The next four questions refer to the situation below.
A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS , with respect to the water. The swimmer travels a distance D in a time tOut . The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS , with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn .
What is tOut in terms of vR, vS, and D, as needed?

Answer: read this hope this helped

Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.

Answer:

Explanation:

I hope it help you;)

Answer:fastest,same,slow down,opposite,slow

Explanation:

A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.

If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).

While moving towards the Earth (along the path from D to A) there is a component of force in the  same direction as the motion; this causes the satellite to speed up.

While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.

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#SPJ2

Answer:

The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".

Explanation:

Formula for calculating the mass in He:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]

Formula for calculating the mass in [tex]N_2[/tex]:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]

by using the temperature balancing the equation:

[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]

    [tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.

Solution:

The relation between electric field and the electric potential is:

E = [tex]\frac{dV}{dr}[/tex]

So, making dV the subject, we have:

dV = E x dr

Integrating the above equation, we get.

V = [tex]\int\limits^_ {} \,[/tex]E x dr      Equation 1

Now, in 2-D

E is inversely proportional to the radius r.

E ∝ 1/r

So, we can write: replacing E ∝ 1/r in the equation 1

V ∝  [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr

Which implies that,

V ∝  log (r)

Hence, distance dependence expected for the electric potential =  ln (r)

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

Explanation:

By using conservation of linear momentum and also by equating work done to kinetic energy,  [tex]V_{2}[/tex]  = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters

Parameters given are :

[tex]m_{1}[/tex] = 19.5 kg

friction, μk = 0.35

distance d = 2.6 m

mass [tex]m_{2}[/tex] = 8.25 kg.

initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.

a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0

Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.

The formula to use will be :

[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]

Substitute all the parameters into the formula above

19.5 x 6.5 = 8.25[tex]V_{2}[/tex]

Make [tex]V_{2}[/tex] the subject of formula

[tex]V_{2}[/tex] = 126.75/8.25

[tex]V_{2}[/tex] = 15.36 m/s

b.) Let us first calculate the work done in by block one.

The K.E = [tex]1/2mU^{2}[/tex]

substitute its mass and velocity into the formula

K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]

K.E = 411.94 Joule

The work done = Kinetic energy

But the resultant Force F = force f - friction

where Frictional force = 0.35 x 19.5 x 9.8

Frictional force = 66.89N

Work done will be the product of resultant Force F and the distance travelled

(F - 66.89) x 2.6 = 411.94

F - 66.89 = 411.94/2.6

F - 66.89 = 158.44

F = 225.3 N

The second block will experience the same force which is equal to 225.3N

Find the kinetic energy of the second block.

K.E =  [tex]1/2mV^{2}[/tex]

K.E = 0.5 x 8.25 x 15.36^2

K.E = 973.2

Using The work done = Kinetic energy

225.3[tex]d_{2}[/tex] = 973.2

[tex]d_{2}[/tex] = 973.2/225.3

[tex]d_{2}[/tex] = 4.32 meters

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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

Answer:

d

Explanation:

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.

Answer:

The magnitude of the potential difference is 900 V.

Explanation:

Given;

mass of the bead, m = 4.0 g = 0.004 kg

charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C

final velocity of the bead, v = 3 m/s

What is the magnitude of the potential difference V?

Apply the principle of conservation of energy;

The electric potential energy at the beginning is equal to kinetic energy of the bead  at the end of the journey.

qV = ¹/₂mv²

[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]

Therefore, the magnitude of the potential difference is 900 V.

The magnitude of the potential difference (V) is equal to 900 Volts.

Given the following data:

To determine the magnitude of the potential difference (V):

We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:

[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]

V = 900 V.

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Answer:

F= m x a

Explanation:

Force (f) = mass (m) x acceleration (a)

Answer:

1767Kw

Explanation:

Velocity of wind = 10 m/s

diameter of the blades= 60m

ρ= air density = 1.25 kg/m3

Acceleration due to gravity= 9.81 m/s^2

Mechanical energy of the wind can be calculated using the expression below

Energy= (e*m)

= ρ V A e............eqn(1)

Where A= area

ρ= air density

e= wind energy per unit mass of air

e= (v^2)/2..........eqn(2)

If we substitute the values into eqn (2) we have

e= [(10)^2]/2

=50J/Kg

But Area=A= (πd^2)/4

Area= ( π× 60^2)/4

Area=2827.8m^2

If we input substitute the values into eqn (1) we have

Energy= 1.25 ×10 × 50×2827.8

=1767145.7W

We can convert to kilo watt

=1767145.7W/ 1000

= 1767Kw

Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw

Answer:

2.5 m/s^2

Explanation:

First, convert 90 km/hr into m/s:

90/3.6 = 25 m/s

vf = final velocity = 25 m/s

vi = initial velocity = 0 m/s

t = time = 10 seconds

a = acceleration, unknown

Then, find a using the following equation:

(vf - vi)/t = a

(25 m/s)/10 s = 2.5 m/s^2

a = 2.5 m/s^2

Hope this helps!! :)

Answer:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Explanation:

Given

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]

Point: [tex](f(t0), g(t0), h(t0))[/tex]

[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information

Required

Determine the parametric equations

[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]

Differentiate with respect to t

[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]

Let t = 1 (i.e [tex]t0 = 1[/tex])

[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + k[/tex]

To solve for x, y and z, we make use of:

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]

This implies that:

[tex]r'(1)t = xi + yj + zk[/tex]

So, we have:

[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]

[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]

By comparison:

[tex]xi = it[/tex]

Divide by i

[tex]x = t[/tex]

[tex]yj = \frac{1}{3}jt[/tex]

Divide by j

[tex]y = \frac{1}{3}t[/tex]

[tex]zk = kt[/tex]

Divide by k

[tex]z = t[/tex]

Hence, the parametric equations are:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Answer:

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Explanation:

Hope it is helpful.....

Answer:

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

AnswerHM???

Explanation:

I dONT KNOW

Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.

Answer:

86.33m/s^2

Explanation:

Acceleration = Force/Mass

= 25.9/0.3

= 86.33

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

Answer:GAS

Explanation:

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

Answer:

We conclude that the total distance traveled by the object in the 4 seconds is 36 m.

Explanation:

Given

To determine

The total distance traveled by the object in the  4.0 seconds is

Important Tip:

We can determine the total distance traveled by the object in the  4.0 seconds by using the equation of motion such as

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

substituting u = 5.0, a = 2, and t = 4 in the formula

[tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]s=\left(5\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(4\right)^2[/tex]

[tex]s=20+16[/tex]

[tex]s=36[/tex] m

Therefore, we conclude that the total distance traveled by the object in the 4 seconds is 36 m.

The total distance traveled by the object is 36 meters.

Given the following data:

To find the total distance traveled by the object, we we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2}at^2[/tex]

Where:

S is the total distance traveled.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

[tex]S = 5(4) + \frac{1}{2}(2)(4^2)\\\\S = 20 + 1(16)[/tex]

Total distance, S = 36 meters.

Therefore, the total distance traveled by the object is 36 meters.

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Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

Answer:

Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.

Answer:

The answer is below

Explanation:

The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m

The resistance (R) of transmission line is given as:

Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4

[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]

Power = [tex]\frac{V_L^2}{R_L}[/tex]

Power = 10 MW = 10 * 10⁶ W

[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]

[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]

a) Since there are two tranmission lines, the power consumed by the lines is:

[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]

b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W

Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%