A charge of 2.308 nC is moved from a position on the y axis of 5.423 cm to a position on the x axis of 1.356 cm while there is a charge 17.119 nC located at the origin. How much work in micro-Joules did it take to move the charge?

Answer: A : 250 is the answer

B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.

Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.

In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is

where  is called Hertz (Hz). So, the correct answer is

Explanation:

#Wavespeed

#1

\(\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s\)

#2

\(\\ \rm\Rrightarrow v=2(0.4)=0.8m/s\)

#Wavelength

#1

\(\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m\)

#2

\(\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m\)

#Frequency

\(\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz\)

#2

\(\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz\)

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The bullet drops "2.76 m" by the time it reaches the target.

First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:

\(s = vt\\\\t = \frac{s}{v}\)

where,

s = distance = 75 m

v = velocity = 100 m/s

t = time = ?

Therefore,

\(t = \frac{75\ m}{100\ m/s}\)

t = 0.75 s

Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.

\(h = v_it+\frac{1}{2}gt^2\)

where,  

h = height dropped = ?  

vi = initial vertical speed = 0 m/s

t = time interval = 0.75 s  

g = acceleration due to gravity = 9.81 m/s²

therefore,

\(h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2\)

h = 2.76 m

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The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer:

1 W = 1 J / sec       Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min    energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

Answer:

120 v

Explanation:

The two resistors have an equivalent of    20 * 30 /(20+30) = 12 ohms

  10 amps of current in the circuit

v = ir

  = 10 * 12 = 120 volts

Here is another way:

The two resistors are in prallel so the voltae across both is the same

  use the one on the right      v = ir     =   4 x 30 = 120 v

The acceleration of the dragster is 2.01 * 10^5 m/s^2

The term acceleration refers to the change in velocity with time. Hence the formula for acceleration is given as;

a = v - u/t

a = acceleration

v = final velocity

u = initial velocity

t = time taken

Now;

v = 100 miles or 160934 meters

u = 0 miles or 0 meters

t = 0.8 seconds

a =  160934 - 0/ 0.8

a = 2.01 * 10^5 m/s^2

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Answer:

Explanation:

Given:

d = 33.9 cm

f = 8.57

___________

F - ?

G - ?

The focal length of the lens:

F = d*f / (d + f) = 33.9*8.57 / (33.9 + 8.57) ≈ 6.84 cm

The magnification of the object:

M = f / d = 8.57 / 33.9 ≈ 0.25

First option is correct.Larry the Rock is going through Static Equilibrium.

In this situation, Larry is at rest and remains stationary despite the forces acting on him. While David and Pancho are exerting equal forces from opposite sides, their forces cancel each other out, resulting in a net force of zero. As a result, Larry does not move or experience any acceleration.

Static equilibrium occurs when an object's forces and torques balance each other, leading to a stable, balanced state. In this case, the forces exerted by David and Pancho are equal in magnitude and opposite in direction, creating a condition where the resultant force is zero. As a result, Larry remains in a state of rest, unable to experience any movement or acceleration.Therefore, the type of equilibrium that Larry the Rock is going through is Static Equilibrium.

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The impedance of the circuit is 200 ohm.

The term reactance refers to the opposition offered to the flow of current by the inductor or capacitor.

a) The inductive reactance is given by;

XL = 2πFL = 2 × (3.14) × 50 × 318.3 × 10^-3 = 100 ohm

b) Z = √R^2 + XL^2

Z = √(200)^2 + (100)^2

Z = 223.6 ohm

c) I = V/Z = 240V/223.6 ohm = 1.1 A

d) P.d across the resistor; IR = 1.1 A × 200 ohm = 220 V

P.d across inductor = IXL = 1.1 A ×  100 ohm = 110 V

e) Phase angle = tan^-1 (X/R) = tan^-1 (100/200) = 27 degrees.

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Answer: C. A large force over a long distance.

Answer:3000

Explanation:

given:u=0v=20m/st= 8sec

thereforea=v-u/t=20-0/8=20/8=5/2 m/som=1200 kg

thereforef=ma=1200*5/2=600*5=3000N

Answer:

\(39000\ \text{kg m/s}\)

West

Explanation:

m = Mass of car = \(1.3\times 10^{3}\ \text{kg}\)

t = Time = 9 seconds

u = Initial velocity = 30 m/s

v = Final velocity = 0

Impulse is given by

\(J=m(v-u)\\\Rightarrow J=1.3\times 10^3(0-30)\\\Rightarrow J=-39000\ \text{kg m/s}\)

The magnitude of the total impulse applied to the car to bring it to rest is \(39000\ \text{kg m/s}\).

The direction is towards west as the sign is negative.

The work done by the given force at the given displacement is \((1812 \times d )\ J\).

The given parameters;

Let the displacement = d

The work done by the given force at the given displacement is calculated as follows;

\(W = Fd \times cos(\theta)\\\\W = 2000 \times d \times cos(25)\\\\W = (1812\times \ d) \ J\)

Thus, the work done by the given force at the given displacement is \((1812 \times d )\ J\).

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The magnitude of the electric field due to the particle at the given distance is \(9\times 10^9 q\).

The electric field strength of a charged particle is the force per unit charge in the given field.

The electric field strength of a charge is given as;

\(E = \frac{F}{q} \\\\E = \frac{kq}{r^2}\)

where;

\(E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q\)

Thus, the magnitude of the electric field due to the particle at the given distance is \(9\times 10^9 q\).

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Answer:

a) ferromagnetic type,  b) superconductor or ferromagnetic

Explanation:

In substances, magnetism is the result of the magnetic moments of the electrons in their orbits and of the spin.

These two let us take interacting magnetic between them, if the sum of these magnetic moments is maximum we have a material of the ferromagnetic type, which is represented in figure a, where all the magnetic moments are aligned in one direction and this type of material is a magnet permanent.

Figure b shows a material where the magnetic moments have a circular shape, so the total moment is zero.The type of substrate material can have several possibilities

* The most test is that the substrate material is a superconductor where the marked areas are the superconducting magnetic areas and the areas without spine are the normal areas

* Another possibility is that the material is ferromagnetic, that is, there are several magnetic sub-networks in different orientation, resulting in a random field, the material is not a permanent magnet

The focal length of the concave mirror is approximately option B. 8.20 cm,

To solve this problem, we can use the mirror equation, which relates the object distance (d_o), image distance (d_i), and the focal length (f) of a concave mirror. The equation is:

1/f = 1/d_o + 1/d_i

Given the object distance (d_o) is 12.2 cm and the image distance (d_i) is 25.0 cm, we can plug in these values to find the focal length (f):

1/f = 1/12.2 + 1/25.0

To solve for f, first find the common denominator and combine the fractions:

1/f = (25 + 12.2) / (12.2 * 25) = 37.2 / 305

Now, take the reciprocal of both sides to isolate f:

f = 305 / 37.2

After calculating, we get:

f ≈ 8.20 cm

Thus, the focal length of the concave mirror is approximately 8.20 cm, Therefore the correct option to option b.

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Answer:

work done on the body is the product of the 3 quantities provided

Explanation:

Raising the object, the work done will be the potential energy.

potential energy equals mgh.

Get the remaining quantities and multiply it out g is usually 10m/s

Answer: 100J

Explanation: K = 1/2mv^2

m=8

v=5

1/2 * 8 * 5^2

1/2 * 8 * 25

4 * 25

100

i answered to the wrong question, apologies.

8.27 x 1013 meres is the orbital radius.

Jupiter's mass, 1.9 x 1027 kg, and the time interval, 16.9 days, are equal to 1.46 x 106 seconds. The radius is needed, thus r. Solution

The moon must be held in its orbit by a gravitational force equal to the centripetal force between Jupiter and the moon.

6.67 x 10⁻¹¹ N/m²kg

2 x 1.9 x 10/27 x 1.46 x 10'6 / 4 r = 6.85 x 102'7 G = 6.67 x 10'11 N/m2kg2 r = 8.27 x 10'7

The second-largest moon in Jupiter's orbit and the third-largest moon in the solar system is called Callisto. Of all the objects in our solar system, its surface has the most craters.

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An incandescent object refers to an object that emits light as a result of being heated to a high temperature.

When an object is heated, its atoms and molecules gain energy, causing them to vibrate and move more rapidly. This increased energy causes the object to emit electromagnetic radiation, including visible light, which makes the object glow.

Incandescence is commonly observed in everyday objects such as incandescent light bulbs, where a tungsten filament is heated by an electric current until it reaches a temperature that causes it to emit visible light.

As the filament gets hotter, it emits light with a higher intensity and shifts towards shorter wavelengths, starting from a warm red glow and progressing towards a brighter, white light.

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Heya!!

For calculate aceleration, let's applicate second law of Newton:

\(\boxed{F=ma}\)

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

\(12\ N = 3\ kg * \textbf{a}\)

\(\textbf{a} = 12\ N / 3\ kg\)

\(\textbf{a} = 4\ m/s^{2}\)

Result:

The aceleration of the object is of 4 m/s²

The boxes acceleration up the ramp is approximately 0.109 m/s².

The forces acting on the box can be resolved into two components: one parallel to the incline (F_parallel) and one perpendicular to the incline (F_perpendicular).

Given:

Mass of the box (m) = 40.7 kg

Coefficient of kinetic friction (μ) = 0.17

The force pulling the box (F_p) = 173 N

Incline angle (θ₁) = 14.5°

Force angle (θ₂) = 25.7°

First, we need to calculate the components of the force pulling the box:

F_parallel = F_p * sin(θ₂)

F_perpendicular = F_p * cos(θ₂)

Next, let's calculate the force of friction:

F_friction = μ * (mass of the box) * g, where g is the acceleration due to gravity (approximately 9.8 m/s²)

The force component parallel to the incline is opposed by the force of friction, so:

Net force parallel to the incline (F_net_parallel) = F_parallel - F_friction

Now, we can calculate the acceleration using Newton's second law:

F_net_parallel = (mass of the box) * acceleration

Rearranging the equation, we get:

acceleration = F_net_parallel / (mass of the box)

Now we can substitute the values into the equations and calculate the acceleration

F_parallel = 173 N * sin(25.7°) ≈ 73.88 N

F_perpendicular = 173 N * cos(25.7°) ≈ 154.37 N

F_friction = 0.17 * (40.7 kg) * 9.8 m/s² ≈ 69.44 N

F_net_parallel = F_parallel - F_friction ≈ 73.88 N - 69.44 N ≈ 4.44 N

acceleration = (4.44 N) / (40.7 kg) ≈ 0.109 m/s²

Therefore, the box's acceleration up the ramp is approximately 0.109 m/s².

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Answer:

whatttttttttttttttttttttttttttt are the items we have to classify into simple machines and compound machine

The part of a road vehicle which must be tested to ensure that there is sufficient friction to stop the vehicle in an emergency is the tyre.

This is referred to as a force that resists the motion of one object against another when they roll or slide against each other.

When dealing with braking, the main factor is to have sufficient friction between the road surface and tyre to bring the vehicle to a standstill. If the tyres are wornout there won't be enough friction to make the vehicle stop during emergencies which is therefore the reason why it was chosen as the correct choice.

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Answer:

You didn't provide a picture of the illustration so unfortunately nobody can help you :(

A field line is drawn at a location that is perpendicular to the net. Therefore, the tangent to the electric field line at any place coincides with the direction of the electric field there.

Each field line has a direction written on it with an arrow, just like gravitational field lines do. This direction indicates the direction of the electric field at all places along the field line. The density of the field lines represents the relative field strength.

Electric field is a vector quantity, hence a vector arrow can be used to depict it. The length of the arrows at any particular position is inversely proportional to the strength of the electric field there, and they all point in that direction.

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Answer:

From the question we are told that

  The length of the rod is  \(L_o\)

    The  speed is  v  

     The angle made by the rod is  \(\theta\)

Generally the x-component of the rod's length is  

     \(L_x =  L_o cos (\theta )\)

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       \(L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }\)

=>     \(L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }\)

Generally the y-component of the rods length  is mathematically represented as

      \(L_y  =  L_o  sin (\theta)\)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e \(L_y \)

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     \(L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}\)

=>  \(L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}\)

=>  \(L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}\)

=>   \(L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}\)

=> \(L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}\)

=> \(L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }\)

=> \(L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }\)

Hence the length of the rod as measured by a stationary observer is

       \( L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }\)

   Generally the angle made is mathematically represented

\(tan(\theta) =  \frac{L_y}{L_x}\)

=>  \(tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }\)

=> \(tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }\)

Explanation:

The special relativity relations allow to find the results for the questions about the measurements made by an observed at rest on the rod are:

     a) The length of the rod is: \(L = L_o \sqrt{1 - \frac{v^2}{c^2} \ cos^2\theta_o }\)  

     b) The angle with respect to the x axis is:   \(tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }\)

Special relativity studies the motion of bodies with speeds close to the speed of light, with two fundamental assumptions.

If we assume that the two systems move in the x-axis, the relationship between the components of the length are:

        \(L_x = L_{ox} \ \sqrt{1- \frac{v^2}{c^2} }\)  

        \(L_y = L_o_y \\L_z = L_{oz}\)  

Where the subscript "o" is used for the fixed observed on the rod, that is, it is at rest with respect to the body, v and c are the speed of the system and light, respectively.

a) They indicate that the length of the rod is L₀ and it forms an angle θ with the horizontal.

Let's use trigonometry to find the components of the length of the rod in the system at rest, with respect to it.

         sin θ = \(\frac{L_{oy}}{L_o}\)  

         cos θ = \(\frac{L_{ox}}{L_o}\)  

         \(L_{oy}\) = L₀ sin θ

         L₀ₓ = L₀ cos θ

Let us use the transformation relations of the length of the special relativity rod.

x-axis

      \(L_x = (L_o cos \theta_o) \ \sqrt{1- \frac{v^2}{c^2} }\)  

y-axis  

      \(L_y = L_{o} sin \theta_o\)  

The length of the rod with respect to the observer using the Pythagorean theorem is:

      L² = \(L_x^2 + L_y^2\)  

      \(L^2 = (L_o cos \theta_o\sqrt{1- \frac{v^2}{c^2} })^2 + (L_o sin \theta_o)^2\)

      \(L_2 = L_o^2 ( cos^2 \theta_o - cos^2 \theta_o \frac{v^2}{c^2} + sin^2\theta_o)\)  

      \(L^2 = L_o^2 ( 1 - \frac{v^2}{c^2} \ cos^2 \theta_o)\)  

      \(L= Lo \sqrt{1- \frac{v^2}{c^2} cos^2 \theta_o}\)  

b) the angle with the x-axis measured by the stationary observer is:

     \(tna \theta = \frac{L_y}{L_x}\)

    \(tan \ theta = \frac{L_o sin \theta_o}{L_o cos \theta_o \sqrt{1- \frac{v^2}{c^2} } }\)  

    \(tan \theta = \frac{tan \theta_o}{\sqrt{1-\frac{v^2}{c^2} } }\)  

In conclusion, using the special relativity relations we can find the results for the questions about the measurements made by an observed at rest on the rod are:

      a) The length of the rod is:  \(L = L_o \sqrt{1- \frac{v^2}{c^2} \ cos^2\theta_o }\)

      b) The angle to the x axis is:  \(tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }\)

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Answer:  54.12 m

Explanation: