Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

Answer:

The resistance is 8.7 ohm.

Explanation:

Voltage, V = 110 V

mass, m = 1 kg

change in temperature, T = 100 - 20 = 80 C

time, t = 4 min = 4 x 60 = 240 s

specific heat, c = 4184 J/kg C

let the resistance is R.

The heat generated by the heater is used to the heat the water.

[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]

Answer:

960 nm

Explanation:

Given that:

wavelength = 640 nm

For the second (2nd) dark spot;  the order of interference m = 1

Thus, the path length difference is expressed by the formula:

[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]

[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]

[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]

dsinθ = 960 nm

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation:

Answer:

rock go down

Explanation:

what comes up must come down.

Answer:

0.01286

Explanation:

In a given single-slit, the central fringe (Y) is 360 times as wide as the slit (a). Then

2Y₁ = 360a

Y₁ = 360a/2

= 180a

The distance D = 14000a

In a given single-slit diffraction, the ratio = [tex]\dfrac{\lambda }{W}[/tex]

and since the angle is infinitesimally small;

sin θ ≅ tan θ = [tex]\dfrac{Y}{D}[/tex]

∴

For the first dark fringe;

Suppose:  [tex]\dfrac{a}{2}sin \theta = \dfrac{\lambda }{2}[/tex]

then,

[tex]\dfrac{a}{2} \ \dfrac{Y_1}{D} = \dfrac{\lambda }{2}[/tex]

[tex]aY_1 = \lambda D[/tex]

[tex]\dfrac{\lambda }{a} = \dfrac{Y_1}{D}\\ \\ \\ \implies \dfrac{180\ a}{14000 \ a} \\ \\ \mathbf{\dfrac{\lambda }{a} = 0.01286}[/tex]

Answer:

Explanation:

v = u + at

v₃ = 100 +30.0(3.00) = 190 m/s

s = vt + ½at²

y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m

x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m

a) v² = u² + uas  

s = (v² - u²) / 2a

ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m

b) t₁ = 3.00 s

   t₂ = (190sin53) / 9.80 = 15.5 s

   t₃ = √(2(1522) / 9.80) = 17.6 s

t = 3.00 + 15.5 + 17.6 = 36.1 s

c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m

Answer:

Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.

Answer:

θ' =  14.44 × [tex]10^{6}[/tex]

Explanation:

given data

total distance is d = 4820

radius = 1.4 cm

solution

we get here total angle by which the wheel rotates traveling is express as

⇒  [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex]     ................1

and

total angle (θ)  and the total distance (d) express as

⇒ d = r × θ       ...............2

here r is radius

and here rotated through some other angle θ' so put value in given equation and find revolutions    

⇒  d = (r+r)θ'      ........3

here  r = d/θ

so

⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]    

so put value and get θ'

⇒  θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]  

Answer:

to adjust from distant to the near objects

Explanation:

Answer:

The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.

Explanation:

The human eye is the optical instrument which works on the refraction of light.

The ability of eye lens to change its focal length is called accommodation power of eye.

The focal length of eye lens is changed by the action of ciliary muscles.

When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.

Answer:

due to production of heat through friction

Explanation:

because of the friction produce between her hands

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

[tex]V=1.4*10^5m/s[/tex]

Explanation:

From the question we are told that:

Electric field [tex]B=1.5*10N/C[/tex]

Distance [tex]d=2 x 10^{-3}[/tex]

At negative plate

Generally the equation for Velocity is mathematically given by

[tex]V^2=2as[/tex]

Therefore

[tex]V^2=\frac{2*e_0E*d}{m}[/tex]

[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]

[tex]V=\sqrt{19.2*10^9}[/tex]

[tex]V=1.4*10^5m/s[/tex]

Answer:

a)  [tex]t=195.948N.m[/tex]

b)  [tex]\phi=13.6 \textdegree[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1.225kg/m^2[/tex]

Velocity of wind [tex]v=14m/s[/tex]

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient [tex]\mu=2.05[/tex]

a)

Generally the equation for Force is mathematically given by

[tex]F=\frac{1}{2}\muA\rhov^2[/tex]

[tex]F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2[/tex]

[tex]F=163.29[/tex]

Therefore Torque

[tex]t=F*r*sin\theta[/tex]

[tex]t=163.29*1.2*sin90[/tex]

[tex]t=195.948N.m[/tex]

b)

Generally the equation for torque due to weight is mathematically given by

[tex]t=d*Mg*sin90[/tex]

Where

[tex]d=sin \phi[/tex]

Therefore

[tex]t=sin \phi*Mg*sin90[/tex]

[tex]195.948=833sin \phi[/tex]

[tex]\phi=sin^{-1}\frac{195.948}{833}[/tex]

[tex]\phi=13.6 \textdegree[/tex]

Answer:

the  speed of the electron at the given position is 106.2 m/s

Explanation:

Given;

initial position of the electron, r = 9 cm = 0.09 m

final position of the electron, r₂ = 3 cm = 0.03 m

let the speed of the electron at the given position = v

The initial potential energy of the electron is calculated as;

[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]

When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;

[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]

Apply the principle of conservation of energy;

ΔK.E = ΔU

[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]

Therefore, the  speed of the electron at the given position is 106.2 m/s

Answer:

Upthrust = 19.6 N

Explanation:

When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.

Upthrust = density of liquid x gravitational force x volume of object

i.e U = ρ x g x vol

Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]

So that;

U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])

   = 19.6 Kg m/[tex]s^{2}[/tex]

U = 19.6 Newtons

The upthrust on the iron is 19.6 N.

Answer:

89.87m/s

Explanation:

Given the velocity function

v(t)=−t²+3t−2

In order to get the displacement function, we will integrate the velocity function as shown:

[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]

at t = 5

[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]

at t = -2

[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]

Required displacement = d(5) - d(-2)

Required displacement = 89.2 - (-0.67)

Required displacement = 89.2 + 0.67

Required displacement = 89.87m/s

The aspect of the US justice system that has its roots in Jewish scripture is:

the idea that all people are subject to the same rules and laws.

It is the doctrine of "equality before the law."  Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc.  Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism.  It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.

Answer:

answer is C

the idea that all people are subject to the same rules and laws

Explanation:

hope this helps!

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

Answer:

b. The heavier one reaches the bottom first.

Answer:

B

Explanation:

The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.

Answer:

the correct one is C

Explanation:

To find the answer, let's propose the solution of the problem

We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.

Initial instant. Before starting to walk

          L₀ = 0

Final moment. After the child is walking

          L_f = I₁ w₁ + m r v₂

where index 1 is used for the platform and index 2 for the child

linear and angular velocity are related

          v₂ = w₂ r

angular momentum is conserved

          0 = I₁ w₁ + m r (w₂ r)

          w₁ =  [tex]- \frac{m r^2}{I1} \ w_2[/tex]

the moment of inertia of the platform bringing it closer to a disk or cylinder

         I₁ = [tex]\frac{1}{2}[/tex] M r²  

sustitute

          w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]

          W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]

from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds

when examining the answers the correct one is C

Answer:

Explanation:

relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system

so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero

so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong

option B is wrong because relative to ground their angular momentum should be equal and opposite

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Answer:

The total surface area is "90 km²".

Explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]

%n,

= 30%

Now,

⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]

then,

⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]

           [tex]=90 \ GW[/tex]

As we know,

⇒ [tex]I=\frac{P}{A}[/tex]

by substituting the values, we get

[tex]1000=\frac{90\times 10^9}{A}[/tex]

    [tex]A = \frac{90\times 10^9}{10^3}[/tex]

        [tex]=90\times 10^6[/tex]

        [tex]=90 \ km^2[/tex]

Answer:

store a charge for later use

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

Answer:

V = V0 + a t

V = 75 - 9.8 * 3 = 45.6 m/s

The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.

The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.

Given, the initial velocity of the projectile, u = 75 m/s

The time taken by the projectile, t = 3 sec

The acceleration due to gravity upward, g = - 9.8 m/s²

From the first equation of motion we can calculate the final velocity of the projectile:

v = u + at

v = u - gt

v = 75 - 9.8 ×(3)

v = 75 - 29.4

v = 45.6 m/s

Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.

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i. [tex]T = 36.8\:\text{N}[/tex]

ii. [tex]a = 2.45\:\text{m/s}^2[/tex]

iii. [tex]x = 1.23\:\text{m}[/tex]

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]

Forces acting on m1:

[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]

Forces acting on m2:

[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

[tex]m_2g - m_1g = m_2a + m_1a[/tex]

or

[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]

Solving for a,

[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]

[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]

Assuming that the two objects start from rest, the distance that they travel after one second is given by

[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]

Answer:

0.98 g/m

Explanation:

Note: Since Tension and frequency are constant,

Applying,

F₁²M₁ = F₂²M₂............... Equation 1

Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.

make M₂ the subject of the equation

M₂ = F₁²M₁/F₂²............... Equation 2

From the question,

Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz

Substitute these values into equation 2

M₂ = 196²(0.31)/110²

M₂ = 0.98 g/m

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

A  5

v=  dt/ dx  =−5+12t

Initial velocity means at t=0, which is −5+0=−5.

Thus, −v=5n