determine the total current on the circuit bellow

[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{Total\: Displacement }{Total\: time}[/tex]

[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{4}{2}=2m/s[/tex]

Answer:

i think its D or B

Explanation:

i just think

On Earth,in atmosphere the hydrogen is found in the less quantity. Option C is correct.

The chemical element hydrogen has the atomic number 1 and the symbol H. The smallest element is hydrogen.

Hydrogen is found in large numbers on Earth in combination with other elements, such as water and hydrocarbons, yet it is only 0.00005 percent present in Earth's atmosphere.

On Earth,in atmosphere the hydrogen is found in the less quantity.

Option C is correct.

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Answer:

F-F(gr) = ma

a= {F-F(gr)}/m =

=(15-10)/15=0.33 m/s² (upward)

Answer

An Achievement test is an assessment of developed knowledge or skill. ... Achievement tests are developed to measure skills and knowledge learned in a given grade level, usually through planned instruction, such as training or classroom instruction. Achievement tests are often contrasted with aptitude tests.

Explanation:

Answer: Even though the genetic susceptibility plays a crucial role on the IQ of the individual, various modifiable environmental factors like education, premature birth, nutrition, pollution, drug and alcohol abuse, mental illnesses, and diseases can have an influence on an individual's IQ.

Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

The magnitude of the force that must be applied to push the book across the desk is 3.97 N.

The given parameters;

The acceleration of the book across the desk is calculated as follows;

a = μg

where;

a = 0.3 x 9.8

a = 2.94 m/s²

The magnitude of the force that must be applied is calculated as follows;

F = ma

F = 1.35 x 2.94

F = 3.97 N

Thus, the magnitude of the force that must be applied to push the book across the desk is 3.97 N.

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Answer:

30.3 meters, 172 degrees

Explanation:

To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.

Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West

Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North

The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.

Applying the Pythagorean theorem leads to the magnitude of the resultant (R).

R2 = (30.0 m)2 + (4.0 m)2 = 916 m2

R = Sqrt(916 m2)

R = 30.3 meters

The angle theta in the diagram above can be found using the tangent function.

tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)

tangent(theta) = 0.1333

theta = invtan(0.1333)

theta = 7.59 degrees

This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.

Explanation:

I think this is it, give it a try

Answer:

anlatamadım herkeze ben sorulara bakamiyorum cumku analmaiyirum yardim edin

bu sorunu nasil cozebilirim

lutfsn...

Answer: C.

Correct me if I'm wrong.

Answer:

138

Explanation:

(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)

1) find the partial derivative of each:

[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]  

 [tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]

2) use partial integrals to make gradient of f:

take whatever you got from partial integral and add them together (if they repeat, just use it once)

[tex]F =[/tex] Vf (V = gradient of)

[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]

3) Evaluate the integrals with given points:

Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)

[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]

= 144 - 6 = 138 units of work

Work done by the force field F in moving an object from A to B = 138 J

Given data :

Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]  

Step 1 : determine the partial derivatives of the vector quantity

Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]

Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]

Equating the partial derivatives :  

[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]  = [tex]6xy^{\frac{3}{2} } + c[/tex]

therefore the gradient of F  i.e. F = vF  = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]

Next step : Determine the work done

Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ]  - [ F(point A) = F( 1,1 ) ]

F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex]  = 144

F( 1,1 )  = 6(1)(1)[tex]^{\frac{3}{2} }[/tex]    = 6

Therefore the work done by the force field = 144 - 6 = 138 J

Hence we can conclude that the work done by the force field F is = 138 J

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Answer:

Newton's second law states that .

Formulically

F=Force

m=mass

a=acceleration

The best example is hitting a tennis ball.

The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.

The magnetic field in the gap at a distance s < a can be computed by using the formula:

[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]

where;

[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]

where;

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]

Making the magnetic flux density the subject, we have:

[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]

[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]

Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]

[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]

Recall that distance in question is said to be (s);

∴

[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

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Answer:

watts is for real power

volt amperes reactive (VA) for reactive power

volt amperes (VA) for apparent power

Answer:

The axis of rotation of the Earth is tilted at an angle of 23.5 degrees away from vertical, perpendicular to the plane of our planet's orbit around the sun. The tilt of the Earth's axis is important, in that it governs the warming strength of the sun's energy.

Explanation:

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The internal energy after heating in terms of U is 100U.

The given parameters;

Assuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.

[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]

Thus, we can conclude that the internal energy after heating in terms of U is 100U.

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Answer:

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Explanation:

The behavior of an object that has been charged is dependent upon whether the object is made of a conductive or a nonconductive material. Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit charge to be transferred across the entire surface of the object. If charge is transferred to the object at a given location, that charge is quickly distributed across the entire surface of the object. The distribution of charge is the result of electron movement. Since conductors allow for electrons to be transported from particle to particle, a charged object will always distribute its charge until the overall repulsive forces between excess electrons is minimized. If a charged conductor is touched to another object, the conductor can even transfer its charge to that object. The transfer of charge between objects occurs more readily if the second object is made of a conducting material. Conductors allow for charge transfer through the free movement of electrons

The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is

∑ F = -f = ma = (4.00 kg) a

Assuming constant acceleration a , the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)

Solve for the acceleration :

a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²

Then the frictional force exerted a magnitude of

-f = (4.00 kg) (-14.8 m/s²)

f ≈ 59.4 N

and was directed opposite the block's motion.

Answer:

OK sure why not!!!!!!!!!!!!

Answer:

4653

4565445655687568677667876

Answer:

F=15N

Explanation:

F=m.a

m=1500g ÷1,000 = 1.5kg

a= 10m/s/s

F =1.5 × 10

F = 15 Newton

Answer:

Critical angle is a angle of incidence, for which refraction is 90 degree, this is a least angle of incidence at which total reflection takes place, please mark me Brainliest

Good morning dear...

Have a beautiful and joyful day ahead.

The center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.

The given parameters:

The center of gravity of the two particles when first particle is fixed is calculated as;

[tex]C_G = \frac{m(0 ) \ +2m(L) }{m+ 2m} \\\\C_G = \frac{2mL}{3m} \\\\C_G = \frac{2L}{3}[/tex]

The center of gravity of the two particles when second particle is fixed is calculated as;

[tex]C_G = \frac{m(L) \ + \ 2m(0)}{m + 2m} \\\\C_G = \frac{mL}{3m} \\\\C_G = \frac{L}{3}[/tex]

Thus, the center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.

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The potential difference between the initial and final point is 3.0 V.

The given parameters:

The potential difference between the initial and final point is calculated as follows;

[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]

where:

[tex]d_2[/tex] is the distance of the electron between the positive and negative plate

[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]

[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]

Thus, the potential difference between the initial and final point is 3.0 V.

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Answer:

Explanation:

All

Answer:

Explanation:

The work of the buoyancy force will convert to gravity potential energy

Fresh Water has a density of 1000 kg/m³

The ball has a density of 370 kg/m³

assume the ball is 1 m³

weight of the ball is 370g

The buoyancy force is 1000g

Assume the buoyancy force drops suddenly to zero when the center of the ball clears the water level.

The Work done on the ball is

W = Fd = 1000g(0.590)

the change in potential energy is

PE = mgh = 370g(0.590 + y)

where y is the height above the water level.

1000g(0.590) = 370g(0.590 + y)

1000(0.590) = 370(0.590 + y)

y = (0.590)(1000 - 370) / 370

y = 1.004594... ≈ 1.00 m

We have that for the Question "(a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor"  

Answer:

From the question we are told

In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field

A) When 86.6\% energy is stored in inductor

[tex]\%[/tex]of energy stored in electric field = [tex]1 - 0.866 = 13.4\%[/tex]

[tex]\frac{V_E}{V} = \frac{\frac{q^2}{2c}}{\frac{Q^2}{2c}} = 0.134\\\\\frac{q}{Q} = \sqrt0.134\\\\\frac{q}{Q} = 0.366\\\\q = 0.366Q_{max[/tex]

B)

[tex]\frac{V_B}{V} = \frac{\frac{Li^2}{2}}{\frac{LI^2}{2}} = 0.866\\\\\frac{i}{I} = \sqrt0.866\\\\\frac{i}{I} = 0.931\\\\i = 0.931I_{max[/tex]

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