A light bulb operating at 110 V draws 1.40 A of current. What is its resistance?

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:

Propulsion system, antigravitational tech

Explanation:

Fuel is extremely inefficient and expensive not to mention it weighs a lot. You really only need to reach escape velocity to leave earth. The rest is just a little amount of boosting to alter course and slow down for landing. I couldn't really think of much. Once we have an antigravitational system then you could say the whole rocket is holding you back because the design would be different. Nobody really knows how to defy gravity but that would be a technolgical limitation for sure.

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:

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:

Explanation:The Mass Of The Left Block M1 = 1.3 Kg And The Mass Of The Right Block M2 = 3.1 Kg. The Angle Between The String And The Horizontal Is ... (10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a ... m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the ...

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:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, [tex]v_s[/tex] = d/t₂

∴ [tex]v_s[/tex] = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, [tex]v_s[/tex] ≈ 5.79 m/s.

Answer:

Eat slowly

Explanation:

If you eat slower, you'll chew your food better, which leads to better digestion. Digestion actually starts in the mouth, so the more work you do up there, the less you'll have to do in your stomach. This can help lead to fewer digestive problems. Less stress.

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:

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:

 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]

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

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: 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: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:

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

Answer:

A.

Explanation:

Answer: A

Explanation:

Answer:

electric current passing through it will be 50mA

Explanation:

electric current = charge / time

current = 0.05 A = 50mA

If 0.5C charge passes through a wire in 10 seconds, then 50mA current is flowing through the wire. Thus, the correct option is C.

Electric current is the flow of electricity in an electronic circuit. It is the amount of electricity flowing through a electronic circuit. It is generally measured in amperes (A). The larger the value in amperes, the more electricity is flowing in that circuit.

The formula for calculation of Electric current is:

I = Q/T

where, I = electric current,

Q = amount of charge,

T = time required

Therefore, the current flowing in the wire is:

I = 0.5C/ 10 seconds

I = 0.05 A or 50mA (1mA = 10⁻³A)

Therefore, the correct option is C.

Learn more about Electric current here:

https://brainly.com/question/2264542

#SPJ6

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:

kekemeeimdeiddnekem

Explanation:

mdjdjdiddmjd jjeneeiej

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

Answer:

I hope this helps.

Explanation:

Answer:

It's A

Explanation:

sound waves are longitudinal they need a medium to travel through

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:

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:

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:

1122.8

Explanation:

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

=1122.786

Rounded=1122.8