Consider the circuit shown in (Figure 1). Assume E = 19 V.
A. What is the equivalent capacitance?
B. What is the charge on 3.0 μF capacitor?
C. What is the charge on 4.0 μF capacitor?
D. What is the charge on 6.0 μF capacitor?

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]

            cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]

            Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]

            Δx = 1.18 m

Answer:

A) F = 21.134 N

B) a = 3180.76 × 10^(24) m/s²

Explanation:

A) We are given;

Mass of alpha particle; m = 4.0026 u

Now, 1u = 1.66 × 10^(-27) kg

Thus; m = 4.0026 × 1.66 × 10^(-27)

Distance apart; r = 6.60 × 10^(−15) m

Charge on the alpha particle is;

q = 2e = 2 × 1.6 × 10^(-19) C

Formula for the force between the two alpha particles is;

F = kq1.q2/r²

k = 8.99 × 10^(9) N.m²/C²

q1 = q2 = 2 × 1.6 × 10^(-19) C

F = 8.99 × 10^(9) × (2 × 1.6 × 10^(-19))²/(6.60 × 10^(−15))²

F = 21.134 N

B) acceleration is given by;

a = F/m

Thus; a = 21.134/(4.0026 × 1.66 × 10^(-27))

a = 3180.76 × 10^(24) m/s²

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

Answer:

6. A

7. C

8. B

9. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Answer:

1. Protons.

2. Electrons.

3. Neutrons.

4. Molecules.

Explanation:

1. Protons: subatomic particles with a positive charge. They are bound together in the nucleus of an atom due to strong nuclear forces.

2. Electrons: subatomic particles with a negative charge. Electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

3. Neutrons: subatomic particles with no charge. The negative charge of the electrons cancels the positive charge of the protons.

4. Molecules: they are made of atoms.

Generally, molecules attach on the inside of a mineral to give it shape. Therefore, the molecule of a mineral is a crystal three-dimensional regular structure (arrangement) of chemical particles that are bonded together and determines its shape.

Due to the fact that these molecules are structurally arranged or ordered and are repeated by different symmetrical and translational operations they determine the shape of minerals.

Answer:

a. of liquid at the time bubbles first emerge slowly from the liquid.

Explanation:

Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.

Answer:

In my textbook's words-

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.

Explanation:

The correct definition of amplitude is that it is a maximum displacement

that occurs on a vibrating body from one point to the other.

The initial point of the wave is regarded as its equilibrium position which is

equal to one-half the length of the vibration path.

Amplitude helps to calculate the peak value of different types of waves

such as water waves and in electrical appliances so as to know the peak

current suitable for it.

Read more on https://brainly.com/question/21632362

Answer:

0.0119502868 kilocalorie

Explanation:

Answer:

increases by 50

Explanation:

Answer:

A.

Explanation:

Answer: A

Explanation:

Answer:

Left

Explanation:

Answer:

Your Left hand is negatively charged, and receives energy. It emits the energy that "allows things to happen''.

Explanation:

did some research

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.

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Answer:

[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]

Or

[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]

Explanation:

We are given that

Force between two charges=0.036 N=[tex]36\times 10^{-3}N[/tex]

Distance between two charges, r=2cm=[tex]2\times 10^{-2}[/tex]m

1m=100cm

Sum of two charges=[tex]10\mu C[/tex]

Let one charge=[tex]q_1=q\mu C=q\times 10^{-6}C[/tex]

[tex]q_2=(10-q)\times 10^{-6} C[/tex]

We know that

Electric force between two charges

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Where [tex]k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}[/tex]

Using the formula

[tex]36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}[/tex]

[tex]\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)[/tex]

[tex]0.0016=10q-q^2[/tex]

[tex]q^2-10q+0.0016=0[/tex]

[tex]10000q^2-100000q+16=0[/tex]

[tex]q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}[/tex]

Using the formula

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]q=9.999[/tex] and [tex]q=0.00016[/tex]

[tex]q_2=10-9.9998=0.0002[/tex]

[tex]q_2=10-0.00016=9.99984[/tex]

Hence, two charges are

[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]

Or

[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]

Answer:

A. pushing a car that isn't moving

Answer:

a)  W = 1.63 10⁻²⁸ J,  b)  W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d)  W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

          W = ΔU = U_∞ -U

the potential energy for point charges is

           U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]

in this case we only have two particles

           U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]

the distance is

           r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]

           r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)

           r₁₂ = √2= 1.4142 m

we substitute

           W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

            W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142

            W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

            U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]

in this case all charges are electrons

             q₁ = q₂ = q₃ = q

             W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]

the distances are

            r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0

            r₁₃ = 3

            r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2

            r₂₃ = √13

            r₂₃ = 3.606 m

let's look for the job

            W = U

let's calculate

            W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]

            W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

             W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]

all charges are equal q₁ = q₂ = q₃ = q₄ = q

             W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]

let's look for the distances

             r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]

             r₁₄ = 5 m

             r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]

             r₂₄ = √13 = 3.606 m

             r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]

            r₃₄ = 4 m

we calculate

           W = 9 10⁹ (1.6 10⁻¹⁹)²  [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]

           W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

            W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]

in this case the charges have the same values ​​but charge 5 is positive and the others negative, so the products of the charges give a negative value

            W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]

we look for distances

            r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2

            r₁₅ = √ 2 = 1.4142 m

            r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]

            r₂₅ = √2 = 1.4142 m

            r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]

            r₃₅ = √5 = 2.236 m

            r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]

            r₄₅ = √13 = 3.606 m

we calculate

           W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]

            W = - 4.93 10⁻²⁸ J

Answer:

a proton because it has a positive charge

Answer:

The answer is

B)  

Answer:

Nutrients the body needs in relatively small amounts are called micronutrients. They include vitamins and minerals. Vitamins are organic compounds that are needed by the body to function properly. ... Vitamins and minerals do not provide energy, but they are still essential for good health

Explanation:

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

Answer:I will say d

Explanation: because Potential energy is the energy stored within an object, due to the object's position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy.

Answer:

3 is the GCF for all these numbers if thats what you're asking

Answer:

1122.8

Explanation:

12.73 kg x 9.8 m/s^2 x 9m

=1122.786

Rounded=1122.8

Answer:

750 J

Explanation:

We have a student that pushes a 50N block  across the floor for a distance of 15m. The question is asking how much work was done to move the block.

To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS

F being the force and S being the distance.

W = FS

W = (50)(15)

W = 750

Therefore, 750 joules is our answer.

Answer:

9

Explanation:

Answer:

All answer are explained below in the explanation section.

Explanation:

1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.

As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.

Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.

2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.

Tire pressure at temperature 18.6 degree C is ~ 35 psi.

Temperature at air pressure of 37 psi is ~26.1 degree C

3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.

3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32

3. c) It is already done in part a of this question.

4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.

5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]

Plugging in the values, we get:

Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:

% Error = 21.33%

6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi

6.  b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.

6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.

7.  a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.

7.  b.) Estimation of temperature from best fit value of line is = 26.64 degree C

7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

[tex]v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\[/tex]

[tex]\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55[/tex]

Answer:

a. 1.93 s b. 3 s

Explanation:

a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?

Since the car accelerates from 0 to 30 mph in 1.00 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 30 mph and t = 1.00s = 1/3600 h

So, substituting the values of the variables into the equation, we have

a = (v - u)/t

a = (30 mph - 0 mph)/ 1/3600 h

a = 30 mph × 3600 /h

a = 108000 mph²

So, we find the time it takes the car to accelerate to 58 mph from 0 mph from

t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 108000 mph²

So, substituting the value of the variables into the equation, we have

t = (v' - u')/a

t = (58 mph - 0 mph)/108000 mph²

t = 58 mph/108000 mph²

t = 5.37 × 10⁻⁴ h

t = 5.37 × 10⁻⁴ × 3600 s

t = 1.93 s

b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

Since the car accelerates from 0 to 29 mph in 1.50 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 29 mph and t = 1.05s = 1/3600 h

So, substituting the values of the variables into the equation, we have

a = (v - u)/t

a = (29 mph - 0 mph)/ 1.5/3600 h

a = 29 mph × 3600/1.5 /h

a = 104400/1.5 mph²

a = 69600 mph²

So, we find the time it takes the car to accelerate to 58 mph from 0 mph from

t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 69600 mph²

So, substituting the value of the variables into the equation, we have

t = (v' - u')/a

t = (58 mph - 0 mph)/69600 mph²

t = 58 mph/69600 mph²

t = 8.33 × 10⁻⁴ h

t = 8.33 × 10⁻⁴ × 3600 s

t = 3 s