For an object to move, a(n) _______ force must be applied. Question 1 options: Balanced Unbalanced

Answer:

 E = - dV / dx

Explanation:

The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.

The electric field and the potential are related by

          E = - dV / dx

therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.

Answer:

C. To draw on an earlier experience.

The question is incomplete, the complete question is;

how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3  m 2  and a plate separation of 2.00 mm.

The capacitance of the capacitor is the quantity of charge stored by the capacitor.

Given that;

C1= εo k * A/d

εo = permittivity of free space

C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3  m 2/2 * 10^-3

= 2.21 * 10^-11 F

C2 = 8.85 x 10-12 * 3 * 5.00×10 −3  m 2/2 * 10^-3

= 6.63 * 10^-11 F

q1 = C1V1 = 2.21 * 10^-11 C * 12 V

= 2.65 * 10^-10 C

q2 = C2V2 = 6.63 * 10^-11 F * 12 V

= 7.96 * 10^-10 C

qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C

qtotal = 1.061 * 10^-9C

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Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4

Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

Iron: 7.86 × 10 37.86 × 10 3

Oak: 7.10 × 10 27.10 × 10 2

Answer:

The distance is  [tex]y = 0.03425 \ m[/tex]

Explanation:

From the question we are told that

   The distance of separation is  [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]

   The wavelength is  [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]

    The distance of the screen is [tex]D = 2.14 \ m[/tex]

Generally the distance of a fringe from the central maxima is mathematically represented as

      [tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]

For the first dark fringe m = 0

             [tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]

             [tex]y_1 = 0.00855 \ m[/tex]

For the second dark fringe m = 1

            [tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]

            [tex]y_2 = 0.0257 \ m[/tex]

So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is

         [tex]y = y_1 + y_2[/tex]

        [tex]y = 0.00855 + 0.0257[/tex]

        [tex]y = 0.03425 \ m[/tex]

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.

MA= Effort Distance / Load Distance

Answer:

well its about our thinking but i do believe in ghost a little

Explanation:

Answer:

Real images are formed where rays of light actually converge, whereas virtual images occur with they are perceived to converge. Real images can be produced by passing light through converging lenses or with a concave mirror of some sort.

Explanation:

hope that helped a little :)

Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.

The converging lens which forms the real images by the intersection of light rays.

Real images are produced by light passing through converging lens or a concave mirror.

The real meeting of the rays coming from the focal point of the rays after reflection. Real images are formed where rays of light actually converge, whereas virtual images occur with they are appear to converge. Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.

Thus,  a convex lens produce real image from virtual object.

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I believe the correct answer is a

The true statement is that, at resonance, current is in phase with the driving voltage.

Resonance occurs in an RLC circuit at the point when the inductive reactance is equal to the capacitive reactance. At the point of resonance, the voltage across the inductor  and capacitor are equal.

It should noted that, at resonance, the current is in phase with the driving voltage. This is the true statement in the list.

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Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

The magnitude of the electric field is 1.25 N/C

But before that the following calculations need to be done.

ε = LB.v = 0.01 m × 5 T × 1 m/s

= 0.05 V

Now

ε = ∫E.ds

here ε = Eds because E is always parallel to the side of the cube

So,

= E∫ds  ∫ds

= 4L so we have 4 sides

Now

= E(4L)

= 4EL

So,4EL = 0.05 V

Now

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

hence, The magnitude of the electric field is 1.25 N/C

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Answer:

18.5 cm

Explanation:

From;

1/u + 1/v = 1/f

Where;

u= object distance = 86cm

image height = 23 cm

Radius of curvature = 37 cm

The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.

Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm

Therefore, the focal length of the mirror is 18.5 cm

Answer:

W = 122.3 J

Explanation:

First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁(A₂/A₁)

where,

F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)

F₁ = 14700 N

F₂ = Force applied to smaller piston = ?

A₁ = Area of larger piston = πd₁²/4

A₂ = Area of smaller piston = πd₂²/4

Therefore,

F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]

F₂ = (14700 N)(d₂²/d₁²)

where,

d₁ = diameter of large piston = 50 cm

d₂ = diameter of small piston = 4 cm

Therefore,

F₂ = (14700 N)[(4 cm)²/(50 cm)²]

F₂ = 94.08 N

Now, for the work done on the car:

Work Done = W = F₂ d

where,

d = displacement of car = 1.3 m

Therefore,

W = (94.08 N)(1.3 m)

W = 122.3 J

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

Answer:

the correct statement is

* atoms cannot be created or destroyed

Explanation:

The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.

Based on the previous descriptive, the correct statement is

* atoms cannot be created or destroyed

Answer:

the Answer is b hope it help

Explanation:

Answer:

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Explanation:

Given:

Mass of truck(m) = 2,100 kg

Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s

Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s

Find:

Change in kinetic energy (ΔKE)

Computation:

Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]

Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]

Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]

Change in kinetic energy (ΔKE) = 1,050[121.9648]

Change in kinetic energy (ΔKE) = 128063.04

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Answer:

[tex]\Large \boxed{\sf kilo}[/tex]

Explanation:

kilo is a prefix that means [tex]1000[/tex] of the base unit.

Answer:

kilo is the correct answer

Explanation:

because my exam says sooo....

Answer:

d.

Explanation:

Since, the capacitance( decreases )

therefore voltage between the plates(increases ).

Hence, option d is correct.

C =εA/d.

d is doubled, therefore  C decrease ( inverse relation).

D) The capacitance decreases and the voltage between the plates increases.

A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly, the capacitance decreases and the voltage between the plates increases.

The capacitance - (decreases)

The voltage between the plates- (increases ).

Thus, the correct answer is D.

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Answer:

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Explanation:

In this problem we have to be clear about the difference between displacement and distance.

The displacement is a vector, that is, it has a modulation and direction, in this case we can draw a vector for the outward trip and another vector for the return trip, both will have the same magnitude, but their directions are opposite, so the resulting vector is zero.

The distance is a scalar and its value coincides with the modulus of the distance vector, in our case the distance is d for the outward journey and d for the return journey, so the total distance is 2d, which is different from zero.

The two students have some reason, but neither complete,

The displacement is zero because it is a vector and

the distance is different from zero (2d) because it is a scalar

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Therefore I agree with both, because each one has a 50% of the reason

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

We're given

[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]

so let's see if we can find a closed form for the n-th term's coefficient.

Notice that

[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]

If the pattern continues, the next few terms are likely

[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]

which leads up to the n-th term,

[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]

where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.

So we have

[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]

Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives

[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]

Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]

but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]

Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :

[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]

This equation is separable:

[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]

From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find

[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{C_1}{x^3} \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2[/tex]

so that you end up with

[tex]\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)[/tex]

But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is

[tex]\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}[/tex]

You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

Answer:

a,)3.042s

b)4.173s

c)3.281s

Explanation:

For a some pendulum the period in seconds T can be calculated using below formula

T=2π√(L/G)

Where L = length of pendulum in meters

G = gravitational acceleration = 9.8 m/s²

Then we are told to calculate

(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?

Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then

use G = 9.8 + 3.0 = 12.8 m/s²

Period T=2π√(L/G)

T= 2π√(3/12.8)

T=3.042s

b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?

G = 9.8 – 3.0 = 6.8 m/s²

T= 2π√(3/6.8)

T=4.173s

C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?

Net acceleration is

g'= √(g² + a²)

=√(9² + 3²)

Then period is

T=2π√(3/11)

T=3.281s

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

Answer:infrared radiation

Explanation:

Infrared radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.

If we arrange   electromagnetic wave with decrease in wavelength, we get:

Radio waves > microwave >  Infrared >  Visible light > Ultraviolet > X-rays > Gamma radiation.

Hence,  Infrared  radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

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Answer:

Explanation:

The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;

B is the magnetic field (in Tesla)

I is the current (in amperes)

L is the length of the wire (in meters)

∝ is the angle that the conductor makes with the magnetic field.

Given parameters

L = 0.56 m

I = 2.6A

F = 0.24N

∝  = 90°

Required

magnitude of the magnetic field (B)

Substituting the given values into the formula given above we will have;

F = BILsin∝

0.24 = B * 2.6 * 0.56 sin90°

0.24 =  B * 2.6 * 0.56 (1)

0.24 = 1.456B

1.456B = 0.24

Dividing both sides by 1.456 will give;

1.456B/1.456 = 0.24/1.456

B ≈ 0.165Tesla

Hence the magnitude of the magnetic field is approximately 0.165Tesla

Answer:

Yes, yes it would

Explanation: