Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
[tex]a_{c}[/tex] = rω²
ω² = [tex]a_{c}[/tex] / r
ω = √( [tex]a_{c}[/tex] / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex] = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
The required revolution per hour is 28.6849
Calculation of revolution per hour:
The expression for the linear acceleration with respect to the angular velocity is
= rω²
So,
ω² = / r
ω = √( / r )
Here r represent the radius of the cylinder
ω represent the angular velocity
Now
diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
And,
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
So,
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
Now
1 rad/s = 9.5493 revolution per minute
So,
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
Now
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Learn more about the cylinder here: https://brainly.com/question/17262276
Plzz answer correctly
Fill in the question
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.
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]
Learn more about the speed of the object here;
https://brainly.com/question/4931057
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
All magnetic fields result from the movement of
A. charged particles
B. electrons only
C. protons only
D. neutrons only
|:Give one word answer
1. An object that allows whole light to pass through it _______
Answer:
Translucent object
Explanation:
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
To learn more about Molecular mass refer to:
https://brainly.com/question/24727597
#SPJ2
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
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
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.
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]
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.
Know more about redox reaction here
https://brainly.com/question/13293425
#SPJ2
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C/m^2 and the bottom plate has a surface charge density of -10C/m^2. Find the total charge on each plate. Find the electric field at the point exactly midway between the plates. Find the electric potential between the two plates. If an electron was in the middle the two plates, find the force on it.
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
[tex]V_{tot} = V_{q1} + V_{q2}[/tex]
[tex]V_q = \dfrac{k \cdot q}{r}[/tex]
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]
The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0
3) The electric field, 'E', between plates is given as follows;
[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e
∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N
Q4. (a) An acre-foot is the volume of water that would cover 1 acre of flat land to a depth of 1
foot. How many gallons are in 1 acre-foot?
Answer:
326,000
Explanation:
One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.
*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
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:
Animals conduct_______.
A. cellular respiration
B. photosynthesis
C. both cellular respiration and photosynthesis
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)
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
convert 0.0345mW
to MW
Answer:
3.45e-11MV
that is ur answer
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]
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:
A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
0.248
Explanation:
Initial speed u = 13.6
Final speed v = 0
Distance s = 38.1
We have umg = ma
We make u subject of the formula
u = a/g
V² = u² + 2as
a = v²-u²/2s
We substitute the values into the above
a = 0-(13.6)²/2*38.1
a = 184.96/76.2
a = 2.427m/sec
Remember that
u = a/g
u = 2.427/9.8
= 0.2476
This is approximately 0.248
This is the minimum coefficient of friction required to keep the crate from sliding.
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:
A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change
Complete question is;
A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change?
Answer:
182.84 %
Explanation:
Formula for rate of heat transfer of an infinite log fin is given as;
q_f1 = (π/2) × (hk)^(½)) × D^(3/2)) × (T_b - T_∞)
Where D is diameter.
Now, if the diameter of the rod is doubled, it means Diameter is now 2D.
Thus;
q_f2 = (π/2) × (hk)^(½)) × (2D^(3/2)) × (T_b - T_∞)
To find how much the rate of heat removal will change, we will calculate as follows;
((q_f2/q_f1) - 1) × 100
Plugging in the relevant expressions, we have;
([[(π/2) × (hk)^(½)) × ((2D)^(3/2)) × (T_b - T_∞)]/[(π/2) × (hk)^(½)) × (D^(3/2)) × (T_b - T_∞)]] - 1) × 100
Upon simplifying, we have;
(((2D)^(3/2))/(D^(3/2)) - 1) × 100
((2^(3/2)) - 1) × 100
This gives;
182.84 %
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;
https://brainly.com/question/800251
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.
Which of the following helps prevent and cope with heat-related conditions?
Drinking water
Wear proper clothing
Rest frequently
all of the above
An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.
What is heat?Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.
This ultimately implies that, heat is most likely to cause dehydration and high body temperature.
In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.
Read more on heat here: brainly.com/question/12072129
