Calculate the force necessary to accelerate a 20,000-lbm rocket vertically upward at a rate of 100 ft/sec2. Assume g = 32.2 ft/sec2.

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

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.

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:

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:

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:

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:

A) P(t) = 2651.25 [ 1 - cos2wt ] W

B)  Real power = 999.79 watts

    Reactive power = 2652.86 VA

c) power factor = 0.3526

Explanation:

Given data:

V(t) = 141.4 cos (ωt)

R(t) = 10 Ω

Inductive reactance XL = ωL = 3.77 Ω

Ir(t) = V(t) / R(t) = 14.14

A) Calculate the instantaneous power absorbed by the resistor and by inductor

By resistor :

Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]

      hence Pr  = 999.698 (cos2ωt + 1) w

By Inductor :

Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)  

                        =  5302.5 [tex]sin^2 wt[/tex]

Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex]   w  =  2651.25 [ 1 - cos2wt ] W

B) calculate the real and reactive power

First we have to determine the power factor

Given that : V(t) = 141.4 cosωt  v ,   Ir(t) = 14.14 cosωt A

IL(t) = 37.5 cos (ωt - 90° )

The phasor representation of the above is :

V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° ,  Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°

Total load current = Ir + IL = 28.35 ∠ -69.35°

power factor = cos -69.35° = 0.3526

Next we will determine the Real and reactive power using the relation below

S = VI = 100 ∠ 0°  * 28.35 ∠ -69.35°

         = 2835 ∠ 69.35°

S = P + jQ = 999.79 + 2652.85 j

Real power  = 999.79 watts

Reactive power = 2652.85 VA

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.

Learn more about motion of satellite here: https://brainly.com/question/25721729

#SPJ2

Answer:

define d first?

you need to list more variables

Answer:

list more valuable unit

Answer:

0.13 m/s

Explanation:

Unfortunately, I don't have an explanation but I guessed the correct answer.

Answer:

F= m x a

Explanation:

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

Answer:

467 volts

Explanation:

Vs/Vp = Ns/Np

Vs = Ns/Np × Vp

Vs = 350/150 × 200 = 7/3 × 200

Vs = 467 volts

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:

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%

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:

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.

Read more on kinetic energy here: https://brainly.com/question/1242059

Answer:

Answer is explained in the explanation section below.

Explanation:

In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.

a) You pass through the highest point:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.

And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.

b) You pass through the lowest point of the ride:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.

And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid

Explanation:

I don't know if I answered correctly, if not I can provide another answer

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:

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:

[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:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

[tex]V_{A}[/tex] = 1 m³

[tex]T_{A}[/tex] = 10°C = 283 K

[tex]P_{A}[/tex] = 350 kPa

[tex]m_{B}[/tex] = 3 kg

[tex]T_{B}[/tex] = 35°C = 308 K

[tex]P_{B}[/tex] = 150 kPa

Now, lets apply the ideal gas equation;

[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]

[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150

[tex]V_{B}[/tex] = 265.188 / 150  

[tex]V_{B}[/tex] = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, [tex]m_{A}[/tex] =  [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221

[tex]m_{A}[/tex]  = 4.309 kg

Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg

Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex]  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

[tex]P_{f}[/tex] =  [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]

given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K

we substitute

[tex]P_{f}[/tex] =  ( 7.309 × 0.287 × 293)  / 2.77

[tex]P_{f}[/tex] =  614.6211119 / 2.77

[tex]P_{f}[/tex] =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Answer:

4. unbalanced and Accelerating

5. balance and rest

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

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.