 5 seconds. How fast will it be going at the end of the 5 seconds? How far did the snail travel?
Answer:
The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.
Explanation:
Given that, the initial velocity of the snail is,
[tex]u=2m/s[/tex]
And the acceleration of the snail is,
[tex]a=1m/s^{2}[/tex]
And the time taken by the snail is,
[tex]t=5 sec[/tex]
Now according to first equation of motion,
[tex]v=u+at[/tex]
Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.
Now substitute all the variables
[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]
Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.
Now according to third equation of motion.
[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]
Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.
Substitute all the variables in above equation.
[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]
Therefore the distance of the snail is 22.5 m.
A certain elevator cab has a total run of 218 m and a maximum speed is 319 m/min, and it accelerates from rest and then back to rest at 1.20 m/s^2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 218 m run, starting and ending at rest?
Answer:
a)11.6m
b)45.55s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
Find the critical angle for total internal reflection for a flint glass-air boundary (you may assume that λ = 580.0 nm). Express your answer to 4 significant figures!
Answer:
the critical angle of the flint glass is 37.04⁰
Explanation:
to calculate the critical angle for total internal reflection.
given,
wavelength of the flint glass = λ = 580.0 nm
= 580 × 10⁻⁹ m
critical angle = sin^{-1}(\dfrac{\mu_a}{\mu_g})
at the wavelength of 580.0 nm the refractive index of the glass is 1.66
refractive index of air = 1
critical angle = sin^{-1}(\dfrac{1}{1.66})
= 37.04⁰
hence, the critical angle of the flint glass is 37.04⁰
Recent findings on the topic of brain based research indicate all of the following except
Recent findings on the topic of brain-based research indicate all of the following except
A. early environments matter.
B. all children are born ready to learn.
C. society isn't addressing the needs of young children. ?
D. the brain stops growing at around age two
How long does it a take a runner, starting from rest to reach max speed, 30 f/s given acceleration 8 f/s^2? After finding the time, calculate the distance traveled in that time.
Explanation:
Given that,
Initial sped of the runner, u = 0
Final speed of the runner, v = 30 ft/s
Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]
Let t is the time taken by the runner. It can be calculated using first equation of motion as :
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{30-0}{8}[/tex]
t = 3.75 seconds
Let s is the distance covered by the runner. Using the second equation of motion as :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]
s = 56.25 feet
Hence, this is the required solution.
Final answer:
To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.
Explanation:
The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.
Calculating Time to Reach Max Speed
To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.
Calculating Distance Traveled
To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding the speed limit? Justify your answer.
Answer: Yes, he is exceeding the speed limit
Explanation:
Hi!
This is problem about unit conversion
1 mile = 1,609.344 m
Then the speed limit v is:
v = 75 mi/h = 120,700.8 m/h
1 hour = 60 min = 60*60 s = 3,600 s
v = (120,700.8/3,600) m/s = 33.52 m/s
38 m/s is higher than the speed limit v.
A spelunker is surveying a cave. She follows a passage 180 m straight west, then 230 m in a direction 45° east of south, and then 280 m at 30° east of north. After a fourth unmeasured displacement, she finds herself back where she started. A: Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
B: Determine the direction of the fourth displacement. Express your answer using two significant figures.
The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.
Explanation:To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.
For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).
To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.
Learn more about Vector Operations here:https://brainly.com/question/20047824
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A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is released at time t = 0 from a height H above the origin. In addition to the vertical component of acceleration due to gravity, a strong wind blowing from the right gives the package a horizontal component of acceleration of magnitude ¼g to the left. Derive an expression for the horizontal distance D from the origin where the package hits the ground.
Answer:
[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]
Explanation:
From the vertical movement, we know that initial speed is 0, and initial height is H, so:
[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]
[tex]0=H-g*\frac{t^{2}}{2}[/tex] solving for t:
[tex]t=\sqrt{\frac{2H}{g} }[/tex]
Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:
[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex] Replacing values:
[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]
Simplifying:
[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]
Assume everyone in the United States consumes one soft drink in an aluminum can every two days. if there are 280 million americans, how many tons of aluminum need to be recycles each year if each can weight 1/15 pound and once ton=2000 pounds?
Answer:
1.708*10^6 tons.
Explanation:
Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.
Total number of people in US= 280,000,000
Total number of cans used by americans.
[tex] = 5.12×10^10 cans[/tex]
Weight of 1 can =1/15 pounds
Weight of all cans used in 1 year
[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]
we know that
1ton=2000pounds.
[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]
An elevator moves downward in a tall building at a constant speed of 5.70 m/s. Exactly 4.95 s after the top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest. (a) At what time does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at t = 0 s.)(b) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume 1 floor congruent 3 m.)
Answer:
a) t = 3.01s
b) 15th floor
Explanation:
First we need to know the distance the elevator has descended before the bolt fell.
[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]
Now we can calculate the time that passed before both elevator and bolt had the same position:
[tex]Y_{b}=Y_{e}[/tex]
[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]
[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex] Solving for t:
t1 = -1.87s t2 = 3.01s
In order to know how the amount of floors, we need the distance the bolt has fallen:
[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex] Since every floor is 3m:
Floors = Yb / 3 = 15 floors.
How do resistors in series affect the total resistance?
Answer:
Explanation:
Resistance in series is given by the sum of all the resistor in series
value of Total Resistance is given by
[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]
Where [tex]R_{th}[/tex] is the total resistance
[tex]R_1,R_2[/tex] are the resistance in series
Current in series remains same while potential drop is different for different resistor
The value of net resistor is always greater than the value of individual resistor.
If a there is a defect in a single resistor then it affects the whole circuit in series.
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground?
(A) The frequency and intensity increase
(B) The frequency decrease and intensity increase
(C) The frequency decrease and intensity decrease
(D) The frequency remains the same, but the intensity decreases.
Answer:
(C) The frequency decrease and intensity decrease
Explanation:
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.
if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.
The frequency is the inverse from the wavelength, so the frequency heard will increase.
The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? What distance does the car travel during the braking period?
Answer:[tex]3.874 m/s^2[/tex]
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec
[tex]60mi/h \approx 26.8224m/s[/tex]
[tex]34mi/h \approx 15.1994 m/s[/tex]
we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]
[tex]a=\frac{15.1994-26.8224}{3}[/tex]
[tex]a=-3.874 m/s^2[/tex]
negative indicates that it is stopping the car
Distance traveled
[tex]v^2-u^2=2as[/tex]
[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]
[tex]s=\frac{488.419}{2\times 3.874}[/tex]
s=63.038 m
A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How many joules of energy must be withdrawn from the water by the refrigerator?
Answer:920.31 J
Explanation:
Given
Volume of water (V)[tex]=14.1 cm^3 [/tex]
mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]
Temperature [tex]=8.4^{\circ} C[/tex]
Final Temperature [tex]=-7.2 ^{\circ}C[/tex]
specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]
Therefore heat required to removed is
[tex]Q=mc(\Delta T)[/tex]
[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]
[tex]Q=920.31 J[/tex]
A uniform electric field of magnitude 4.9 ✕ 10^4 N/C passes through the plane of a square sheet with sides 8.0 m long. Calculate the flux (in N · m^2/C) through the sheet if the plane of the sheet is at an angle of 30° to the field. Find the flux for both directions of the unit normal to the sheet.
1)unit normal with component parallel to electric field (N · m^2/C)
2)unit normal with component antiparallel to electric field (N · m^2/C)
Answer:
1. 1.568 x 10^6 N m^2 / C
2. - 1.568 x 10^6 N m^2 / C
Explanation:
E = 4.9 x 10^4 N/C
Side of square, a = 8 m
Area, A = side x side = 8 x 8 = 64 m^2
Angle between lane of sheet and electric field = 30°
Angle between the normal of plane of sheet and electric field,
θ = 90°- 30° = 60°
The formula for the electric flux is given by
[tex]\phi = E A Cos\theta[/tex]
(1) [tex]\phi = E A Cos\theta[/tex]
By substituting the values, we get
Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C
(2) [tex]\phi = E A Cos\theta[/tex]
By substituting the values, we get
Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C
Meredith walks from her house to a bus stop that is 260 yards away. If Meredith is 29 yards from her house, how far is she from the bus stop? 231 Correct yards If Meredith is 204.8 yards from her house, how far is she from the bus stop? 55.2 Correct yards Let the variable x represent Meredith's varying distance from her house (in yards). As Meredith walks from her house to the bus stop, the value of x varies from 0 Correct to 260 Correct . How many values does the variable x assume as Meredith walks from her house to the bus stop? 3 Incorrect
Answer:
a) 231 yards
b) 55.2 yards
c) 0 yards to 260 yards
d) Infinite values
Explanation:
This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:
x=Final Position - Initial Position
or
[tex]x=x_{f} - x_{i}[/tex]
a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:
[tex]x=260 y - 29 y=231 y[/tex]
b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:
[tex]x=260 y - 204.8 y=55.2 y[/tex]
c) If we take Meredith's initial position at her house [tex]x_{i}=0 y[/tex] and her final position at the bus stop [tex]x_{f}=260 y[/tex], the value of [tex]x[/tex] varies from 0 yards to 260 yards.
d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]
The answers to the possible distance covered by Meredith at the various distances from her house are;
A) distance = 231 yards
A) distance = 231 yardsB) distance = 55.2 yards
A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260
A) We are told that meredith walks from her house to a bus stop that is 260 yards away.
After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.
Distance left to reach bus stop = 260 - 29 = 231 yards
B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;
Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.
C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.
Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260
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A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 18.1 m/s at an angle of 49.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Answer:
16.17 m/s
Explanation:
h = 3.2 m
u = 18.1 m/s
Angle of projection, θ = 49°
Let H be the maximum height reached by the ball.
The formula for the maximum height is given by
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]
[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]
The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m
Let v be the velocity of ball with which it strikes the ground.
Use third equation of motion for vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]
here, uy = 0
So,
[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]
vy = 11.13 m/s
vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s
The resultant velocity is given by
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]
v = 16.27 m/s
If a marathon runner averages 9.39 mi/h, how long does it take him or her to run a 26.22-mi marathon? Express your answers in h, min and s.
Answer:
[tex]t=2.8h[/tex]
[tex]t=10080s[/tex]
[tex]t=168 min[/tex]
Explanation:
From this exercise we have velocity and distance. Using the following formula, we can calculate time:
[tex]v=\frac{d}{t}[/tex]
Solving for t
[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]
[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]
[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]
A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identical mass and spring next to the first set is also set into motion. At what time t should the second system be set into motion so that the phase difference in oscillations between the two systems is pi/2?
Answer:
The second system must be set in motion [tex]t=0.70s[/tex] seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later
Suppose a Southwest Airlines passenger plane took three hours to fly 1800 miles in the direction of the Jetstream. The return trip against the Jetstream took four hours. What was the plane’s speed (as read on the plane’s speedometer) in still air and the Jetstream’s speed? How can applying matrices and linear systems help solve this problem?
Answer:
plane speed: 525mph, jetstream speed=75mph, in explanation it is solved with a linear equations system
Explanation:
First lets name each speed
vs:=speed of the jetstream
vp:=speed of the plane
Now when in the jetstream direction the speeds are added and on the opposite direction are subtracted, then we get these equations, that are linear.
1800 mi=(vp+vs)*3h
1800 mi=(vp-vs)*4h
which is a linear equation system equivalent to:
600 mph=vp+vs (1)
450 mph=vp-vs (2)
Now from (2) vp= 450mph+vs (3), replacing this in (1) we get:
600mph=(450mph+vs)+vs=450mph+2*vs, then 2*vs=150mph or vs=*75mph, this is the jetstream speed, replacing this in (3) we get the plane speed too vp=450 mph +75mph = 525 mph
A piece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. It is illuminated with light of wavelength 680 nm. Determine the minimum thickness of the coating that will result in no reflection
Answer:
Thickness = 123.19 nm
Explanation:
Given that:
The refractive index of the glass = 1.50
The refractive index of thin layer of magnesium fluoride = 1.38
The wavelength of the light = 680 nm
The thickness can be calculated by using the formula shown below as:
[tex]Thickness=\frac {\lambda}{4\times n}[/tex]
Where, n is the refractive index of thin layer of magnesium fluoride = 1.38
[tex]{\lambda}[/tex] is the wavelength
So, thickness is:
[tex]Thickness=\frac {680\ nm}{4\times 1.38}[/tex]
Thickness = 123.19 nm
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
Answer:
306500 N/C
Explanation:
The magnitude of an electric field around a single charge is calculated with this equation:
[tex]E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}[/tex]
With ε0 = 8.85*10^-12 C^2/(N*m^2)
Then:
[tex]E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}[/tex]
E(0.89) = 306500 N/C
Which of the following is not a unit of torque? O pound-foot 0 kilogram-newton Newton-meter O pound-inch
Answer: Kilogram- newton is wrong unit for Torque
Idem pound-inch is also wrong unit for Torque
Explanation: As it is well known the torque is defined as :
Τorque= F x R
so its UNITS are Newton*meter (SI)
or in the Imperial System is often use Pound-foot
The driver of a sports car traveling at 10.0m/s steps down hard on the accelerator for 5.0s and the velocity increases to 30.0m/s. What was the average acceleration of the car during the 5.0s time interval?
Answer:
[tex]a=4m/s^{2}[/tex]
Explanation:
From the concept of average acceleration we know that
[tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} }[/tex]
From the exercise we know that
[tex]v_{2}=30m/s\\v_{1}=10m/s\\t_{2}=5s\\t_{1}=0s[/tex]
So, the average acceleration of the car is:
[tex]a=\frac{30m/s-10m/s}{5s}=4m/s^{2}[/tex]
Comment on why the acceleration due to gravity is less for the plastic ball. Why do the other two balls (steel ball and golf ball) not have such a low value for the acceleration?
Answer and Explanation:
The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:
[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]
where
[tex]m_{o}[/tex] = mass of the object
G = Gravitational constt
Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.
This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.
Final answer:
The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.
Explanation:
The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.
The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.
If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.
What happens to the width of the central diffraction pattern (in the single slit experiment) as the slit width is changed and why?
Answer:
width of fringes are increased
Explanation:
The width of central maxima is given by the following expression
Width = 2 x Dλ / d
D is distance of screen from source , d is slit width and λ is wavelength of light source. Here we see , on d getting decreased , width will increase because d is in denominator .Due to increased width , position of a fringe moves away from the centre.
A car is making a 40 mi trip. It travels the first half of the total distance 20.0 mi at 18.00 mph and the last half of the total distance 20.0 mi at 56.00 mph. What is the car’s average speed in mph for the entire second trip?
Answer: The average speed is 27,24 mph (exactly 1008/37 mph)
Explanation:
This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:
1h--->18mi
X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h
1h--->56mi
X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h
Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.
Oppositely charged parallel plates are separated by 4.67 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? ________ N/C (b) What is the magnitude of the force on an electron between the plates? ________ N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.00 mm from the positive plate?_________ J
Answer:
a) 1.28 *10^5 N/C
b)2.05 *10^{-14} N
c) 4.83 *10^{-17} J
Explanation:
Given Data:
Distance between the plates, d = 4.67 mm
[tex]= (4.67) *10^{-3} m[/tex]
[/tex]= 4.67 *10^{-3} m[/tex]
Potential difference, V = 600 V
Solution:
(a) The magnitude of the electric field between the plates is,
[tex]E = \frac{V}{d}[/tex]
[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]
[tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]
(b) Force on electron btwn the plates is,
F = q E
[tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]
[tex]= 2.05 *10^{-14} N[/tex]
(c) Work done on the electron is
W = F * s
[tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]
[tex]= 4.83 *10^{-17} J[/tex]