QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!

Mass remains mass no matter what you do to it.

Answer:

a. Same � the electric fields point in opposite directions and therefore cancel at some midpoint.

b. bee 1 has a magnitude of charge less than bee 2

Explanation:

a. Do the bees have the same or opposite signs of charge?

They have the same charge. This is because since same charges would produce electric fields in opposite directions, that is the only way they can cancel out at some point. So, the charges are the same and the electric fields point in opposite directions and therefore cancel at some midpoint.

b. Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?

Let q be the charge on bee 1 and r its distance from the neutral electric field point. So, it electric field E = kq/r².

Also, let q' be the charge on bee 2 and d its distance from the neutral electric field point. So, it electric field E' = kq'/d².

Since E = E' at the neutral point.

kq/r² = kq'/d²

q/q' = r²/d² = (r/d)²

Given that r < d, so r/d < 1 and (r/d)² < 1

So, q/q' < 1

q < q'

So, the charge on bee 1 is less than that on bee 2  

Answer:

distance = 0.2330142 miles

= 0.2330142 mi

Explanation:

not sure if its right though....

Answer:a

Explanation:

Because its has to use tihs potential energy to overcome the atmosphere so the shuttle will not go back down

Answer:

add them

100+150 = 250N same direction

that's the resultant

same direction

Answer:

1399.2 ft

Explanation:

The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s

[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

Answer:

meteor

Explanation:

or some people call it a shooting star

Answer:

the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.

Answer:

A

Explanation:

I did the test on ap3x

Answer:

Only the energy of the wave travels through the medium. In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. ... In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle.

No. The object has to have motion for it to have kinetic energy.

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

Answer:

Sodium (K)

Explanation:

Answer:

a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]

Explanation:

From the information given;

Using the work-energy theorem

ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]

K = [tex]\mathbf{ F_f \times r}[/tex]

∴

[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]

Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)

Then;

[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]

[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]

Answer: 134 = 143 = 151 = 166 = 176

Hope this helps!!

Sorry if it's incorrect!!

:'(

In comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.

The given parameters;

in group A, distance traveled by the red block, s = 15 ± 3 feet

in group B, distance traveled by the green block, s = 25 ±  4 feet

To find:

Since both groups performed the experiment at equal times, we assume the time for both motion = t

Also, assume the initial velocity of both blocks = 0

For group A, we set-up the equation of motion as follows;

[tex]s = v_0t + \frac{1}{2} at^2 \\\\15+ 3 = 0 + 0.5\times a_1t^2\\\\18 = 0.5a_1t^2\\\\t^2 = \frac{18}{0.5a_1} \\\\t^2 = \frac{36}{a_1}[/tex]

For group B, we set-up the equation of motion as follows;

[tex]25 + 4 = v_0t + \frac{1}{2} a_2t^2\\\\29 = 0.5\times a_2t^2\\\\t^2 = \frac{29}{0.5a_2} = \frac{58}{a_2}[/tex]

Solve the first equation and the second equation together;

[tex]\frac{36}{a_1} = \frac{58}{a_2} \\\\\frac{a_2}{a_1} = \frac{58}{36} \\\\\frac{a_2}{a_1} = 1.61[/tex]

The ratio of error margin of both experiments;

[tex]\frac{4}{3} = 1.33[/tex]

The resulting parameter for comparison;

[tex]parameter, t' = 1.33 \times 1.61 = 2.14 \approx 2.0[/tex]

Thus, in comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.

Learn more here: https://brainly.com/question/12891261

Speed is the time rate of an object moving from one place to another, while velocity is the rate and direction of the object's movement. They are very similar but they don't mean the same thing.

Answer:

V = 0.48 m/s

Explanation:

In this case, we need to analyze the given data by parts.

At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.

With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:

X = 6 - 4 = 2 m

And as it's traveling at constant speed then:

X = V.t

We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.

To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:

y = gt²/2 ---> solving for t

2y/g = t²

t = √2y/g

If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:

t = √2*84.5 / 9.8

t = 4.15 s

So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:

V = X/t

V = 2 / 4.15

Hope this helps

Answer:

a)   f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b)   Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo[tex]\frac{c+v}{c}[/tex]

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   [tex]\frac{c}{c-v}[/tex]

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 [tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]

                f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]

                f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]

leave the linear term

               f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

Answer:

The answer is below

Explanation:

a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.

Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.

Hence, the microbes would collect on the negatively charged electrodes.

b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.

Answer:

Explanation:

The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .

Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .

When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .

After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.

Answer:

19600 N

Explanation:

weight = mass x gravity

We know that gravity = 9.8 m/s^2 and mass = 2000 kg.

w = m x g

w = 2000 kg x 9.8 m/s^2

w = 19600 N

The weight of the object is 19600 N (newtons).

Answer:

the answer i 2000kg

Explanation:

Answer:

a) t = 0.25 s,  b)  x = 0.075 m

Explanation:

a) For this exercise we will use kinematic relationships in one dimension

         v = v₀ + a t

in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²

          t = [tex]\frac{v -v_o}{a}[/tex]

we calculate

         t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]

         t = 0.25 s

b) We can also find the distance traveled during this acceleration

         v² = v₀² + 2a x

         x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]

let's calculate

         x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]

         x = 0.075 m

Answer:

the neutron is located in the nucleus of an atom

Answer:

nucleus

Explanation:

Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.

Answer:

Option B, Fix the piston in place so the volume of the pas remains constant

Explanation:

As we know

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

The effect on variable due to another variable can be studied by keeping the third variable constant.

Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.

Hence, option B is correct

The speed will be 10 m/s after the 1st and 20 m/s after the 2nd for an average of 15 m/s. So it will travel 15 m during that 2nd second

Mark as brainlist

The object, which  undergoes constant acceleration after starting from rest, will go in the next second​ 15 m.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Given parameters:

initial velocity of object: u = 0.

time = 1 second.

distance travelled: d= 5 m.

So, acceleration of the object: a = 2d/t² = (2×5)/1² m/s² = 10 m/s².

Hence,   it will go in the next second​ = 1/2×a(2²-1²) m

= 1/2×10×3 m.

= 15 m.

Learn more about acceleration here:

brainly.com/question/12550364

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Answer:

       y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Explanation:

Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation

             λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]

Let's use conservation of energy for speed

starting point

             Em₀ = U = e V

final point

             Em_f = K = ½ m v²

             Em₀ = Em_f

             eV = ½ m v²

             v =[tex]\sqrt{\frac{2eV}{m} }[/tex]

we substitute

             λ=  [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]

the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum

            a sin θ = m λ

first zero occurs at m = 1, also these experiments are performed at very small angles

            sin θ = θ

            θ = λ / a

This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain

           θ = 1.22 λ / D

where D is the diameter of the opening

we substitute

          θ = [tex]\frac{1.22}{D}[/tex]   \sqrt{ \frac{h^2 m}{2eV}}  

this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians

            θ = y / L

when substituting

where y is the minimum distance that can be resolved for this acceleration voltage

            y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Answer:

MGH=energy

3500*9.8*h=90000

h=90000/34300

h=2.62m

Answer:

 rₙ = 1,325 10⁻⁹ m

Explanation:

To solve this problem we use the bohr atomic model

           Eₙ = -13.606 /n²     [eV]

the brackets indicate that the units are in electron volts.

let's reduce the photon energy to eV

        E = 4.5 10-19J (1 eV / 1.6 10⁻¹⁹ eV) = 2.8125 eV

This energy is in the visible range, so the transition must occur in this range, this is for the Balmer series whose initial number is n₀ = 2

for an atomic transition on two levels

         ΔE = Eₙ - E₀ = [tex]\frac{-13.606}{n^2} + \frac{13.606}{2^2}[/tex]

         2.8125 = [tex]\frac{-13.606}{n^2} + 3.4015[/tex]

         [tex]\frac{13.606}{n^2}[/tex] = 3.4015 - 2.8125 = 0.589

           n² = 13.606 / 0.589

           n² = 23.1

           n = 4.8

as n must be an integer

           n = 5

taking the quantum number as far as the electron goes, we substitute in the equation for the radius

           rn = n² a₀

where ao is the radius of the lowest level a₀ = 5.3 10⁻¹¹ m

            rₙ = 5 2 5.3 10⁻¹¹

            rₙ = 132.5 10⁻¹¹ m

            rₙ = 1,325 10⁻⁹ m

Answer is C

Explanation:

Answer:

C. wavelength

good luck, i hope this helps :)