A 0.015 kg marble sliding to the right at 0.225 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 0.180 m/s. After the collision, the first marble moves to the left at 0.180 m/s. Find the velocity of the second marble after the collision.

Answer:

C. Velocity and Position

Explanation:

The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.

The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.

The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.

Therefore, the correct option is b. velocity and position.

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The question is incomplete. Your most probably complete question is given below:

Velocity and acceleration

velocity and position

velocity only

velocity, position, and acceleration

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.

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Answer:

its b it must be charge

Explanation:

Answer:

non economic benefits to international trade means not being too much profit not selling too much things /items in international trade

Answer:

See Below

Explanation:

I am not sure what you exact question is to know what direction it is but here is how you do it.

The direction of the net force is the direction of the largest force.

for example if you were to push a box forward with 100 newton's of force and someone pushed at the same time 50 newton's backwards on the box, the box would move forwards because the was a greater force on the box in a forward direction. hope this helps

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

Age of the universe; [tex]t = \ ?[/tex]

We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:

[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]

Now,

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

We know that;

[tex]1\ light\ years = 9.46*10^{15}m[/tex]

so

[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]

Therefore;

[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]

Now, we input this Hubble's constant value into our equation;

[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.

Answer:

thx for the points

Explanation:

no need brainliest

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.

Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.

Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.

Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.

The following is an accurate matching of the ions;

11 protons, 12 neutrons, 10 electrons = +1 - Na^+

31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+

Group 2, period 5 = +2 - Sr^2+

Silver is a lustrous metal.

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Answer:

If your answers are the green check marks all of them are correct

Explanation: I took the test

Hi there!

We know the following kinematic equation:

vf = vi + at

Where:

vf = final velocity

vi = initial velocity

a = acceleration

t = time

In this instance, the ball is experiencing a constant acceleration of that of gravity, thus:

vf = 12 + 10(1) = 22 m/s (if downward is considered positive in this instance)

Answer:

Ammonium perchlorate NH4ClO4

Ammonium Nitrate

Calcium Cyanamide

When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.

Hopefully this helps :)

this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see

Explanation:

a solar eclipse means when the moon goes infront of the sun and the earth turns dark

The rate of heat energy exhausted to the river is 600.96 MW.

The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.

Given parameters:

Efficiency of the power plant; η = 31 %

Output  electric power; O = 270 MW.

We know that, Efficiency of the power plant;

η  = (Output  electric power/ input power)× 100%

⇒ input power = (Output  electric power × 100)/η

⇒ input power = (270 × 100)/31 MW

= 870.96 MW.

So, the rate of heat energy exhausted to the river that cools the plant =  Input power- output power

= (870.96 - 270) MW

= 600.96 MW.

Hence, heat energy exhausted to the river that cools the plant  is 600.96 MW.

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Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

Answer:

3.9m/s² is the acceleration

The acceleration of the object is 3.9m/s².

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

When you fall off a bridge. The car turning at the corner is an example of acceleration because the direction is changing. The quicker the turns, the greater the acceleration.

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5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get

[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]

Taking the square root of r^2, we then get the distance as

[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]

The values are given as follows:

[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]

[tex]g = 9.8\:\text{m/s}^2[/tex]

[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]

[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]

Putting in all of these values in our equation for r,

[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]

Answer:

I don't understand your language

The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.

The given parameters;

The normal force on the wooden box is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]

The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;

[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]

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Answer:

Explanation:

Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.

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

Explanation:

Answer:

The answer is D

Explanation:

The chloroplast absorbs sunlight for energy not the cell wall

Answer:

the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.

Answer:

Explanation:

Ignoring friction and assuming one ramp end rests on level ground.

gravity acceleration acting parallel to the ramp is gsinθ.

F = mgsinθ

mg = 250 N,  sinθ = 3/25

F = 250(3/25)

F = 30 N

A 25 meter long ramp strong enough to hold a person plus a 250 N refrigerator would weigh much more than 250 N.

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

[tex]F_{1}[/tex] = 250

[tex]A_{1}[/tex] = Area of the small piston = π[tex]r^{2}[/tex]

[tex]A_{1}[/tex] = 22/7 x [tex]0.4^{2}[/tex]

[tex]A_{1}[/tex] = 0.5 [tex]m^{2}[/tex]

[tex]F_{2}[/tex] = ?

[tex]A_{2}[/tex] = Area of the large piston = π[tex]r^{2}[/tex]

[tex]A_{2}[/tex] = π x 1

[tex]A_{2}[/tex] = 3.14 [tex]m^{2}[/tex]

To calculate the force on the large piston, we will use the below formula

[tex]F_{1}[/tex]/ [tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]

Substitute all the parameters into the equation

250/0.5 =  [tex]F_{2}[/tex]/3.14

[tex]F_{2}[/tex] = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

[tex]F_{1}[/tex]/ 0.25[tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

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The total reactance of the inductor and the capacitor is 470.1 ohms.

The given parameters;

The inductive reactance of the coil is calculated as follows;

[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]

The capacitive reactance of the capacitor is calculated as follows;

[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]

Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.

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answer: what language is this???

Answer:

Frictional force is produced when two bodies are rubbed against each other. It is the force that oppose the motion and therefore it stops or slow down a moving body.It depends upon the roughness or smoothness of the surface of the body in contact.Rough surface have more friction that the smooth surface. Similarly, the heavier body produces more friction than a lighter body. Frictional force acts in the opposite direction of the motion of the body.

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904