can someone help me answer this question? A student connected an ammeter as shown in this picture. Did the student connect the ammeter correctly? Explain

Answer:

  1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

  5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

  10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

  (5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

  (10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

  v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

Answer:

False..

Explanation:

An electoMagnets attract iron due to the influence of their magnetic field upon the iron. ...

Explanation:

[tex]v = u + at \\ 0 = u + ( - 10 { ms}^{ - 2}) \times 4.2s \\ u = 42 {ms}^{ - 1} [/tex]

The motion of the bouncy mushroom can be described as a simple

harmonic motion, SHM.

Reasons:

The given parameters are;

Mass of the mushroom cap, m = 30 g = 0.03 kg

Mass of the bird = 50 g = 0.05 kg

The spring constant, K = 20 N/m

a) The equilibrium height of the mass spring system, is given as follows;

F = -K·x

[tex]x = \dfrac{F}{K}[/tex]

The applied force, F = The weight of the bird

∴ F = (0.05 kg) × 9.81 m/s² = 0.4905 N

[tex]x = \dfrac{0.05 \, kg \times 9.81 \ m/s^2}{20 \, N/m} = 0.024525 \, m[/tex]

The equilibrium height of the mushroom is 0.024525 m below its initial height.

b) The frequency of oscillation of a spring, ω, is given as follows;

[tex]\omega = \sqrt{\dfrac{K}{m} }[/tex]

Therefore;

[tex]\omega = \sqrt{\dfrac{20 \, N/m }{80 \, kg} } = \dfrac{1}{2} \, Hz[/tex]

The frequency of resulting oscillation, ω = [tex]\dfrac{1}{2} \, Hz[/tex] = 0.5 Hz

c) The applied force, F = The weight of the bird and the mushroom cap

F = (0.03 kg + 0.05 kg) × 9.81 m/s² = 0.7848 N

[tex]x = \dfrac{0.7848 \, N}{20 \, N/m } = 0.03924 \, m[/tex]

The maximum compression of the mushroom = 0.03924 m

d) The motion of the mushroom is a Simple Harmonic Motion, SHM.

The equation of a SHM as a function of time is; x(t) = A·cos(ω·t + Φ)

For the mushroom, we have;

The amplitude, A = 0.03924 m - 0.024525 m = 0.014715 m

Ф = The phase angle

When t = 0, cos(ω × 0 + Φ) = 1

cos(Φ) = 1

Ф = arcos(1) = 0

The equation is therefore;

x(t) = 0.014715·cos(0.5·t)

Equation of the oscillation of the mushroom is;  [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]

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Answer: a, c, d

Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it

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Explanation:

inertia ...

you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.

please consider : we are only talking about moving. not about acceleration.

so, yes, (a) is true.

(c) is true.

(d) is true.

The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;

As we know; the periodic table is an array of elements in order of their atomic number.

In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.

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Gravitational pull downward on the object:  7.8 x 9.8 = 76.44N

A. 7.8 x 1.5 = 11.7N upward force

Tension = 76.44 + 11.7 =  88.14 N

Answer: 88.14 N

B. 76.44 - 11.7 = 64.74N

Answer: 64.74 N

Answer:

88.14 N

64.74 N

Explanation:

Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]

These combinations of frequency produce 4 beats per sound.

i.e.

[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]

[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]

[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]

When it is altered, the beats first diminish and increase again by 4.

i.e.

[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]

[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]

If we equate both equations (1) and (2) together, we have:

[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

∴

[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]

By squaring both sides, we have:

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]

[tex]\implies \mathbf{273 +T =306.726912 }[/tex]

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

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The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]

The distance between the glass is calculated as follows;

[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

Answer:

I'm pretty sure it true sorry if I'm wrong

The motion of the rope which is perpendicular to the direction of the

propagation of the wave is a transverse wave motion.

Reasons:

The given function for the wave speed is presented as follows;

[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]

Where;

[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]

Taking the mass of the rope as, m = 2.00 kg

The length of the rope, L = 80.0 m

The mass hanging on the rope, M = 20.0 kg

We have;

T = 20.0 kg × 9.81 m/s² = 196.2 N

[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]

Therefore;

Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;

v = f × λ

Therefore;

v = 7.9 Hz × 7.9 m = 62.41 m/s

Which gives;

[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]

T = 62.41² × 0.025 = 97.3752025

[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]

Where;

g = The acceleration due to gravity which is approximately 9.81 m/s²

[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]

Therefore;

The mass of the box, m ≈ 9.93 kg

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The parameters obtained from a similar question online are;

[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]

Length of the rope, L = 80.0 m

Mass of the rope, m = 2.0 kg

Frequency of a point on the rope, f = 20 Hz

The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

The given parameters;

The magnitude of the magnetic field inside the solenoid is calculated as;

[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]

where;

[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A

[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]

Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

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Answer:

1) Dependent variable: Type of separation method used

   Independent variables: Substance that separated out

  Constants: Sand, pepper, and salt

2) you can use the properties of matter to decide on a method to separate out a particular substance based on its unique properties. For example, knowing pepper has a very small mass allows you to use the charge of the static electricity to separate out those particles. Additionally, knowing the solubility of salt allows you to add water to the mixture to be able to remove it from the sand, since the salt dissolves in the water which is poured away.

Answer:

Increases with increase in mass

Explanation:

gravity is proportional to mass and inversely proportional to the square of the distance between them

F = GMm/d²

I think 1980is the answer because you add???

Identify:

In each case the forces are constant and the displacement is along a straight line, so

[tex]$$W=F s \cos \phi \text {. }$$[/tex]

Set-Up:

In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls  [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car  [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.

Execute:

(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].

When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].

(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],

So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].

(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.

Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork

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Explanation:

→ p = 3 kg × 3 m/s

→ p = 9 kg.m/s

Option D is correct.

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

    This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

    Would adding another question to an already broad question help? No.

O Use the very first source you find for your project

    If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question

O Change the scope of your project​    

    You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.

Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:

[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]

Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:

[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]

Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:

[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]

Finalmente, reemplazamos los valores para obtener:

[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]

Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.

Aprende más:

Answer:

600 N

Explanation:

The force exerted from the engine to maintain this forward acceleration is 600 N.

Force is defined in physics as: the push or pull on an object with mass that causes it to change velocity. Force is an external agent that can change the state of rest or motion of a body.

The term "force" has a specific meaning in science. At this level, it is perfectly acceptable to refer to a force as a push or a pull.

A force is not something that an object possesses or possesses. Another object applies a force to another.

We know that,

F = ma

Here, it is given that

Acceleration = 3 m/s

Mass = 200Kg.

So, by keeping the value

F = 200 x 3

F = 600 N

Thus, the force is 600 N.

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The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;

       x = 0.026 m

Given parameters

To find

The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. With maximum compression.

        Em₀ = U = ½ k x²

Final point. Where the kinetic and potential energy are equal.

        [tex]Em_f = K +U[/tex]  

Since the mechanical energy is constant at this point K = U, therefore we can write the energy.

        [tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]

Energy is conserved.

        [tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf

        ½ k x² = 2 (½ k xf²)

        [tex]x_f = \frac{x_o}{2}[/tex]  

let's calculate.

        [tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]  

In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;

       x = 0.026 m

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h =(3.7 - .58)m  = 3.12m

Now put PE into KE and we have to use the formula:

√2gh (g = gravity and h = height) therefor:

√2 x 9.8 x 3.12

= 7.82m/s

I hope this helps!

Answer:

The answer is D

Explanation:

The chloroplast absorbs sunlight for energy not the cell wall

Answer:

b

Explanation:

when the wavelength increase it doesnt affect the frequency of a wave.

Answer: The wavelength increases by a factor of 3

Explanation:

Explanation:

I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.

Answer:

a = 6.22 m//s²

Explanation:

F = ma

a = F/m

a = 573.0 / 92.1

a = 6.221498...

Answer:

[tex]\boxed {\boxed {\sf 6.22 \ m/s^2}}[/tex]

Explanation:

We are asked to find the acceleration due to gravity on another planet.

Weight is the measure of the force of gravity. Therefore, we can use the following version of the force formula:

[tex]F_g=mg[/tex]

In this formula, [tex]F_g[/tex] is the weight, m is the mass, and g is the acceleration due to gravity.

The object weights 573.0 Newtons (or 573.0 kg*m/s²) on the planet. The object has a mass of 92.1 kilograms.

Substitute these values into the formula.

[tex]573.0 \ kg*m/s^2 = 92.1 \ kg * g[/tex]

We are solving for g, so we must isolate the variable. It is being multiplied by 92.1 kilograms. The inverse of multiplication is division, so divide both sides of the equation by 92.1 kg.

[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}= \frac{92.1 \ kg*a}{92.1 \ kg}[/tex]

[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}=a[/tex]

The units of kilograms cancel.

[tex]6.22149837 \ m/s^2=a[/tex]

The original measurements of weight and mass have 4 and 3 significant figures. Our answer must have the least number of sig figs, or 3. For the number we found, that is the hundredth place. The 1 in the thousandths place tells us to leave the 2 in the hundredth place.

[tex]6.22 \ m/s^2=a[/tex]

The acceleration due ot gravity on planet Xyleneer is approximately 6.22 meters per second squared.

Answer:

thanks for the points dia