Answer:
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}, t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
[tex]v_{sg 1} = v_{sr} + v_{rg}[/tex]
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
[tex]v_{sg1}[/tex] = D / [tex]t_{out}[/tex]
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
[tex]v_{sg 2} = v_{sr} - v_{rg}[/tex]
[tex]v_{sg 2}[/tex] = D / [tex]t_{in}[/tex]
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
[tex]v_{sg1} t_{out} = v_{sg2} t_{in}[/tex]
t_{out} = t_{in}
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / [tex]v_{sg2}[/tex]
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA?
The question is incomplete. Here is the complete question.
Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].
Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.
Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.
Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Explanation: First, let's write an equation of motion for each car.
Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:
[tex]x=x_{0}+vt[/tex]
where
[tex]x_{0}[/tex] is initial position
v is velocity
t is time
Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].
The equation will be:
[tex]x_{A}=D_{A}+v_{A}t[/tex]
Car B started at the starting line. So, its equation is
[tex]x_{B}=v_{B}t[/tex]
Part A: When they meet, both car are at "the same position":
[tex]D_{A}+v_{A}t=v_{B}t[/tex]
[tex]v_{B}t-v_{A}t=D_{A}[/tex]
[tex]t(v_{B}-v_{A})=D_{A}[/tex]
[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.
Part B: With the meeting time, we can determine the position they will be:
[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]
[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.
The distance traveled by the car A and car B should be equal to the as they meet at the same position.
The time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
How to calculate the distance traveled by body?The distance is the product of the speed of the body and the time taken to travel the distance.
Given information-
Car A has a head start and is a distance DA beyond the starting line at,
[tex]t=0[/tex]
Car A travels at a constant speed [tex]v_A[/tex].
Car B travels at a constant speed [tex]v_B[/tex].
The distance is the product of the speed of the body and the time taken to travel the distance.
The position equation from the motion for car A can be given as,
[tex]x_A=v_At+D_A[/tex]
The position equation from the motion for car B can be given as,
[tex]x_B=v_Bt[/tex]
The distance traveled by the car A and car B should be equal to the as they meet at the same position. Thus,
[tex]x_A=x_B[/tex]
Put the values,
[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]
Hence the time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
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Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.
1) What is Cole's acceleration?
2) What is his velocity when he reaches school?
Explanation:
this is the answer for your question. if you have any doubt.
you can send your doubt to:6369514784(what's app)
Which list of reaction types are all redox reactions?
A.
Synthesis, decomposition, single-replacement, combustion
B.
Synthesis, double-replacement, combustion, decomposition
C.
Acid-base, single-replacement, double-replacement, synthesis
D.
Decomposition, double-replacement, acid-base, synthesis
A. Synthesis, decomposition, single-replacement, combustion
List of reaction types are redox reactions:
Synthesis, decomposition, single-replacement, combustion.What are redox reaction ?"Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states." The term 'redox' is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes – a reduction process and an oxidation process.
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What can you conclude about the electric potential and the field strength at the two noted points between the two electrodes? What can you conclude about the electric potential and the field strength at the two noted points between the two electrodes? The potential is greater at point B; the field strength is greater at point A. The potential is greater at point B; the field strength is greater at point B. The potential is greater at point A; the field strength is greater at point B. The potential is greater at point A; the field strength is greater at point A
Answer: The potential is greater at point B; the field strength is greater at point B.
Explanation:
The thing that can be concluded about the electric potential and the field strength at the two noted points between the two electrodes is that the potential is greater at point B; the field strength is greater at point B.
We should note that electrodes are used in the provision of current which typically takes place through a nonmetal objects.
Which two statements below are central ideas in the article, "How Gross Is Your Bathroom"?
a. What you can't see might hurt you.
b. Different numbers of bacteria are hiding on various surfaces around your bathroom.
c. Most bacteria are harmless, and some are even good for you.
d. Your bathroom is filled with germs that you might not know anything about, including
viruses and bacteria.
Two people each do 100 joules of work by pushing a crate to the right. During this process, 50 joules of heat is generated from the
friction between the floor and the crate. How much energy is gained by the crate during this process?
Animals conduct_______.
A. cellular respiration
B. photosynthesis
C. both cellular respiration and photosynthesis
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0 m/s , and the distance between them is 52.0 m . After t1 = 3.00 s , the motorcycle starts to accelerate at a rate of 4.00 m/s^2. The motorcycle catches up with the car at some time t2.
Required:
a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?
b. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?
Answer:
a) 5.09 seconds
b) 107.07 meters
Explanation:
a) As we know
[tex]t_2- t_1 = \sqrt{\frac{2 X}{a} }[/tex]
Substituting the given values we get
[tex]t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09[/tex]
It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car
b)
[tex]X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07[/tex]
Explain the difference in the function of plant and animal cell organelles, including cell membrane, cell wall, nucleus, cytoplasm, mitochondria, chloroplast, and vacuole
Answer:
Plant cell Animal cell
2. Have a cell membrane. 2. Have no chloroplasts.
3. Have cytoplasm. 3. Have only small vacuoles.
4. Have a nucleus. 4. Often irregular in shape.
5. Often have chloroplasts
containing chlorophyll. 5. Do not contain plastids.
What is the mass of 9.11 moles of
ozone, 03?
The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
What is molecular mass?Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule. The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.
To compute the Molecular Mass of [tex]$O_{3}[/tex]
Atomic mass of oxygen(O) = 16
As [tex]$O_{3}[/tex] contains 3 atoms,
The molecular mass of [tex]$O_{3}[/tex]
= (16 x 3) = 48g/mol
Hence the mass of 9.11 moles O3
= 9.11 mol x 48g/mol
= 437.28g
= 0.43728kg
Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
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|:Give one word answer
1. An object that allows whole light to pass through it _______
Answer:
Translucent object
Explanation:
how could you get the pancake to cook faster without burning it?
Answer:
put it in medium to full heat and set an alarm so that it doesn't burn
Answer:
when you're cooking your pancakes, use clarified butter and cook on medium heat.
All magnetic fields result from the movement of
A. charged particles
B. electrons only
C. protons only
D. neutrons only
*How much energy is
transferred in lifting a 5 kg
Mass 3m
Answer:
147 J
Explanation:
The energy transferred to potential energy is :
U = m * g * h = (5 kg) * (9.8 m/s^2) * (3 m) = 147 J
On a car trip you drive for 2 hours and 41 minutes on a highway at a speed of 107.0 km/h. Then you stop at a gas station to fill up your tank. You also eat a quick lunch. The whole break lasts 23 minutes. After the break you start your engine up and you switch to a state road. You drive for another 3 hours and 31 minutes at a speed of 67.0 km/h before you arrive to your destination. What was your average speed for the whole trip with the lunchbreak included
Answer:
v = 79.3 km/h
Explanation:
By definition, the average speed, is the quotient between the total distance traveled and the time needed to travel that distance.The total time, is the sum of three times: one while driving before stopping at the gas station (t₁), the time spent there (t₂) and the time since leaving the gas station until reaching the final destination (t₃) .Let's convert these times to seconds first:[tex]t_{1} = 161 min* \frac{60s}{1min} = 9660 s (1)[/tex]
[tex]t_{2} = 23 min* \frac{60s}{1min} = 1380 s (2)[/tex]
[tex]t_{3} = 211 min* \frac{60s}{1min} = 12660 s (3)[/tex]
[tex]t_{tot} =t_{1} +t_{2} +t_{3} = 9660s + 1380s + 12660s = 23700s (4)[/tex]
In order to find the total distance traveled, we need to add the distance traveled before stopping at the gas station (x₁) and the distance traveled after leaving it (x₂).Applying the definition of average speed, we can find these distances as follows:[tex]x_{1} = v_{1} * t_{1} (5)[/tex]
[tex]x_{2} = v_{2} * t_{3} (6)[/tex]
where v₁ = 107.0 km/h, and v₂= 67.0 km/hAs we did with time, let's convert v₁ and v₂ to m/s:[tex]v_{1} = 107.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 29.7 m/s (7)[/tex]
[tex]v_{2} = 67.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 18.6 m/s (8)[/tex]
Replacing (7) and (1) in (5) we get x₁, as follows (in meters):[tex]x_{1} = v_{1} * t_{1} = 29.7 m/s * 9660 s = 286902 m (9)[/tex]
Doing the same for x₂ with (3) and (8):[tex]x_{2} = v_{2} * t_{3} = 18.6 m/s * 12660 s = 235476 m (10)[/tex]
Total distance traveled is just the sum of (9) and (10):[tex]x_{tot} = x_{1} +x_{2} = 286902 m + 235476 m = 522378 m (11)[/tex]
As we have already said, the average speed is just the quotient between (11) and (4), as follows:[tex]v_{avg} =\frac{\Delta x}{\Delta t} = \frac{522378m}{23700s} = 22.0 m/s (12)[/tex]
Converted back to km/h:[tex]v_{avg} = 22.0 m/s*\frac{1km}{1000m}*\frac{3600s}{1h} = 79.3 km/h (13)[/tex]
Need help on another homework question
Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?
A. blue
B. black
C. red
D. white
Answer:
a
Explanation:
Earth's magnetic field is approximately 1/2 gauss, that is 50 micro-tesla because the SI field unit of a tesla is 10,000 gauss. Earth's north geographic pole is close to its south magnetic pole, and magnetic field is directed from the north to the south poles of a magnetic dipole so it goes from Earth's south geographic pole towards its north. Suppose you have wire carrying a large DC current from the south wall of a building to its north wall and that it is horizontal, on the floor. If Earth's field is parallel to the ground and does not dip, what force if any would the wire experience
Answer:
F = 0
Explanation:
The magnetic force is described by two expressions
for a moving charge
F = q v x B
for a wire with a current
F = I L xB
bold indicates vectors
let's write this equation in module form
F = I L B sin θ
where the angle is between the direction of the current and the direction of the magnetic field
In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current
The magnetic field of the Earth goes from the south to the north and in this part it is horizontal
Therefore the current and the magnetic field are parallel, the angle between them is zero
sin 0 = 0
consequently the magnetic force is zero
F = 0
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attached to hanging blocks. The block attached to string 1 has a mass of 20 kg and the block attached to string 2 has a mass of M. Listeners hear a beat frequency of 2 Hz when string 1 is excited at its fundamental frequency and string 2 is excited at its third harmonic. What is one possible value for mass M
Answer:
2.18 kg
Explanation:
The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.
For string 1, its fundamental frequency f is when n = 1. So,
f = 1/2L√(T/μ) = 1/2L√(mg/μ)
Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f = 1/2L√(mg/μ)
f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)
f = 1/1 m√(196 kgm/s²/0.005 kg/m)
f = 1/1 m√(39200 m²/s²)
f = 1/1 m × 197.99 m/s
f = 197.99 /s
f = 197.99 Hz
f ≅ 198 Hz
For string 2, at its third harmonic frequency f' is when n = 3. So,
f' = 3/2L√(T/μ) = 3/2L√(mg/μ)
Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f' = 3/2L√(Mg/μ)
f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)
f' = 3/1 m√(M1960 m²/s²kg)
f' = 3/1 m√M√(1960 m²/s²kg)
f' = 3/1 m √M × 44.27 m/s√kg
f' = 132.81√M/s√kg
f' = 132.81√M Hz/√kg
Since the frequency of the beat heard is 2 Hz,
f - f' = 2 Hz
So, 198 Hz - 132.81√M Hz/√kg = 2 Hz
132.81√M Hz/√kg = 198 Hz - 2 Hz
132.81√M Hz/√kg = 196 Hz
√M Hz/√kg = 196 Hz/138.81 Hz
√M/√kg = 1.476
squaring both sides,
[√M/√kg] = (1.476)²
M/kg = 2.178
M = 2.178 kg
M ≅ 2.18 kg
A boat is moving in a river with a current that has speed vW with respect to the shore. The boat first moves downstream (i.e. in the direction of the current) at a constant speed, vB , with respect to the water. The boat travels a distance D in a time tOut . The boat then changes direction to move upstream (i.e. against the direction of the current) at a constant speed, vB , with respect to the water, and returns to its original starting point (located a distance D from the turn-around point) in a time tIn .
1) What is tOut in terms of vW, vB, and D, as needed?
2) What is tIn in terms of vW, vB, and D, as needed?
3) Assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is vB, the speed of the boat with respect to the water?
4) Once again, assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is tOut, the time it takes the boat to move a distance D downstream?
Answer:
Explanation:
Current has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat with respect to shore
vW + vB .
Distance travelled with respect to shore by boat = D
time ( tout ) = distance / speed with respect to shore
tOut = D / ( vW + vB )
When the boat travels upstream , its velocity with respect to shore
= ( vB - vW ) , vB must be higher .
tin = D / ( vB - vW )
3 ) tin = D / ( vB - vW )
170 = 120 / (vB - 0.3 )
(vB - 0.3 ) = 12 / 17 = .706
vB = 1.006 m / s
4 )
tOut = D / ( vW + vB )
= 120 / ( .3 + 1.006 )
= 92.26 s
Time taken by a body is ratio of the distance traveled by it to the speed.
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.What is upstream and downstream speed?
The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.
Given information-
The speed of the boat with respect to shore is [tex]v_w[/tex].
The speed of the boat in downstream with respect to water is [tex]v_B[/tex].
The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].
Time taken by a body is ratio of the distance traveled by it to the speed.
1) The net speed of the boat is upstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex]. Thus,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2) The difference of the speed of the boat is downstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{in}[/tex]. Thus,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,
3) The speed of the boat with respect to the water-Put the values in the formula obtains from the 2nd part of the problem,[tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]
Hence the speed of the boat with respect to the water is 1.006 m/s.
4) The time it takes the boat to move a distance D downstream-Put the values in the formula obtains from the 1st part of the problem,[tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]
Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.
Thus,
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.Learn more about the upstream and downstream speed here;
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Which of the following is the BEST explanation for why oceans have two different types of currents?
Answer:
sddww
Explanation:
szsswa
4. What is the acceleration of the car in each section?
b
с
d
a
Answer:
0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.
Explanation:
Plzz answer correctly
Bill is walking to the store and he walks the first 500m in 60s. He then runs 1000m in 90s. After stopping for 45s, he was the remaining 450m to the store in 50s. What is the average velocity for Bills entire
trip?
Answer:
letra A segundo o couculo a divisão e completa
Fill in the question
Earth’s atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by Earth’s atmosphere?
Answer:
Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.
Explanation:
The low-frequency speaker of a stereo set has a surface area of and produces 1W of acoustical power. What is the intensity at the speaker
Answer:
I = [tex]\frac{1}{4\pi \ r^2}[/tex]
we see the intensity decreases with the inverse of the distance squared
Explanation:
Intensity is defined as power per unit area,
I = P / A
in this case we have that the sound is emitted in a spherical form therefore the area is
A = 4 pi r2
therefore the intensity is
I = [tex]\frac{1}{4\pi \ r^2}[/tex]
as we see the intensity decreases with the inverse of the distance squared
Which describes increasing the efficiency of an energy resource?
switching from coal to solar
making cars that get better gas mileage
using biomass instead of coal
turning off the lights when leaving a room
Answer:
making cars that get better gas mileage
Answer: Making cars that get better gas
Explanation: Gradpoint
Explain the working and principle of perisocope.
Answer:
a periscope use total internal reflection to allow us to see things
the reflection happens at 45°
Explanation:
pls help me this is a major SOS pls help pls btw this is IXL
Explanation:
the object with the higher temperature has greater thermal energy
So the answer is
the stick of butter with less thermal energy.
Hope it will help :)
Answer:
The stick of butter with less thermal energy
Explanation:
I am pretty sure
convert 0.0345mW
to MW
Answer:
3.45e-11MV
that is ur answer
