The Earth, which has an equatorial radius of 6380 km, makes one revolution on its axis every 23.93 hours. What is the tangential speed of Fayetteville, NC, whose latitude is 35o N

Answer:

T= (m2a2)/2

Explanation:

Using Newton's second law

F21+F22+ F33.....= M2a2

Where F21 and F22 are forces acting on M2

Thus T = ( M2a2)/2 this is tension on the string

Based on the magnitude of the horizontal acceleration and the mass, the tension will be (m₂ x a₂) / 2.

In the relevant diagram, there are two strings so the tension will be multiplied by 2.

The tension on the string can be found by:

2 x Tension on string = m₂ x a₂

Tension = (m₂ x a₂) / 2

In conclusion, the tension is (m₂ x a₂) / 2.

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Answer:

The force is  [tex]F = 8*10^{-12} \ N[/tex]

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]

       The energy provided by a single ATP is  [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]

       The velocity of the kinesin is  [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]

The power provided by the ATP in one second is  mathematically represented as

       [tex]P = E_{ATP} * R[/tex]

substituting values

       [tex]P = 80 * 0.8*10^{-19 }[/tex]

       [tex]P = 6.4 *10^{-18}J/s[/tex]

Now  this power is mathematically represented as

       [tex]P = F * v[/tex]

Where  F  is  the force the kinesin is exerting

  Thus  

          [tex]F = \frac{P}{v}[/tex]

substituting values

            [tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]

           [tex]F = 8*10^{-12} \ N[/tex]

The force exerted by the kinesin  is 8  × 10-12 N.

Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.

Since;

Energy  =  0.8 × 10-19 J/molecule

Number ATP molecules transported per second = 80 ATP molecules/s

Power =  0.8 × 10-19 J/molecule × 80 ATP molecules/s

Power = 6.4  × 10-18 J

Again, we know that;

Power = Force × Velocity

Velocity of the ATP molecules =  800 nm/s or 8 × 10-7 m/s

Force = Power/velocity

Force =  6.4  × 10-18 J/ 8 × 10-7 m/s

Force = 8  × 10-12 N

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Answer:

q/6Eo

Explanation:

See attached file pls

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

Explanation:

[tex]1.2 \mathrm{N} ; 2 \mathrm{N}[/tex]

2.[tex]200 \mathrm{N} ; 200 \mathrm{N}[/tex]

4.[tex]2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}[/tex]

[tex]5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}[/tex]

[tex]6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}[/tex]

[tex]8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}[/tex]

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of [tex]256 \mathrm{m} / \mathrm{s}[/tex]

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

Answer:

current I = 38 mA

Explanation:

given data

distance r = 7.2 cm

repel each other force per unit length \frac{F}{l} = 4.2 nN/m

solution

we know 2  wire is parallel and when current flow through these wire they exert force each other due to magnetic field

and current I(1) = I(2)   ................1

so

[tex]\frac{F}{l} = \frac{\mu _o}{2\pi } \times \frac{I(1) \times I(2)}{r}[/tex]     ..................2

put here value

4.2 × [tex]10^{-9}[/tex] = [tex]\frac{4\pi \times 10^{-7}}{2\pi } \times \frac{I^2}{7.2\times 10^{-2}}[/tex]    

solve it we get

I = 0.038884 A

current I = 38 mA

The current flow in the wires will be:

"38 mA".

According to the question,

Distance, r = 7.2 cm

Force per unit length, [tex]\frac{F}{l}[/tex] = 4.2 nN/m

Current passes, when wire is parallel:

→ I₁ = i₂

We know the relation,

→           [tex]\frac{F}{l}[/tex] = [tex]\frac{\mu_0}{2 \pi}\times \frac{I_1\times I_2}{r}[/tex]

By substituting the values, we get

4.2 × 10⁻⁹ = [tex]\frac{4 \pi\times 10^{-7}}{2 \pi}\times \frac{I^2}{7.2\times 10^{-2}}[/tex]

hence,

The current will be:

              I = 0.038884 A or,

                = 38 mA

Thus the above answer is appropriate.

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Answer:

5. The amount of work is the same for both.

Explanation:

Work done is a measure of change in kinetic energy of each egg

For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy

For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.

So work done is same for both eggs since they have same change in kinetic energy.

Given that,

Charge per unit length = λ

Point (x, y)=(0. d) parallel to the z axis

We know that,

The electric field due to the infinitely long wire is

[tex]E=\dfrac{\lambda}{2\pi\epsilon_{0}y}\hat{y}[/tex]

The electric potential is

[tex]V=-\int_{d}^{r}{\dfrac{\lambda}{2\pi\epsilon_{0}y}dy}[/tex]....(I)

Here, [tex]r=\sqrt{x^2+y^2}[/tex]

We need to calculate the potential due to this line charge

Using equation (I)

[tex]V=-\int_{d}^{r}{\dfrac{\lambda}{2\pi\epsilon_{0}y}dy}[/tex]

On integratinting

[tex]V=-\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{r}{d})[/tex]

[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{d}{r})[/tex]

Put the value of r

[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{d}{\sqrt{x^2+y^2}})[/tex]

[tex]V=\dfrac{\lambda}{4\pi\epsilon_{0}}ln(\dfrac{d^2}{x^2+y^2})[/tex]

Hence, The potential due to this line charge is [tex]\dfrac{\lambda}{4\pi\epsilon_{0}}ln(\dfrac{d^2}{x^2+y^2})[/tex]

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Answer:

198.9 x 10^-16

Explanation:

E = hc/ wavelength

E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)

E = 198.9 x 10^-16

Answer:

Explanation:

Beats are produced when two sound waves from two  different sources having slightly different frequencies  interfere . Due to both constructive and destructive interference , high and low intensity sound is heard at some regular interval . This is called rate of beat formation . Amplitudes of sound may be same or different . High intensity sound heard per unit time is called beat . It is equal to difference of frequencies of the sound waves interfering

each other .

Answer:

The angular displacement is  [tex]\theta = 28.33 \ rad[/tex]

Explanation:

From the question we are told that

     The speed of the driver is  [tex]v =13 \ m/ s[/tex]

     The acceleration of the driver is  [tex]a = 1.02 \ m/s^2[/tex]

      The time taken is [tex]t = 0.70 \ s[/tex]

      The radius of the tire is  [tex]r = 33 cm = 0.33 \ m[/tex]

The distance covered by the car during this  acceleration can be  calculated using the equation of motion as follows

        [tex]s = v*t +\frac{1}{2} * a * t^2[/tex]

Now substituting values  

       [tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]

      [tex]s = 9.35 \ m[/tex]

Now the angular displacement of the car with respect to the tire movement can be  represented mathematically as

      [tex]\theta = \frac{s}{r}[/tex]

substituting values

      [tex]\theta = \frac{9.35}{0.33}[/tex]

      [tex]\theta = 28.33 \ rad[/tex]

Answer: c. it is moving from high potential to low potential and gaining electric potential energy.  

Explanation: As an electron moves in the direction the electric field lines it is moving from high potential to low potential and gaining electric potential energy.

As an electron moves in the direction the electric field lines "it is moving from high potential to low potential and gaining electric potential energy."

When a charge and system of charges were brought from infinity to the current configuration without being accelerated, the entire work performed because of an external agent would be referred to as the electric potential energy of that particular charge and system of charges.

The electron should be starting to move from high potential energy to reduced potential energy also anyway since the direction of the electric field might be from positive to negative. However, since the electron would be an opposite charges particle, its own energy tends to increase as it tries to move from high potential energy to low potential energy.

Hence, the correct option is c.

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Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

V = 0.45 Volts

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

Answer:

(a) ω = 12.57 rad/s

(b) a = 189.5 m/s²

(c) T = 9.47 N

Explanation:

(a)

The speed of rotation is given by the formula:

ω = θ/t

where,

ω = speed of rotation = ?

θ = angular displacement = (1 rotation)(2π rad/1 rotation) = 2π rad

t = time taken = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.57 rad/s

(b)

The centripetal acceleration of the object is given by the formula:

a = v²/r

where,

a = Centripetal Acceleration = ?

v = linear speed of object = rω

r = length of rope = 1.2 m

Therefore,

a = (rω)²/r

a = rω²

a = (1.2 m)(12.57 rad/s)²

a = 189.5 m/s²

(c)

The tension required to maintain the motion is equal to the centripetal force:

Tension = Centripetal Force

T = ma

where,

m = mass of object = 50 g = 0.05 kg

Therefore,

T = (0.05 kg)(189.5 m/s²)

T = 9.47 N

Answer:

D

Explanation:

The power equation is P= V^2/R

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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

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Answer:

is this a company name.? java is a computer software right..

Answer:

q = 2.997*10^-4C

Explanation:

In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.

You use the second Newton law to sum both electrical and gravitational forces:

[tex]F_e-W=ma\\\\qE-mg=ma[/tex]             (1)

Fe: electric force

W: weight of the penny

q: required charge = ?

m: mass of the penny = 3g = 0.003kg

E: magnitude of the electric field = 100N/C

g: gravitational acceleration = 9.8m/s^2

a: acceleration of the penny = 0.19m/s^2

You solve the equation (1) for q, and replace the values of the other parameters:

[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]

It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2

Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]

For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

[tex]I_{c} = \int {J} \, dA[/tex]

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]

[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]

For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) Ampere's Law to calculate magnetic field B is given by:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

(i) First, first find [tex]I_{c}[/tex] for r ≤ R:

[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]

[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]

[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]

Calculating B(r), using Ampere's Law:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0

B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0

B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

(ii) For r ≥ R:

[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]

So, as calculated before:

[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]

[tex]I_{c} =[/tex] I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0

For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.

A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:

The Radius of the cross-section of the wire R

Current passing through the wire I

Current Density [tex]J = \alpha r[/tex]

Constant [tex]\alpha[/tex]

Distance of the point from the center [tex]r[/tex]

For part a)

Consider a circular strip between two concentric circles of radii r and r+dr.

Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]

[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]

Integration

[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]

For part b)

The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.

We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as

[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]

where ‘I’ is the threading current through the circle of radius ‘r’

[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]

(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where

[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]

[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)

[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]

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Answer:

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Answer:

The force exerted on a given area.

Option B is the right option.

Explanation:

Pressure is defined as force acting per unit area. It's SI unit is Pascal or N/m^2.

Formula to find pressure is:

[tex] \frac{thrust(f)}{area(a)} [/tex]

The force acting perpendicular on a surface is called the thrust. It is a vector quantity.

One Pascal pressure

One Pascal pressure is defined as the pressure exerted in a surface of area m^2 by the thrust of 1 N

Hope this helps ..

Good luck on your assignment.....

Answer:

I hope it will help you....

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.