With the sun and the earth back in their regular positions,consider a space probe with mass mp = 125 kg launched from the earth towardthe sun. When the probe is exactly halfway between the earth andthe sun along the line connecting them, what is the direction ofthe net gravitational force acting on the probe? Ignore the effects of other massive objectsin the solar system, such as the moon and other planets.
A. The force is toward the sun.
B. The force is toward the earth.
C. There is no net force because neither the sunnor the earth attracts the probe gravitationally at themidpoint.
D. There is no net force because the gravitationalattractions on the probe due to the sun and the earth are equal insize but point in opposite directions, so they cancel each otherout.

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5

Answer:

The work must be negative or positive depending on direction of motion of the object

Explanation:

This is because

W= force x distance

And = F x S cosစ

And this is positive when

Theta is less than π/2 and negative when theta is less than or equal to π/2

If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.

A force is indeed an influence which can alter an angular velocity. An object can alter its velocity, or speed, as a result of a force. Naturally, force can be characterized as a push or just a pull.

If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.

Find out more information about force here:

https://brainly.com/question/26115859?referrer=searchResults

Complete Question

A  voltaic cell is  constructed with two [tex]Zn^{2+}[/tex][tex]- Zn[/tex] electrodes. The  two cell compartment have  [tex][Zn^{2+}] =  1.6 \ M[/tex] and  [tex][Zn^{2+}] =  2.00*10^{-2} \  M[/tex] respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is   [tex]E =  0.06 V[/tex]

Explanation:

Generally from the question we are told that

   The  concentration of [tex][Zn^{2+}][/tex] at the cathode is  [tex][Zn^{2+}]_a =  1.6 \ M[/tex]

    The  concentration of [tex][Zn^{2+}][/tex] at the anode is [tex][Zn^{2+}]_c =  2.00*10^{-2} \  M[/tex]

Generally the the cell emf for the concentration is mathematically represented as

     [tex]E =  E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}[/tex]

Generally the [tex]E^o[/tex]is the standard emf of a cell, the value is  0 V

So

      [tex]E =  0  -  \frac{0.0591}{2}  * log[\frac{ 2.00*10^{-2}}{1.6} ][/tex]

=>      [tex]E =  0.06 V[/tex]

Answer:

(a) [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j[/tex] , (b) [tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex], (c) [tex]W = -51[/tex]

Explanation:

The statement is incomplete:

The force on an object is [tex]\vec F = -17\,j[/tex]. For the vector [tex]\vec v = 2\,i +3\,j[/tex]. Find: (a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex], (b) The component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex], and (c) The work [tex]W[/tex], done by force [tex]\vec F[/tex] through displacement [tex]\vec v[/tex].

(a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex] is determined by the following expression:

[tex]\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}[/tex]

Where [tex]\hat{v}[/tex] is the unit vector of [tex]\vec v[/tex], which is determined by the following expression:

[tex]\hat{v} = \frac{\vec v}{\|\vec v \|}[/tex]

Where [tex]\|\vec v\|[/tex] is the norm of [tex]\vec v[/tex], whose value can be found by Pythagorean Theorem.

Then, if [tex]\vec F = -17\,j[/tex] and [tex]\vec v = 2\,i +3\,j[/tex], then:

[tex]\|\vec v\| =\sqrt{2^{2}+3^{3}}[/tex]

[tex]\|\vec v\|=\sqrt{13}[/tex]

[tex]\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)[/tex]

[tex]\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j[/tex]

[tex]\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)[/tex]

[tex]\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}[/tex]

[tex]\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)[/tex]

[tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex]

(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:

[tex]\vec F_{\perp} = \vec F - \vec F_{\parallel}[/tex]

Given that [tex]\vec F = -17\,j[/tex] and [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex], the component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex] is:

[tex]\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)[/tex]

[tex]\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j[/tex]

[tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex]

(c) The work done by  [tex]\vec F[/tex] through displacement [tex]\vec v[/tex] is:

[tex]W = \vec F \bullet \vec v[/tex]

[tex]W = (0)\cdot (2)+(-17)\cdot (3)[/tex]

[tex]W = -51[/tex]

Given :

Number of operations move through a pocket calculator during a full day's operation , [tex]n=1.34 \times 10^{20}[/tex] .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in  one electron is :

[tex]e^-=-1.6\times 10^{-19}\ coulombs[/tex]

So , charge on n electron is :

[tex]C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C[/tex]

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

Answer:

2km

Explanation:

Answer:

projectiles

electromagnetic

Answer:

Explanation:

física cuántica y  Quantum Moves

Answer:

Explanation:

A vector is parallel to the y axis .

Let its magnitude be A . So the vector can be represented as A j .

where i and j are unit vectors in x and y axis direction .

The x component of A j will be dot product of A j with i

The x component of A j = A j . i

= A x 0      [  Since j . i = 0 ]

= 0

Answer:

124 just subtract only one

Explanation:

Answer:

213 nA

2.13 mA

851e^-t μA

Explanation:

We have a pretty straightforward question here.

Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as

V = IR, since we need I, we can write that

I = V/R

a) at V = 1 mV

I = (1 * 10^-3) / 4.7 * 10^3

I = 2.13 * 10^-7 A or 213 nA

b) at V = 10 V

I = 10 / 4.7 * 10^3

I = 0.00213 A or 2.13 mA

c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

Answer:

Copernicus rediscovered Aristarchus’s heliocentric model

Explanation:

Answer:

Explanation:

1) Acceleration is the change in velocity of a body with respect to time.

Acceleration a = vf - vi/t

vf is the final velocity

vi is the initial velocity.

t is the time taken

Since the body accelerates from rest, vi = 0m/s

a = 27-0/3

a = 27/3

a = 9m/s²

2) Given

u = 0m/s (accelerates from rest)

a = 190m/s²

t = 2.4seconds

v = ?

Using v = u+at

v = 0+190(2.4)

v = 190×2.4

v = 456m/s

The final velocity of the car is 456m/s

3) Given

u = 15m/s

a = 3.5m/s

t = 5seconds

Using the relationship

S = ut+1/2at²

S is the distance covered by the car.

S = 15(5)+1/2(3.5)×5²

S = 75+25×3.5/2

S = 75+43.75

S = 118.75m

4) Given

initial velocity u = 23m/s

Deceleration a = -0.25m/s²(negative acceleration)

Final velocity v = 0m/s

Using the relationship

V² = u²+2as

0² = 23²+2(-0.25)s

-23² = -0.5S

23² = 0.5S

S = 529/0.5

S = 1058m

The distance required for the train to stop is 1058m.

5) Given

initial velocity u = 12m/s

Final velocity v = 26m/s

time = 14sec

Acceleration a = v-u/t

a = 26-12/14

a= 14/14

a = 1m/s²

For Distance covered

v² = u²+2as

26² = 12²+2(1)S

676 = 144 +2S

2S = 676-144

2S = 532

S = 532/2

S = 266m

Distance that the car will cover is 266m

6) Given

Initial velocity u = 0m/s (person starts from rest)

acceleration a = 3.2m/s²

time t = 3.0s

To get the distance;

S = ut + 1/2at²

S = 0(3)+1/2(3.2)×3²

S = 0+1.6×9

S= 9×1.6

S = 14.4m

The distance that the person covered in 3.0s is 14.4m

7) initial velocity of train u = 12m/s

Distance covered S = 541m

Final velocity = 0m/s (on stopping)

acceleration a= ?

Acceleration will be negative since the train is coming to a stop (decelerating)

Using the formula v² = u²+2as

0² = 12² - 2a(541)

-12² = -1082a

144 = 1082a

a = 1082/144

a = 7.51m/s²

Hence the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m is 7.51m/s²

Answer:

Given that

I = 1340 W/m²

µo = 4πx 10^-7 Tm/A

sc = 3 x 10^8 m/s

So to find the peak values of B we use

I = (c Bm²) / (2 µo)

1340 = (3 x 10^8 x Bm²) / (2 x 4πx 10^-7)

1340 = (3 x 10^8 x Bm²) / (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 1340 x (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 3.366 x 10^-3

Bm² = 3.366 x 10^-3 / 3 x 10^8

Bm² = 1.122 x 10^-11

Bm = 3.4 x 10^-6 T

Also to find the peak values of E we

use Em = c x Bm

Em = (3 x 10^8) (3.4 x 10^-6)

Em = 1.005 N/C

Answer:

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

Explanation:

The magnitude on the charge at the bottom-left corner due to the charge on the top vertex of the triangle will act along the +ve x-axis and the +ve y-axis.

From Coulomb's law the magnitude of the forces on the charge at the bottom-left corner, due to the charge on the top vertex of the triangle are [tex]\frac{kq^{2} }{a^2}[/tex]cos 60, and

The magnitude on the charge due to the charge at the bottom-right corner will only act in the -ve x-axis, since they repel each other (like charges repel). The magnitude is  [tex]\frac{kq^{2} }{a^2}[/tex]

The angle made by the upper charge to the charge we're considering is 60° with the horizontal.

The total force on the charge along the x-axis is

[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{a^2}[/tex]cos 60 -

[tex]F_{x}[/tex]  = [tex]\frac{kq^{2} }{2a^2}[/tex] - [tex]\frac{kq^{2} }{a^2}[/tex]

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

For the y-axis, we have

[tex]F_{y}[/tex] = [tex]\frac{kq^{2} }{a}[/tex]sin 60

[tex]F_{y}[/tex] = [tex]\frac{\sqrt{3}* kq^{2} }{2a^2}[/tex]

The resultant force is

[tex]|F| = \sqrt{F_{x}^{2}+ F_{y}^2 }[/tex]

The common factors between the two x-axis force, and the y-axis force is

[tex]\frac{kq^{2} }{2a^2}[/tex], we put this outside the square root (squaring this and square rooting will give us the initial value)

[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3} }{2})^2 }[/tex]

[tex]|F|[/tex] =  [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{\frac{1}{4} +\frac{3}{4} }[/tex]

==>  [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{1}[/tex]

the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is

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

Complete Question

The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Answer:

The  value  is  [tex]P = 3.270960 *10^{6} \  W[/tex]

Explanation:

From the question we are told that

   The  electric field strength is  [tex]E = 4.0 \ M \ V/m = 4.0 *10^6 \ V/m[/tex]

    The  diameter is  [tex]d = 1.4 \ cm = 0.014 \ m[/tex]

Generally the radius is mathematically represented as

        [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.014}{2}[/tex]

=>    [tex]r = 0.007 \ m[/tex]

Generally the cross-sectional area is mathematically represented as

        [tex]A =  \pi r^2[/tex]

        [tex]A =  3.142 *  (0.007)^2[/tex]

       [tex]A = 0.000154 \ m^2[/tex]

Generally the maximum power that can be delivered is mathematically represented as

        [tex]P = \frac{c *  \epsilon_o  *  E^2 *  A }{2}[/tex]

Here c is the speed of light with value  [tex]c =  3.0*10^{8} \  m/s[/tex]

        [tex]\epsilon_o[/tex] is the permittivity of free space with value  [tex]\epsilon_o  =  8.85 *10^{-12}  \ m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2[/tex]

     [tex]P =  \frac{3,0*10^8 *  8.85*10^{-12} *  (4 *10^6)^2 * 0.00154}{ 2}[/tex]

       [tex]P = 3.270960 *10^{6} \  W[/tex]

Answer:

0.004394Joules

Explanation:

The question is incomplete. Find the complete question in the attachment below;

Energy stored in a capacitor = 1/2CV² where;

C is the capacitance of the capacitor

V is the potential difference across the plates

Given parameters

C = 13.0μF = 13 * 10⁻⁶F

V = 26.0V

Required

Energy stored in the capacitor

Substituting the values into the equation given;

E = 1/2 * 13 * 10⁻⁶ * 26²

E = 1/2 * 13 * 10⁻⁶ * 676

E = 1/2* 8788 * 10⁻⁶

E = 4394  * 10⁻⁶

Hence the Energy stored in the capacitor in Joules is 0.004394Joules

Answer:Nuclear fusion is the process by which two or more atomic nuclei join together, or “fuse,” to form a single heavier nucleus. During this process, matter is not conserved because some of the mass of the fusing nuclei is converted to energy, which is released.

Explanation:

Atoms of different elements can combine to make new substances. A molecule is formed when two or more atoms join together chemically. If atoms combine that are of two or more different elements, we call that a compound. ... When two hydrogen atoms combine with one oxygen atom, it becomes the compound water.

Answer:

nuclear fusion

Explanation:

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

Answer:

Hello your question is incomplete attached is the missing diagram and solution

One elevator arrangement includes the passenger car, a counterweight, and two large pulleys, as shown in (Figure 1). Each pulley has a radius of 1.2 m and a moment of inertia of 410 kg⋅m2. The top pulley is driven by a motor. The elevator car plus passengers has a mass of 3000 kg, and the counterweight has a mass of 2600 kg.

a) If the motor is to accelerate the elevator car upward at 1.8 m/s2, how much torque must it generate? Express your answer to two significant figures and include appropriate units.

Answer : 6030 N.m ≈ 6000 N.m

Explanation:

To determine how much torque the motor accelerating at 1.8 m/s^s will generate we will have to determine T1 ( tension  generated for upward acceleration in the rope  ) and Tg ( tension  generated in the rope when there is a counterweight accelerating downwards)

Torque in the pulley (T2) = (T1 - Tg ) r

torque generated by motor = T1 + T2 = 1230 + 4800 = 6030 N.m to two significant figures = 6000 N.m

r = 1.2

a = 1.8

attached below is the remaining part of the solution

Answer:

We know that the second equation of motion is

S= ut + 1/2a²

And S is displacement and u is initial velocity

So in the case of Haley lets take downwards as positive Y-axis

S = 2h and

initial velocity = v

a = g (acceleration due to gravity = 9.8)

Substituting

2h = vt + 1/2gt²

And for Joe we take ownwards as positive Y-axis

S = h and

initial velocity = 0 (since the ball is dropped from rest)

a = g

h = 0x t + 1/2gt2²

t2= √ 2h/g

Now since both balls reach ground at same time: t1=t2

So

putting value of t2 in Hayley's equation:

2h= v(√2h/g) + 1/2 g( √2h/g)²

So v= √gh/2

Answer:

The acceleration will be a

Explanation:

Using F = ma

But we are given that F2= 2F1 and

m2= 2m1

So

a2= F2/m2= F1/m1

and

F1/m1=a

Answer:

when an x-ray passes trough matter, it undergoes a process called attenuation

Answer:

The wavelength of the waves created in the swimming pool is 0.4 m

Explanation:

Given;

frequency of the wave, f = 2 Hz

velocity of the wave, v = 0.8 m/s

The wavelength of the wave is given by;

λ = v / f

where;

λ is the wavelength

f is the frequency

v is the wavelength

λ = 0.8 / 2

λ = 0.4 m

Therefore, the wavelength of the waves created in the swimming pool is 0.4 m

To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.

Answer:

Explanation: Thermals are created by the irregular heating of Earth's surface from solar radiation, and are an example of convection, specifically atmospheric convection.

At this point, air is converging on low pressure and rising air forms clouds.

Answer:

Explanation:

over a million times

Answer:

b. henry must have stopped for one hour during his trip

Answer:

Convex lens and convex mirrors are similar that

1. They have the same image characteristics at various object positions

2. They possess a positive focal length

3. Both their ray lines converge to a particular focal point

Convex lens and concave mirror have certain similarities that are explained below.

The shape of concave mirror is spherical.

Convex lens is the combination of two concave mirrors.

When the ray lines are drawn at the convex lens, then the coverage of the rays leads parallel to its principal axis.

When the ray lines are drawn at the concave mirror, then the coverage of the rays leads parallel to its focal point.

The focal length of concave mirror and convex lens is positive.

Concave mirror and convex lens both project the real and inverted image of the object.

For more details, follow the link given below.

https://brainly.com/question/14295029.

Answer:

The options are

A) isochoric.

B) isothermal.

C) adiabatic.

D) isobaric.

The answer is C. Adiabatic

Adiabatic process involves zero loss or gain of heat in a system. This is usually depicted as Q= 0.

An ideal gas being compressed in a well-insulated chamber using a well-insulated piston involves the use of adiabatic process. The insulated chamber and piston helps to prevent heat loss or gain of heat. This is because insulators are poor conductors of heat and electricity.

The process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.

When an ideal gas in compressed adiabatically, the heat lost is zero and the work is done it which causes increase its temperature.

Q = 0

The well-insulated chamber and piston helps to prevent heat loss by conduction and convection. The results in a zero heat lost to the surroundings.

Thus, we can conclude that the process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.

Learn more about adiabatic process here: https://brainly.com/question/17185468

The question is incomplete,so the complete question is as follows:

According to Einstein, increasing the brightness (or intensity) of a beam of light without changing its color will increase (circle all that apply):

A) the number of photons per second traveling in the beam.

B) the energy of each photon.

C) the speed of the photons.

D) the frequency of the light.

E) the wavelength of the photons

Answer:

A)  

Explanation:

According to Einstein, an increase in the intensity or brightness of light beam will increase the number of photons over a given time interval.

It means that the energy emitted will depend on the energy of individual photon rather than the intensity of the incoming light.

Hence, the correct option is "A)"