A kite 100 ft above the ground moves horizontally at a speed of 7 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. [tex]x=x_{0}+vt[/tex]

2. [tex]x=x_{0}+v_{0}t+\frac{at^{2}}{2}[/tex]

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

[tex]vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}[/tex]

We simplify the fraction present and rearrange for our formula so that it equals 0:

[tex]0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0[/tex]

In the very last step we factored a common factor t. There is two possible solutions to the equation at [tex]t=0[/tex] and:

[tex]0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s[/tex]

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at [tex]t=0 s[/tex] (when the SUV passed the police car) and [tex]t=15.56s[/tex](when the police car catches up to the SUV)

Answer:

18.3 m , 25.4°

Explanation:

d1 = 6 m, θ1 = 40°

d2 = 8 m, θ2 = 30°

d3 = 5 m, θ3 = 0°

Write the displacements in the vector form

[tex]d_{1}=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}[/tex]

[tex]d_{2}=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}[/tex]

[tex]d_{3}=5\widehat{i}[/tex]

The total displacement is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]

[tex]d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}[/tex]

[tex]d=16.53\widehat{i}+7.86\widehat{j}[/tex]

magnitude of resultant displacement is given by

[tex]d ={\sqrt{16.53^{2}+7.86^{2}}}=18.3 m[/tex]

d = 18.3 m

Let θ be the angle of resultant displacement with + x axis

[tex]tan\theta =\frac{7.86}{16.53}=0.4755[/tex]

θ = 25.4°

Answer:

protons and neutrons.

Explanation:

protons and neutrons are the largest particles in an atom. Electrons are not included in the mass because they are too small to make a difference in the mass.

The mass number of an atom of carbon represents the total number of protons and neutrons in the atom. This is because the mass number, or atomic mass, is calculated by adding the number of protons and neutrons in an atom.

The mass number of an atom of carbon (C) represents the total number of option C) protons and neutrons in the atom.

The mass number (also known as atomic mass) in an atom is calculated by adding the number of protons and neutrons, both of which are located in the nucleus of the atom. Electrons, on the other hand, contribute negligible mass to an atom and so are not counted towards the mass number. Accounting for only the protons would give us the atomic number (which is unique for each element), but the mass number includes both protons and neutrons.

For example, a Carbon-12 atom would have 6 protons and 6 neutrons as atomic mass is 12 (6 protons and 6 neutrons).

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The magnitude of the magnetic field at the center of the solenoid is 2π x 10^-5 Tesla, found using the formula B = μ₀nI and assuming the length of the solenoid can be estimated from the diameter of the wire and the number of turns.

To calculate the magnetic field inside a solenoid, we use Ampere's law, which is in the form B =
μ₀nI when the core is air. The term μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Just by looking at the problem, we don't need to use the diameter of the wire, as the number of turns per unit length (n) is the critical value needed in the formula and they are given directly.

The number of turns per unit length can be found by dividing the number of turns by the length of the solenoid. However, the length of the solenoid is not provided, so we would typically divide the number of turns (200) by the solenoid's length. Assuming that the coils are tightly wound with adjacent coils touching and the diameter of the wire is 2.0 mm, the length can be estimated as the diameter times the number of turns (0.002 m * 200 = 0.4 m).

So, substituting the values into the equation B = μ₀nI, where μ₀ = 4π x 10^-7 T·m/A, n = 200 turns / 0.4 m = 500 turns/m, and I = 0.10 A, we get:

B = (4π x 10^-7 T·m/A) * (500 turns/m) * (0.10 A) = 2π x 10^-5 (T)

Thus, the magnitude of the magnetic field at the center of the solenoid is 2π x 10^-5 Tesla.

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

Answer:

Final velocity will be 556.8 m/sec

Explanation:

We have given the average rate of change of velocity that is [tex]a=5.8\times 10^5m/sec[/tex]

Time [tex]t=9.6\times 10^{-4}sec[/tex]

Initial velocity of bullet will be zero that is u=0

Now according to first law of motion

v=u+at, here v is final velocity, u is initial velocity and t is time

So [tex]v=0+5.8\times 10^5\times 9.6\times 10^{-4}=556.8m/sec[/tex]

Answer: 384,300,000 m

Explanation:

Speed [tex]S[/tex] is mathematically expressed as:

[tex]S=\frac{d}{t}[/tex]

Where:

[tex]S=c=3(10)^{8} m/s[/tex] is the speed of light

[tex]d[/tex] is the distance traveled

[tex]t=\frac{2.562 s}{2}=1.281 s[/tex] since [tex]2.562 s[/tex] is the time it takes to the pulse to go the moon and return.

Finding [tex]d[/tex]:

[tex]d=c t[/tex]

[tex]d=(3(10)^{8} m/s)(1.281 s)[/tex]

Finally:

[tex]d=384,300,000 m[/tex]

Answer:

The average speed will be 2.54 Km/h

Explanation:

Given that

A to B

Distance = 175175 m=175.175 KM

B to C

Distance = 7070 m=7.07 Km

Lets take time taken from B to C is t then A to B will be t+66

We know that

Distance = velocity x time

175.175 =  V x (t+66)              -------------1

7.070 = V x t                      -------------2

From equation 1 and 2

t=2.77 hr

Now by putting the values in equation 2

7.07 = V x 2.77

V = 2.54 Km/h

So the average speed will be 2.54 Km/h

Answer:

38.77 kg

Explanation:

The weight of an object is given by:

Weight = mg,  g = acceleration due to gravity ( 9.8 m/s square)

and m is the mass.

So, in this case:

Weight of dog is 382 N, so, by putting the values of weight and g into the equation we will get the mass of dog.

382 = mass multiply with 9.8

so, the mass of the dog will be 38.77 kg

Answer:

[tex]\frac{dx}{dt}=7.14m/s[/tex]

Explanation:

As is showed at the figure annexed, we can solve this problem finding the relation between the girl displacement and the shadow displacement.

Relation the triangles (see figure annexed):

[tex]\frac{x}{H}=\frac{x-y}{h}\\x=\frac{H}{H-h}y[/tex]

We derive in order to find the speed of the shadow, because:

dx/dt: shadow's speed

dy/dt: girl's speed

[tex]\frac{dx}{dt} =\frac{25}{25-4}*6=7.14m/s[/tex]

Answer:

various parts have been answered

Explanation:

Inverse square for light is [tex]I_1r_1^2=I_2r_2^2[/tex]

initial distance from sun to earth is[tex]r_1=150\times10^6 [/tex]

and intensity or apparent brightness of sun is [tex]I_1=1300\ W/m^2[/tex]

a)

If distance from sun to earth is [tex]r_2=r_1/2=\frac{150\times10^6}{2}[/tex]

then apparent brightness is  [tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(r_1/2)^2}=5200[/tex]

b)

If distance from sun to earth is  [tex]r_2=2r_1[/tex]

then apparent brightness is  [tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(2r_1)^2}=325\,W/m^2[/tex]

c)

If distance from sun to earth is  [tex]r_2=7r_1[/tex]

then apparent brightness is

[tex]I_2=\frac{I_1r_1^2}{r_2^2}=\frac{1300\times r_1^2}{(7r_1)^2}=26.5\,W/m^2[/tex]

Using the inverse square law, Earth's fraction of intercepted solar energy and the Sun's total power output are calculated based on the given distance and apparent solar brightness. The result shows Earth intercepts a minuscule part of the Sun's total energy output.

This question concerns the application of the inverse square law for light and the calculation of the Sun's total power output based on its apparent brightness from Earth. Given that Earth is approximately 150 million kilometers (or 1.5 × 1011 meters) away from the Sun and the apparent brightness of the Sun (also known as solar constant) at Earth is about 1300 watts/m2, we can determine several key aspects:

Fraction of the Sun’s energy striking Earth: Using Earth’s diameter of about 12,756 km (or 1.2756 × 107 meters), the cross-sectional area of Earth (which is a circular area) can be calculated as AEarth = π(1.2756 × 107/2)2 ≈ 1.28 × 1014 m2.

Total power output of the Sun: Applying the inverse square law, which states that the total power output P of the Sun is distributed over a sphere of radius equal to Earth’s orbit (1.5 × 1011 meters). The surface area of this sphere is A = 4π(1.5 × 1011)2 ≈ 2.83 × 1023 m2. The total power output (luminosity) L of the Sun can be calculated by multiplying this surface area by the apparent brightness (1300 W/m2), yielding L = (1300 W/m2) × (2.83 × 1023 m2) ≈ 3.68 × 1026 watts.

In essence, Earth intercepts a tiny fraction of the Sun’s total energy, owing to the vast difference in scales between the size of Earth and the distance to the Sun.

Explanation:

Given that,

Speed with which the bullet is fired, u = 136 m/s

Time, t = 1.3 s

We need to find the displacement of the bullet after 1.3 seconds. It can be calculated using second equation of motion as :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Here, a = -g

[tex]s=ut-\dfrac{1}{2}gt^2[/tex]

[tex]s=136\times 1.3-\dfrac{1}{2}\times 9.8\times (1.3)^2[/tex]

s = 168.51 meters

So, the displacement of the bullet after 1.3 seconds is 168.51 meters. Hence, this is the required solution.

The density of the shuttle is 0.0491 mol/L, and the specific volume is 20.4 L/mol.

To find the density, we can use the ideal gas law, which states that the density of an ideal gas is equal to its pressure divided by its temperature and the ideal gas constant. The formula for density (ρ) is: ρ = P / (R * T), where P is the pressure, R is the ideal gas constant, and T is the temperature in Kelvin. In this case, the pressure is 1.2 atm and the temperature is 300 K. Using the ideal gas constant for air (R = 0.0821 L * atm / (mol * K)), we can calculate the density:

ρ = 1.2 atm / (0.0821 L * atm / (mol * K) * 300 K) = 0.0491 mol / L

To find the specific volume, we can use the reciprocal of density. The specific volume (v) is equal to 1 / density. So, the specific volume for the given conditions is:

v = 1 / 0.0491 mol / L = 20.4 L / mol

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Answer:

B. Nebular Hypothesis

Explanation:

It is important to read the statement carefully so we can rule out options. The text is referring to solar system formation, it means we should discard all options which are not related to that. Big bang theory is a concept to explain the origin of the whole universe, not the solar system only, so it is discarded. Plate tectonics describes inner earth layers that are essential parts of the planet movemements like earthquakes and they are not related to solar system either. Discarded. Helioentric theory is the idea that defines the sun as the center of the solar system, but again it does not explain anything about its origin, so let me explain a litte in detail about the right choice: Nebular Hypothesis.

Nebular Hypothesis

This is the way astronomers accept the solar system was created as well as other other plantetary systems. Million of years after Big Bang, there were gases and small dust particles all over the universe and they started to join and concentrate into bigger and bigger amount of gases turning into enourmous clouds made of all the necessary elements to begin the formation of stars, planets and eventually the solar system.

To find the speed of the object after it has fallen 4.0 cm, we can use the principles of conservation of energy and Hooke's law. The potential energy stored in the spring when it is stretched by 4.0 cm is given by U = 1/2kx^2, where k is the spring constant and x is the displacement. The gravitational potential energy of the object when it is at a height of 4.0 cm can be calculated as mgx, where m is the mass and g is the acceleration due to gravity. Equating these two expressions for potential energy and solving for the speed of the object, we find v = sqrt((2gh - kx^2)/m), where h is the distance from the equilibrium position to the lowest point of the object. Plugging in the given values, we have v = sqrt((2 * 9.8 m/s^2 * 0.04 m - 200 N/m * 0.04 m^2) / 2 kg) = 1.6 m/s.

To find the speed of the object after it has fallen 4.0 cm, we can use the principles of conservation of energy and Hooke's law. The potential energy stored in the spring when it is stretched by 4.0 cm is given by U = 1/2kx^2, where k is the spring constant and x is the displacement. The gravitational potential energy of the object when it is at a height of 4.0 cm can be calculated as mgx, where m is the mass and g is the acceleration due to gravity. Equating these two expressions for potential energy and solving for the speed of the object, we find v = sqrt((2gh - kx^2)/m), where h is the distance from the equilibrium position to the lowest point of the object. Plugging in the given values, we have v = sqrt((2 * 9.8 m/s^2 * 0.04 m - 200 N/m * 0.04 m^2) / 2 kg) = 1.6 m/s.

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Answer:

0.36

Explanation:

The maximum force of friction exerted by the surface is given by:

[tex]F_f = \mu N[/tex] (1)

where

[tex]\mu[/tex] is the coefficient of friction

N is the normal reaction

The shed's weight is 2200 N. Since there is no motion along the vertical direction, the normal reaction is equal and opposite to the weight, so

N = 2200 N

The horizontal force that is pushing the shed is

F = 800 N

In order for it to keep moving, the force of friction (which acts horizontally in the opposite direction) must be not greater than this value. So the maximum force of friction must be

[tex]F_f = 800 N[/tex]

And substituting the values into eq.(1), we can find the maximum value of the coefficient of friction:

[tex]\mu = \frac{F_f}{N}=\frac{800}{2200}=0.36[/tex]

The largest coefficient of friction required to keep the ice fishing shed moving east with a force of 800N, when it weighs 2,200N, is roughly 0.3636.

To determine the largest coefficient of friction required to keep the ice fishing shed moving, you need to understand the balance between the applied force (the wind) and the resistive force due to friction. The force of friction is calculated using the formula:

Ffriction = μ * N

where μ is the coefficient of friction and N is the normal force. The normal force on the shed is equal to its weight when the surface is flat, which is given as 2,200N. The applied force is to the east with 800N.

To just keep moving indicates that the forces are balanced, so the frictional force equals the applied wind force. Therefore, to find the largest coefficient of friction (μ), we set the frictional force equal to the applied force (800N):

μ * 2,200N = 800N

μ = 800N / 2,200N

μ = 0.3636

Thus, the largest coefficient of friction required to keep the shed moving is approximately 0.3636.

Answer:

Energy of UV light [tex]=4.95\times 10^{-19}j[/tex]

Energy of green light [tex]=3.6\times 10^{-19}j[/tex]

Energy of infrared light [tex]=2.2\times 10^{-19}j[/tex]

Explanation:

We have given the wavelength of UV light = 400 nm [tex]=400\times 10^{-9}m[/tex] , wavelength of green light = 550 nm and wavelength of infrared = 900 nm

Speed of light [tex]c=3\times 10^8m/sec[/tex]

Plank's constant [tex]h=6.6\times 10^{-34}[/tex]

Energy of the signal is given by [tex]E=h\nu =h\frac{c}{\lambda }[/tex]

So energy of UV light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{400\times 10^{-9}}=4.95\times 10^{-19}j[/tex]

Energy of green light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{550\times 10^{-9}}=3.6\times 10^{-19}j[/tex]

Energy of infrared light [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{900\times 10^{-9}}=2.2\times 10^{-19}j[/tex]

Answer:

0.79 s

Explanation:

The initial velocity of the ball is, [tex]u=7.79 m/s[/tex].

And now we know that first equation of motion.

[tex]v=u+at[/tex]

Here, v is the final velocity, u is the initial velocity, t is time taken by the object, a is the acceleration.

Given that, final velocity is zero and ball is inb upward direction means opposite to acceleration due to gravity which means a=-g. and the value of g is [tex]9.8 m/s^{2}[/tex]

Therefore,

[tex]0=u-gt\\t=\frac{u}{g}\\ t=\frac{7.79}{9.8}\\ 0.79 s[/tex]

Therefore, it is the required minimum time that player must waiot to touch the ball.

Answer:

Uniform

Directed from the negative to the positive plate

Explanation:

The field inside a charged parallel-plate capacitor is __________.

• zero

• directed from the negative to the positive plate

• parallel to the plates

• uniform

A parallel plate capacitor is made of two opposite charged plates.

As a universal rule the electric field is directed from the negative charge to the positive charge. So in the capacitor the field is directed from the negative plate to the positive plate

Also the electric field inside the capacitor is constant irrespective of the location.

[tex]E = \frac{\sigma }{ \epsilon_{0} \epsilon_{r} } \\[/tex]

Where

[tex]\sigma[/tex] is the surface charge density

[tex]\epsilon_{0}[/tex] is the permittivity of free space

[tex]\epsilon_{r}[/tex] is the relative permittivity of the material inside the plates

The field inside a charged parallel-plate capacitor is uniform, directed from the negative to the positive plate, and parallel to the plates, with magnitude E = σ / ε₀.

The field inside a charged parallel-plate capacitor is uniform, directed from the negative to the positive plate, and parallel to the plates. At points far from the edges of the plates, the electric field is perpendicular to the plates and is constant in both magnitude and direction, as described in Essential Knowledge 2.C.5. This uniform field can be used, for example, to produce uniform acceleration of charges between the plates, such as in the electron gun of a TV tube.

Note that the electric field outside the region between the two plates is essentially zero because the fields from the positive and negative plates point in opposite directions and cancel each other out, except near the edges of the plates. The magnitude of the electrical field in the space between the parallel plates can be expressed as E = σ / ε₀, where σ denotes the surface charge density on one plate and ε₀ is the permittivity of free space.

Answer:

b)

Explanation:

Rotational inertia of the sphere: [tex]\frac{2}{5}mr^2[/tex]

Rotational inertia of hollow sphere: [tex]\frac{2}{3}mr^2[/tex]

Rotational inertia of flat disk: [tex]\frac{1}{2}mr^2[/tex]

The largest value is 2/3 mr², therefore, hollow sphere's inertia is largest.

The hollow sphere has the largest rotational inertia among the flat disk, solid sphere, and itself because its mass is distributed farther from its axis of rotation.

The question asks which among a flat disk, a solid sphere, and a hollow sphere - each with the same mass and radius and an axis of rotation passing through their center - has the largest rotational inertia. To solve this, we consider the formula for the moment of inertia (rotational inertia) for each shape. A flat disk has a rotational inertia of ½ MR², the solid sphere has ¾ MR², and the moment of inertia for a hollow sphere is ¾ MR². Considering these formulas, the moment of inertia is directly influenced by how the mass is distributed relative to the axis of rotation. In objects where mass is distributed farther from the axis, like the hollow sphere, the rotational inertia increases. Therefore, the hollow sphere has the largest rotational inertia, making (b) The hollow sphere has the largest rotational inertia the correct answer.

Answer:

19.2 m/s

Explanation:

The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added

18 + 1.2 = 19.2

If you were walking backwards (west) your velocity with respect to the ground would be

18 - 1.2 = 16.8

Answer:angle of ramp

Explanation:

The acceleration of a cart rolling down a ramp depends on the angle of the ramp and not on the length of the ramp.

Let the car have wheel in form of thin cylinder, therefore its acceleration is given by

[tex]a=\frac{gsin\theta }{1+\frac{I}{mr^2}}[/tex]

where

a=acceleration

g=acceleration due to gravity

I=moment of inertia of the wheel

m=mass of wheel

r=radius of wheel

[tex]\theta [/tex]=angle made by ramp with the horizontal

In above term there is no sign of length of ramp thus it is independent of it.

The acceleration of a cart rolling down a ramp is primarily influenced by the angle of the incline and gravitational acceleration, but in practical scenarios, friction and other factors also play a role. The final velocity as the cart leaves the ramp is a function of the acceleration and the distance over which it acts.

The acceleration of a cart rolling down a ramp depends on several factors, including the angle of the incline and the presence of friction. When considering a cart on a frictionless inclined plane, the acceleration only depends on the angle of the ramp and gravitational acceleration. However, in real-world scenarios, factors such as friction, the mass of the cart, and the initial velocity may also play a role. Acceleration is directly related to the sine of the angle of the incline. Furthermore, the final velocity of the cart as it leaves the ramp would depend on both the acceleration of the cart down the ramp and the distance over which the acceleration acts. In the absence of air resistance, all objects slide down a frictionless incline with the same acceleration if the angle is the same. Also, linear and angular accelerations are directly proportional, with the radius of wheels also affecting the acceleration.

Answer:

See Below

Explanation:

28. In ionic compounds the metal is always at the left.

MgBr2

Cu(SO4)2

CaCO3

FeO

Li2O

etc

29.

Elements from same group are more likely to have similar properties.

Since Y and Z have one valence electron they are from the same group, therefore are more likely to have similar properties

Answer:

[tex]Ep_x = 288.97*10^3\frac{N}{C}[/tex]

[tex]Ep_y = 2770.6*10^3\frac{N}{C}[/tex]

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

[tex]r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m[/tex]

[tex]\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o[/tex]

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

[tex]Ep = Ep_x i + Ep_y j[/tex]

Ep₁ₓ = 0

[tex]Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}[/tex]

[tex]Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}[/tex]

[tex]Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}[/tex]

Calculation of the electric field components at point P

[tex]Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}[/tex]

[tex]Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}[/tex]

Answer: [tex]38\ m[/tex]

Explanation:

For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.

Then:

[tex]d^2=\triangle x^2+\triangle y^2[/tex]

You can observe that the square of the displacement is equal to the sum of the square  of the horizontal displacement and the square of the vertical displacement.

Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:

[tex]\triangle x=37.0\ m\\\triangle y=8.5\ m[/tex]

Substituting values and solving for "d", you get:

[tex]d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m[/tex]

Answer:

Answered

Explanation:

[tex]V_A= -v_1j[/tex] ( in the south direction)

[tex]V_B= -v_2i[/tex] (in the west direction)

[tex]V_{AB}= -V_B-V_A[/tex]

[tex]= -v_2i-(-v_1j)[/tex]

so direction of v_1 and v_2 are  north and west respectively.

we used the formula for relative velocity

[tex]V_{AB}= V_B- V_A[/tex]

is the velocity of B with respect to A.

Final answer:

From Jim's perspective, Samantha's plane appears to move in a primarily westward direction but slightly towards him, resembling a southwest direction, due to the orthogonal paths of their movements.

Explanation:

Two private airplanes are taxiing at a small airport. Jim is in plane A rolling due south with respect to the ground. Samantha is in plane B rolling due west with respect to the ground. Samantha is in front of Jim and to his left. In what direction(s), relative to himself, does Jim see Samantha’s plane moving?

Since Jim is moving south and Samantha is moving west, from Jim’s perspective, Samantha’s plane would be seen moving to his front-left but more towards the left due to the orthogonal (right angle) relationship between their movement paths. Therefore, Samantha’s plane would appear to be moving in a direction that is primarily westward but slightly towards Jim, making it seem to be moving in a southwest direction relative to Jim's position.

Answer:

The answer to your question is below

Explanation:

Data

light speed = 300 000 km/s

a) Express it in scientific notation

to do it, we just move the decimal point 5 places to the left

         300 000 = 3.0 x 10 ⁵ km/s

b) Convert this value to meters per hour

  (300 000 km/s)(1000 m/1 km)(3600 s/1 h) = 300000x1000x3600 / 1x1x1

                                                                        = 1.08 x 10¹² m/h

c) What distance in centimeters does light travel in 1 s?

data

v = 300 000 km/s

d = ?

t = 1 s

 formula   v = d/t      we clear distance     d = vxt

                                  d = 300000 x 1 = 300000 km

                                   d = 300000000 m = 30000000000 cm

A. Expression of 300000 km/s in scientific notation.

Scientific notation is a most reliable way of writing a very large or very small number in a convenient way.

To express 300000 km/s in scientific notation, move the decimal point from the last zero number to the non zero number. Include a multiple of 10 raised to power of the number of move as shown below:

Number to express = 300000 Km/s

The Non zero number is => 3

Number of moves = 5

Therefore, the scientific notation of 300000 Km/s is 3.0 × 10⁵ Km/s

B. Converting 300000 Km/s to m/h

We'll begin by converting 300000 Km/s to m/s. This can be obtained as follow :

1 Km/s = 1000 m/s

Therefore,

[tex]300000 Km/s =\frac{300000 Km/s * 1000 m/s}{1 Km/s } \\[/tex]

Finally, we shall convert 300000000 m/s to m/h

1 m/s = 3600 m/h

Thus,

[tex]300000000 m/s = \frac{300000000 m/s * 3600 m/h}{1 m/s}[/tex]

Therefore,

C. Determination of the distance (in cm)  travelled in 1 second.

We'll begin by converting 300000 km/s to cm/s.

This can be obtained as follow:

1 km/s = 100000 cm/s

Therefore,

[tex]300000 km/s = \frac{300000 km/s * 100000 cm/s }{1 km/s} \\[/tex]

Finally, we shall determine the distance (in cm). This can be obtained as follow:

Speed = 3×10¹⁰ cm/s

Time = 1 s

[tex]Speed = \frac{Distance}{Time}\\\\3*10^{10} = \frac{Distance}{1}[/tex]

Cross multiply

Distance = 3×10¹⁰ × 1

Therefore, light travels a distance of 3×10¹⁰ cm in 1 second.

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Answer:

Acceleration will increase.

Explanation:

The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

F = ma

We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

[tex]a=\frac{F}{m}[/tex]

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.

Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.

So, the acceleration of the truck will increase.

As the truck loses mass due to leaking sand, while a constant driving force is maintained, the acceleration of the truck increases according to Newton's second law of motion.

The question is about the changes in the acceleration of a truck as it loses mass due to leaking sand. From a physics perspective, the truck's acceleration will actually increase as it loses mass. This is because the force propelling the vehicle remains constant while its mass decreases. According to Newton's second law of motion, which states that force equals mass times acceleration (F=ma), if the force remains constant and mass decreases, the acceleration must increase to keep the equation balanced.

For example, with initial values, if the force propelling the truck is 30 newtons and the truck (with its sand) has a mass of 15 kilograms, the acceleration would be 2 m/s² (from F=ma, or 30N=15kg*a). Now suppose the sand leaks out and the truck’s mass drops to 10kg. If the driving force stays the same at 30N, when plugging these values back into F=ma, the new acceleration would be 3 m/s². Thus, the acceleration of the truck has increased as it has lost mass.

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Answer:

Part 2: 4.30 m/s

Part 3: 1503 m/s^2

Explanation:

First you already had -5.29 m/s

Part 2

With what velocity does it leave the ground ?

it is the same process as question 1, but in this case the height is 0.945 m

[tex]v =\sqrt{2gh}[/tex]

[tex]v =\sqrt{2*9.8*0.945} \\v =4.30\ m/s[/tex]

Part 3

This is a simple formula of :

[tex]v_{f}-v_{i}=at \\where \\v_{f}: final\ speed \\v_{i}: initial\ speed \\a: acceleration\\t: time[/tex]

Final speed: is the speed the ball leaves the ground = 4.30 m/s

Initial speed: is the speed the ball hits the ground = -5.29 m/s

4.30 - (-5.29) = 0.00638a

a = 1503 m/s^2

Answer:

[tex]3,666,500\ m^3 = 3.666\times 10^9\ Liters[/tex]

Explanation:

Given that

Volume

[tex]V=3,666,500\ m^3[/tex]

As we know that

[tex]1\ m^3 = 1000\ Liters[/tex]

[tex]1\ m^3 = 10^3\ Liters[/tex]

So

[tex]3,666,500\ m^3 = 3,666,500\times 10^3\ Liters[/tex]

[tex]3,666,500\ m^3 = 3,666,5\times 10^5\ Liters[/tex]

[tex]3,666,500\ m^3 = 3.666\times 10^9\ Liters[/tex]

So we cab say that [tex]3,666,500\ m^3[/tex] is equal to [tex]3.666\times 10^9\ Liters[/tex].