What voltage is required to move 6A through 5Ω?

Answer:

The ratio is  [tex]\frac{B_1}{B_2} = 1.265[/tex]

Explanation:

From the question we are told that

   The density of fresh water is  [tex]\rho__{f}} = 998 \ kg/m^3[/tex]

     The density of ethanol is  [tex]\rho_{e} = 789 \ kg /m^3[/tex]

Generally speed of a wave in a substance is mathematically represented as

    [tex]v = \sqrt{\frac{B}{\rho} }[/tex]

Here B is the adiabatic bulk modulus of the substance  while [tex]\rho[/tex]  is the density of the substance

So at constant wave speed

     [tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]

=>   [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]

=>   [tex]B_1 \rho_2 = B_2\rho_1[/tex]    

=>  [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]

Here  [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex]   and  [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]

So

  =>  [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]    

  =>  [tex]\frac{B_1}{B_2} = 1.265[/tex]    

When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.

The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.

A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.

The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.

To know more about the pitch:

https://brainly.com/question/12244387

#SPJ2

Answer:

gravitational force

Explanation:

Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects

Answer:

T = 102.3 ≈ 100

Explanation:

r(pedal) = 17 cm = 0.017 m

r(gear) = 16 cm / 2 = 8 cm = 0.008 m

θ = 180°-110° = 70°

F = 48 N

T = ?

τ₁ = r(pedal)*F*cosθ = (0.017 m)(48 N)cos(70°) = 0.28 N*m (*negative because it is applied clockwise)

τ₂ = -τ₁ = -(-0.28 N*m) = 0.28 N*m

τ₂ = r(gear)*T*cosθ

T = τ₂÷(r(gear)*cosθ) = 0.28 N*m ÷ (0.008 m * cos(70°))

T = 102.3 ≈ 100 ( 2 sig figs)

Answer:the answer is D

Explanation: I just took the quiz

Answer:

An External Stimulus is a stimulus that comes from outside an organism. Examples: You feel cold so you put on a jacket. When you sweat, the external stimulus is either you're anxious or hot.

Explanation:

hope it helps! <3

Answer:

B. Boats need to float.

Answer:

Explanation:

horizontal component of velocity of throw = 20 cos20 = 18.8 m /s

vertical downwards component = 20 sin20 = 6.84 m /s

time to displace by height 30 m = t  , initial velocity u = 6.84 m /s

h = ut + 1/2 gt²

30 = 6.84 t + .5 x 9.8 t²

4.9 t² + 6.84 t - 30 = 0

t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9

=  - 6.84 ±√( 46.78 + 588 ) / 9.8

=  - 6.84 ±√(634.78 ) / 9.8

= - 6.84 ±25.2 / 9.8

= 1.87 s

horizontal displacement in 1.87 s

= 18.8 x 1.87

= 35.15 m .

Answer:

K must be an acceleration

Explanation:

Recall that unit of energy consists the following dimensions:

mass * length^2 / time^2

then if we have :  mass * K = energy / length

then the units of K can be obtained by replacing the units of energy by (mass * length^2 / time^2), and then dividing by units of mass:

K = (mass * length^2 / time^2)/ (length * mass)

where we cancel out units of mass, and one unit of length resulting in units of length divided units of time squared. and such are units of acceleration. Therefore K must be an acceleration.

Answer: 8280

Explanation: This is the answer that my online school curriculum said was right.

God loves you!!!

Answer:

This isn't an answer, but rather a question for the answer that was given (which was correct! Thank you so much!) I was trying to figure out how to comment because I am in no way trying to steal points or anything like that, but I have a genuine question that I couldn't figure out any other way to ask...

How did you come up with 8280? Whenever I do it I get 8289....

I would like to know the process for future reference. Thank you for your help and again, I apologize for taking these points with no answer to contribute.

Answer:

The workdone is  [tex]W = 1.712 *10^{-20 } \ J[/tex]  

Explanation:

From the question we are told that

    The potential difference is  [tex]V = 107 mV = 107 *10^{-3} \ V[/tex]

Generally the charge on  [tex]Na^{+}[/tex] is  [tex]Q_{Na^{+}} = 1.60 *10^{-19 } \ C[/tex]

 Generally the workdone is mathematically represented as

         [tex]W = Q_{Na^{+}}V[/tex]

=>     [tex]W = 1.60 *10^{-19 } * 107 *10^{-3}[/tex]    

=>     [tex]W = 1.712 *10^{-20 } \ J[/tex]    

Answer:

15 m/s

Explanation:

v = u+ at

35 = u + 20

35-20 = u

u= 15 m/s

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

Answer:

Explanation:

Step one:

given data

acceleration a=4m/s^2

final velocity v=35m/s

time t=5 seconds

Step two:

Required

initial velocity u=?

Applying the first equation of motion

v=u+at

substituting we have

35=u+4*5

35=u+20

u=35-20

u=15 m/s

The initial velocity is 15m/s

Answer:

Q' = 320 J

Explanation:

The arrangements are given in attachment. The Fourier's Law of Heat Conduction states that:

Q = KAΔT/L

where,

Q = Heat Transferred

K = Constant (Conduction Coefficient)

A = Surface Area of Heat Transfer

ΔT = Difference of Temperature between two surfaces

L = Length between surfaces

For Arrangement AL

Q = 80 J

Therefore,

80 = KAΔT/L   ------------- equation (1)

Now, for arrangement B:

A' = 2 A (As, the rods are now connected in parallel with each other)

L' = L/2

Therefore,

Q' = K(2A)ΔT/(L/2)

Q' = 4 KAΔT/L

using equation (1)

Q' = 4(80 J)

Q' = 320 J

Answer:

200 meters per minute

Explanation:

You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.

Answer:  see attachment

Explanation:

Answer:

b

Explanation:

i took the quiz i think its right

Answer:

  C = A - B

Explanation:

The addition of vectors takes into account the magnitude of each vector and its direction, so when adding two vectors, the result depends on the direction of the vectors.

* If the vectors have the same direction the result is maximum

            C = A + A

* if the vectors have 90 between them, the magnitude of the result is given by the Pythagorean Theorem

            C = √(A² + B²)

* if the vectors have 180º between them the result is minimal

            C = A - B

We can also perform this sum graphically, where the resulting vector goes from the origin of the first vector to the tip of the last one, it can clearly be seen that when the vectors are antiparallel (180º angle) the magnitude is minimal

Answer:

Find the letter below.

Explanation:

                                                                                   5183 Richmond Avenue,

                                                                                   Houston City,

                                                                                   Texas, U.S.A.

                                                                                   16th November 2020.

Dear Jane,

I received with joy the news of your admission into Texas City High School, my alma mater. This is a big feat and evidence of your hard work and commitment to your academics. To excel in this school, there are certain points which I would like you to note.

First, this is a new phase of your life and it would be important for you to clearly keep your purpose for going to school in mind. Your foremost purpose is to study and gain the needed knowledge that would help you in life.

Secondly, the school is a very sociable environment with lots of interesting activities such as sports, dancing, literary groups, and so many others. Find a group that appeals to you and participate actively in it. Extracurricular activities would help you socialize with your mates and teachers. It would also help you meet interesting people who might later serve some useful purpose in your life after graduation.

The teachers at Texas City High School are very kind and willing to help you excel in your academics and other chosen fields. Be sure to respect them and meet them when you have difficulties in any area of concern. There are also counselors that guide you and help you to remain on the path to excellence.

My dear Jane, life in Texas City High School would not always be rosy as there are bullies in school who would want to intimidate you or get you into fights. Avoid them by all means. Be cordial to all. Report any cases of harassment to the relevant authorities for they are seasoned in handling such matters maturely.

The cafeteria is also an interesting feature of the school. There you can find different kinds of snacks that I know you would love given how much you love snacks. I would only recommend that you make healthier choices so that you do not become overweight or fall ill due to overindulgence in unhealthy meals.

My dear baby girl, be assured of my love and kind wishes as you move on to this next stage of your life. Always know that I am here to give you my advice when you need it. Be a good girl my dear sister. Do have a lovely time at school.

Yours lovingly,

Queen Walters.

Answer:

775 N  due North.

Explanation:

If the crate is pulled South with 350 N force, and the net force on the crate results into 425 N due North, then the other force (F) acting must be larger than the 350 N, and pointing North:

F - 350 N = 425 N

F = 425 N + 350 N = 775 N  due North.

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J  whose closest answer is 1,595 J

The change in the internal energy of the gas 1,595 J

the emitter of x-rays rotates around the patient and the detector, placed in diametrically opposite side, picks up the image of a body section (beam and detector move in synchrony)

i believe that should be ur answer =) good luck !

Answer:

a = 11.03 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force = 160000 [N]

m = mass = 14500 [kg]

a = acceleration [m/s²]

160000 = 14500*a

a = 11.03 [m/s²]

Answer: A plane mirror always forms a virtual image (behind the mirror). The image and object are the same distance from a flat mirror, the image size is the same as the object size, and the image is upright.

Explanation:

Answer:

a) they both will appear to have the same loudness

Explanation:

When we talk about how loud a sound is , we describe it in decibels (dB). The value of this unit on measurement in this question is what we are going to build our answer on.

Their sound intensity level is the same at 60db even though each of these separate sounds have different frequencies of 40Hz and 100Hz respectively.

The both of them will therefore have the same loudness actually as their sound intensity level has no difference bat 60db each.

Explanation:

Total energy (kinetic + potential)

1/2×mv^ + mgh

1/2×5×8×8+ 5×9.8×10

1/2×320 + 5×98

320/2 + 480

320/2 + 980/2 (multipling to make the denominator equal)

1300/2 = 650 joules

Hope it helps, if you don't understand try doing it yourself with the formula.

:)

Answer:

a

[tex]H =212.6 \ J[/tex]

b

[tex]v = 7.647 \ m/s[/tex]

Explanation:

From the question we are told that

   The child's weight is  [tex]W_c = 287 \ N[/tex]

    The length of the sliding surface of the playground is  [tex]L = 7.20 \ m[/tex]

    The coefficient of friction is  [tex]\mu = 0.120[/tex]

      The angle is [tex]\theta = 31.0 ^o[/tex]

      The initial  speed is  [tex]u = 0.559 \ m/s[/tex]

Generally the normal force acting on the child is mathematically represented as

=>    [tex]N = mg * cos \theta[/tex]

Note  [tex]m * g = W_c[/tex]

Generally the frictional force between the slide and the child is    

         [tex]F_f = \mu * mg * cos \theta[/tex]

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      [tex]F =m* g sin(\theta) - F_f[/tex]

Here  F is the resultant force and it is represented as  [tex]F = ma[/tex]

=>   [tex]ma = m* g sin(31.0) - \mu * mg * cos (31.0)[/tex]

=>   [tex]a = g sin(31.0)- \mu * g * cos (31.0)[/tex]

=>  [tex]a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)[/tex]

=>[tex]a = 4.039 \ m/s^2[/tex]

So

   [tex]F_f = 0.120 * 287 * cos (31.0)[/tex]

=> [tex]F_f = 29.52 \ N[/tex]

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     [tex]H = F_f * L[/tex]

=>  [tex]H = 29.52 * 7.2[/tex]

=>  [tex]H =212.6 \ J[/tex]

Generally from kinematic equation we have that

    [tex]v^2 = u^2 + 2as[/tex]

=>  [tex]v^2 = 0.559^2 + 2 * 4.039 * 7.2[/tex]

=>  [tex]v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}[/tex]

=>  [tex]v = 7.647 \ m/s[/tex]