How much work does the force F(x)=(−2.0/x)N do on a particle as it moves from x=2.0m to x=5.0m?

Answer:

Current in the circuit= 2 A

Explanation:

Current in a Series Connection

When two or more resistors are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.

We know a voltmeter connected to R2=3Ω measures 6V. Applying Ohm's law:

[tex]V=R.I[/tex]

We can calculate the current through R2:

[tex]\displaystyle I=\frac{6}{3}=2\ A[/tex]

Since the current in R2 is the same in the rest of the resistors, the current of R1 and R3 is also 2 A.

Answer:

The value is [tex]d = 3.66 *10^{10} \ m [/tex]

Explanation:

From the question we are told that

The orbital period is [tex]T = 31.3 \ days = 31.3 * 86400 = 2704320 \ s[/tex]

Generally for the centripetal force acting one of the stars will be equal to the gravitational force between the stars and this can be mathematically represented as

[tex]F_g = F_c[/tex]

=> [tex]\frac{Gm*m}{d^2} = \frac{mv^2}{r}[/tex]

Here m represents the mass of each star given that they have same mass and this equal to the mass of the sun whose value is

[tex]m = 1.989*10^{30} \ kg[/tex]

d(2 * r) is the diameter of the orbit which is distance between each star

r is the radius of the orbit

G is the gravitational constant with value [tex]G= 6.67*0^{-11} \ m^2/kg \cdot s^2[/tex]

v is the velocity of the stars which can be mathematically represented as

[tex]v = \frac{2 \pi r}{T}[/tex]

So

[tex]\frac{Gmm }{(2 * r)^2} = \frac{m (\frac{2\pi r}{T} )^2}{r}[/tex]

=> [tex]r =[ \frac{Gm T^2}{16 \pi ^2} ]^{\frac{1}{3} }[/tex]

=> [tex]r = [\frac{6.67 *10^{-11} * (1.989 *10^{30}) * 2704320^2}{16 * (3.142)^2} ]^{\frac{1}{3} }[/tex]

=> [tex]r = 1.83 *10^{10} \ m[/tex]

Generally the distance between the star is

[tex]d = 2 * r[/tex]

=> [tex]d = 2 * 1.83 *10^{10} [/tex]

=> [tex]d = 3.66 *10^{10} \ m [/tex]

The expression for the initial velocity at point A is:

0 = (velocity at point A - 0) / time

Simplifying the equation, we find:

Velocity at point A = 0

The initial velocity at point A is zero, indicating that the object starts from rest before sliding on the horizontal plane AB.

To write an expression for the initial velocity at point A, we need to analyze the forces acting on the object and apply the principles of motion.

Given:

Mass of the object, m

Radius of the semi circle, R

Coefficient of kinetic friction, μ[tex]_k[/tex] = 0.5

Force applied at point C, F = 3mg

The object is initially at rest.

Let's break down the motion into two parts: the motion on the horizontal plane AB and the motion along the semi circle BCD.

1. Motion on the horizontal plane AB:

The only force acting on the object on the horizontal plane is the force of kinetic friction. The frictional force can be calculated using:

Frictional force, f = μ[tex]_k[/tex]* Normal force

The normal force is equal to the weight of the object, which is mg.

Normal force, N = mg

Frictional force, f = μ[tex]_k[/tex] * mg

The frictional force acts in the opposite direction to the motion, so its magnitude is negative. Thus, the net force on the object on the horizontal plane is:

Net force = -f = -μ[tex]_k[/tex]* mg

Using Newton's second law, we can relate the net force to the acceleration:

Net force = mass * acceleration

-μ[tex]_k[/tex] * mg = m * acceleration

The acceleration can be expressed as the rate of change of velocity:

Acceleration = (final velocity - initial velocity) / time

Since the object is initially at rest, the initial velocity is zero.

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Answer:

9 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 3m/s²

t = 3 s

Find: v

v = at + v₀

v = (3 m/s²) (3 s) + 0 m/s

v = 9 m/s

Answer:

The answer is "Speed".

Explanation:

In the question, the Speed is the correct answer because it can be viewed as its rate from, which a length is covered via an object. It is also divided by time into other components of distance. Its SI unit of speed seems to be the meter/second, but still, the kilometer/hour or, in the Us UK, miles per hour is the most basic fundamental of speed in everyday use.

Physical fitness is the body's ability to efficiently work together to enable good health while performing daily living tasks

The component of physical fitness James Yap applied is agility

The reason agility is the correct physical fitness component is as follows:

Question: The possible options in the question are;

Agility

Speed

Coordination

Therefore, the component of physical fitness James Yap applied to run fast while dribbling the ball, changing phase (pace), is agility

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Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.

Any speculative construct is used to represent phases or steps in a time-based process, such as a belief that development proceeds in discontinuous phases indicated by functional changes.

Each stage corresponds to a specific age the statements do not represent assumptions that stage theories share.

The assumptions for the stage theory are as follows;

1. Every stage serves as a building block for the next.

2. The phases are followed in the same sequence by everyone.

3. Each step is distinct in terms of quality. It is a change in nature, not merely a change in amount.

Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.

Hence option B is the right option.

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Answer:

oxygen

Explanation:

as it is in the air it can't be depleted or used up

Answer:

Deplete definition is - to empty of a principal substance. How to use deplete in a ... to reduce in amount by using up The soil was depleted of minerals. deplete.

Explanation:

I would say the decision not to have children or to have only one child is becoming more common.

Answer:

The horizontal distance traveled by the rock from the wall after 2 seconds is 6m.

Explanation:

Given;

initial speed of the rock, u = 3 m/s

height of the wall, h = 25 m

time of travel, t = 2 seconds

The horizontal distance is determined as follows;

X = ut + ¹/₂gt²

where;

g is acceleration due to gravity, horizontal distance is not affected by gravity, g = 0

X = ut

X = 3 x 2

X = 6 m

Therefore, the horizontal distance traveled by the rock from the wall after 2 seconds is 6m.

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east  for  10 meters,  

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required  diagram

Answer:

a

  The  value at an inside point is Zero  

b

The electric field is [tex]E = 2.7*10^{6} \ N/C[/tex]

Explanation:

From the question we are told that

The magnitude of the charge is [tex]q = 3.0 \mu C = 3.0 *10^{-6} \ C[/tex]

The radius of the spherical ball is [tex]r = 5.0 \ mm = 0.005 \ m[/tex]

Generally according to law postulated by Gauss the magnitude of charge enclosed inside a conducting material is zero which implies that the electric field inside the spherical ball is zero

Generally the electric field out side the spherical ball is mathematically represented as

[tex]E = \frac{kq}{ a^2}[/tex]

Here a is the position outside the spherical ball that is been considered and the value is [tex]a = 10 \ cm = \frac{10}{100} = 0.1 \ m[/tex]

and k is the coulombs constant with value

[tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

=> [tex]E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}[/tex]

=> [tex]E = 2.7*10^{6} \ N/C[/tex]

Answer:

1) an observer in B 'sees the two simultaneous events

2)observer B sees that the events are not simultaneous

3)  Δt = Δt₀ /√ (1 + v²/c²)

Explanation:

This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems

1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events

Consequently an observer in B 'sees the two simultaneous events

2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.

Consequently observer B sees that the events are not simultaneous

3) let's calculate the times for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ is the time in the system S' which is at rest for the events

Answer:

I don't know sorry I hope you find what your looking

Answer:

A

Explanation:

edge 2020

The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero. Option B is correct.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.

The projectile's motion is divided into two parts: horizontal and vertical motion.

The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero.

Hence, option B is correct.

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Answer:

waste

Explanation:

Nuclear power produces radioactive waste that can harm the human body. The material produced (usually from either U - 235 or U - 238 or some radioactive isotope) can be just as destructive to cells as in outer space.

These wastes don't directly emit greenhouse gases, but the material waste mist be disposed properly or it'll harm humans and other organisms.

Answer:

Higher Pitched! (b)

Explanation:

The Doppler effect makes it seem high pitched, as the sound waves are mushed together at a higher frequency, while moving away, the sound waves spread out like a lower frequency wave!

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = [tex]v_{oy}[/tex] t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = [tex]v_{oy}^2[/tex] - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

First compute the resultant force F:

[tex]\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N[/tex]

[tex]\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N[/tex]

[tex]\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N[/tex]

Then use Newton's second law to determine the acceleration vector [tex]\mathbf a[/tex] for the particle:

[tex]\mathbf F=m\mathbf a[/tex]

[tex](10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a[/tex]

[tex]\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}[/tex]

Let [tex]\mathbf x(t)[/tex] and [tex]\mathbf v(t)[/tex] denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so [tex]\mathbf v(0)=0[/tex]. Then the particle's velocity vector at t = 10.4 s is

[tex]\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du[/tex]

[tex]\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}[/tex]

[tex]\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}[/tex]

If you don't know calculus, then just use the formula,

[tex]v_f=v_i+at[/tex]

So, for instance, the velocity vector at t = 10.4 s has x-component

[tex]v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}[/tex]

(b) Compute the angle [tex]\theta[/tex] for [tex]\mathbf v(10.4\,\mathrm s)[/tex]:

[tex]\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ[/tex]

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

[tex]\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du[/tex]

[tex]\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j[/tex]

Then using the fundamental theorem of calculus again, we have

[tex]\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du[/tex]

where [tex]\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m[/tex] is the particle's initial position. So we get

[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du[/tex]

[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}[/tex]

[tex]\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m[/tex]

So over the first 10.4 s, the particle is displaced by the vector

[tex]\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m[/tex]

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

Answer:

We study motion becuase it is the best way of determining the best method of doing the work by scrutinizing the motions made by the worker or the machine

Hope this helps :)

Answer:

its just built different

I could give you a big, fancy scientific answer, but it often comes down to the fact that the way that most instructors teach math is not the way the student(s) can most easily understand it.

Answer:

A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

Answer:

There is our potential energy in climbing stairs.

Answer:

Me and I have to work on my yum new drink I think ima was my birthday I think we can get a good drink or something haha was that I

Explanation: is the red one is not the one that is the one that was killed and I killed the red light and I did the same for me and it is not the proof that you are

I did the test

604.8 kg

Hi there !

d = m/V => m = d×V

V = l×w×h = 10m×12m×4m = 480 m³

m = 1.26kg/m³×480m³ = 604.8 kg

Good luck !

Answer:

I think it is Q

=

−

53

,

796.6

J

Explanation:

I note a 100% sure ok

Answer: A ([tex]\sqrt{61}[/tex],309.8°)

              B (2[tex]\sqrt{2}[/tex], 315°)

             C ([tex]3\sqrt{5}[/tex], 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

[tex]r=\sqrt{x^{2}+y^{2}}[/tex] and [tex]\theta=tan^{-1}(\frac{y}{x})[/tex]

For point A:

[tex]r=\sqrt{(-5)^{2}+6^{2}}[/tex]

[tex]r=\sqrt{61}[/tex]

[tex]\theta=tan^{-1}(\frac{6}{-5})[/tex]

[tex]\theta=tan^{-1}(-1.2)[/tex]

[tex]\theta=-50.2[/tex]°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

[tex]\theta=360-50.2[/tex]

[tex]\theta=[/tex] 309.8°

Polar coordinates for point A is ([tex]\sqrt{61}[/tex], 309.8°)

For point B:

[tex]r=\sqrt{2^{2}+(-2)^{2}}[/tex]

[tex]r=\sqrt{8}[/tex]

[tex]r=2\sqrt{2}[/tex]

[tex]\theta=tan^{-1}(\frac{-2}{2} )[/tex]

[tex]\theta=tan^{-1}(1)[/tex]

[tex]\theta=-45[/tex]°

Point B is in IV quadrant, so:

[tex]\theta=360-45[/tex]

[tex]\theta=[/tex] 315°

Polar coordinates for point B is ([tex]2\sqrt{2}[/tex], 315°)

For point C:

[tex]r=\sqrt{(-6)^{2}+(-3)^{2}}[/tex]

[tex]r=\sqrt{45}[/tex]

[tex]r=3\sqrt{5}[/tex]

[tex]\theta=tan^{-1}(\frac{-3}{-6} )[/tex]

[tex]\theta=tan^{-1}(0.5)[/tex]

[tex]\theta=[/tex] 26.56°

Polar coordinates for point C is ([tex]3\sqrt{5}[/tex], 26.56°)