What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L

Answers

Answer 1

Answer:

2.78 cm³/s

Explanation:

From the question,

Q = v/A'.................... Equation 1

Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.

Given: v = 100 km/h, A' = 10 km/L

Substitute this value into equation 1

Q = 100/10

Q = 10 L/h.

Now, we convert L/h to cm³/s.

Since,

1 L = 1000 cm³, and

1 h = 3600 s

Therefore,

Q = 10(1000/3600) cm³/s

Q = 2.78 cm³/s


Related Questions

a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resistance r/9 what is the diameter of the second wire g

Answers

Answer:

d₂ = 3dThe diameter of the second wire is 3 times that of the initial wire.

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense.
(a) For points OUTSIDE the loop, the magnetic field caused by this current:________.
a. points straight up.
b. circles the loop in a clockwise direction.
c. circles the loop in a counterclockwise direction.
d. points straight down.
e. is zero.
(b) For points INSIDE the loop, the magnetic field caused by this current:________.
a. circles the loop in a counterclockwise direction.
b. points straight up.
c. points straight down.
d. circles the loop in a clockwise direction.
e. is zero

Answers

Answer:

D &B

Explanation:

Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force

g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision

Answers

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

with a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.​

Answers

Answer:

120000    kgxm/s

Explanation:

momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000  kgxm/s

Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?

Answers

Answer:

The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.

A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter of 50 cm. The speed of the top of the wheel is

Answers

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]

Then the addition of these two velocities equals:

[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?

Answers

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?

Answers

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
A. The time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels

Answers

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

[tex]R = \frac{MV}{qB}[/tex]

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

[tex]a_c = \frac{V^2}{R}[/tex]

where;

[tex]a_c[/tex] is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

[tex]T = \frac{2\pi M}{qB}[/tex]

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]

Therefore, the correct options are "C" and "D"

A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true?

a. the mass is initially released from rest from a point 1 meter below the equilibrium position
b. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 11 m/s

Answers

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

[tex]x^\prime[/tex] would thus represent the velocity of this mass.[tex]x^{\prime\prime}[/tex] would represent the acceleration of this mass.

Constructing the ODE

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

The restoring force of the spring: [tex]F(\text{spring}) = -k\, x[/tex], where [tex]k[/tex] denotes the constant of this spring.The damping force: [tex]F(\text{damping}) = - 11\,x^\prime[/tex] according to the question. Note the negative sign in this expression- the damping force should always oppose the direction of motion.

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Size of this mass: [tex]m = 1\; \rm kg[/tex].Spring constant: [tex]k = 18\; \rm N \cdot m^{-1}[/tex].

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

Finding the general solution to this ODE

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

[tex]m_1 = -2[/tex], and[tex]m_2 = -9[/tex].

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Finding the particular solutions to this ODE

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

The position at time [tex]t = 0[/tex] would be [tex]x(0) = c_1 + c_2[/tex].The velocity at time [tex]t = 0[/tex] would be [tex]x^\prime(0) = -2\, c_1 - 9\, c_2[/tex].

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Which two types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light?

radio waves and infrared light
visible light and X-rays
microwaves and gamma rays
gamma rays and X-rays

Answers

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

WHAT ARE ELECTROMAGNETIC WAVES?

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Radio wavesInfraredUltravioletVisible lightX-raysGamma raysmicrowave

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Learn more about electromagnetic waves at: https://brainly.com/question/8553652

Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have

Answers

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Will give brainliest ASAP! Please help (1/10 questions, will mark 5 stars and brainliest for all answers if correct)

Answers

Answer:

Option (A)

Explanation:

A 20 kg boy chases the butterfly with a speed of 2 meter per second.

Angle at which he runs is 70° North of West.

Therefore, Horizontal component (Vx) directing towards West will be,

Vx = v(Cos70°)

Vy = v(Sin70°)

Since momentum of a body is defined by,

Momentum = Mass × Velocity

Therefore, Westerly component of the momentum will be,

Momentum = 20 × (v)(Cos70°)

                   = 20 × 2Cos70°

                   = 13.68

                   ≈ 13.7 kg-meter per second

Therefore, Option (A) will be the answer.

7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?

Answers

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

9.09 × 10³ capacitors must be connected in parallel.

How to calculate the number of capacitors connected in parallel?

Given C = 1.00μF = 1 × 10⁻⁶ F

          q = 1.00 C

          V = 110 V

The equivalent capacitance is given by

Ceq = q/V

where q = total charge on all the capacitors

             V = potential difference

For N number of identical capacitors in parallel,

Ceq = NC

Therefore,

NC = q/V

N = q/VC

Putting on the values in the above formula,

N = 1/ (110)(1 × 10⁻⁶)

   = 1 / 110 × 10⁻⁶

   = 9.09 × 10³

Learn more about the capacitors here:

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Assume that a lightning bolt can be represented by a long straight line of current. If 15.0 C of charge passes by in a time of 1.5·10-3s, what is the magnitude of the magnetic field at a distance of 24.0 m from the bolt?

Answers

Answer:

The magnitude of the magnetic field is 8.333 x 10⁻⁷ T

Explanation:

Given;

charge on the lightening bolt, C = 15.0 C

time the charge passes by, t = 1.5 x 10⁻³ s

Current, I is calculated as;

I = q / t

I = 15 / 1.5 x 10⁻³

I = 10,000 A

Magnetic field at a distance from the bolt is calculated as;

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷

I is the current in the bolt

r is the distance of the magnetic field from the bolt

[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T

What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?

Answers

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

wrench is to Hammer as ​

Answers

Answer:

Pencil is to pen

Step by step explanation:

They are similar items, as they are both tools, but are different as to how they function.

Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.

Answers

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.

Answers

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

 

The formula of the pressure of the static fluid

P1 = P2+ρgh

Where,

P1 -  pressure at depth 'h'

P2 -  atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] =  1670900 Pa

h - Depth = 160m  

ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]

g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]

The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]  

 

[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]  

 

 For an ideal gas,  

PV =nRT  

[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]  

[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]

So,

 

[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]  

Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

To know more air bubble volume,

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An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg

Answers

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

23.15. Can an object carry a charge of 2.0 10-19 C?​

Answers

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)

Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?

Answers

Answer:

Explanation:

Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction ,  so their relative velocity

= u + v = 2.3 m /s ( given )

We shall apply conservation of momentum law for the movement of astronaut and metal plate

mu  = M v where m is mass of astronaut , M is mass of metal plate

71 u = 230 x v

71 ( 2.3 - v ) = 230 v

163.3 = 301 v

v = .54 m / s

u = 1.76 m / s

honeycomb will be at rest  because honeycomb surface  is frictionless . Plate will slip over it . Over plate astronaut is walking .

a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s

b ) velocity of astronaut relative to honeycomb will be + .54 m /s

Here + ve direction is assumed to be the direction of astronaut .  

what is electric field strength​

Answers

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

Please mark me as the brainliest!!!

Thanks!!!

For every particle there is a corresponding ______________.

Answers

Answer:

Anti-Particle

Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m

Answers

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.

Answers

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum

Answers

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Which of the following represents a concave mirror? +f,-f,-di,+di

Answers

Answer:

fully describes the concave mirror is + f

Explanation:

A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f

This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di

Answer:

+f

Explanation:

because you have to be really dumb to get an -f

What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c

Answers

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the work done by the force?

Answers

Answer: W.D = 1/2mv^2

Explanation:

If an external force or a single force is acting on a body. Just like the first law of thermodynamics, the force acting on the body will cause work done on the system.

Work done = force × distance

And the work done on the body will cause the molecules of the body to experience motion and thereby producing kinetic energy.

The work done will be converted to kinetic energy.

W.D = 1/2mv^2

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