what is a four-step energy transformation chain from mechanical to electrical? (something that we use in everyday life)

Answer:

When speed reduces, acceleration decreases

Explanation:

From first equation of motion;

\({ \tt{v = u + at}}\)

When time is constant:

\({ \tt{v = 0 + kt}} \\ { \tt{v = kt}} \\{ \tt{v \: \alpha \: \: a}}\)

When speed is decreasing, velocity and acceleration are in opposite directions.

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is  50 N upward normal force

From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be

Weight W = mg

let g = 10 m/\(s^{2}\)

W = 5 x 10

W = 50 N

The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,

Normal force N = 50 N

The block will remain on the table because the normal force balances with the weight of the block.    

Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.

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Answer:

True

Explanation:

the independent variable would be like if the psychologist is experimenting the impact of sleep deprivation, sleep deprivation would be the independent variable that is being manipulated or changing its level systematically in the experiment.

- The resistance of the first resistor (R₁) is 12 Ω.

- The electromotive force (EMF) of the battery is 18 V.

- The internal resistance of the battery is 12 Ω.

To solve the given problem, we can apply Kirchhoff's laws and Ohm's law to determine the resistance of the first resistor (R₁) and the electromotive force (EMF) and internal resistance of the battery.

Let's start by calculating the resistance of the first resistor (R₁):

1. Apply Ohm's law to find the voltage drop across the external circuit:

  V = I * R

  V = 1 A * 6 Ω

  V = 6 V

2. The voltage drop across the external circuit is equal to the EMF minus the voltage drop across the internal resistance of the battery:

  V = E - Ir

  6 V = E - (1 A * r)   (where r is the internal resistance of the battery)

3. We also know that the EMF of the battery is the sum of the voltage drops across each cell in the battery:

  E = 6 cells * 3 V/cell

  E = 18 V

4. Substitute the value of E in the equation from step 2:

  6 V = 18 V - r

  r = 12 Ω

Therefore, the resistance of the first resistor (R₁) is 12 Ω.

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According to the give value of the mass of lemonade and change in the temperature, the specific heat capacity of the lemonade is 3.81 J/kg/k.

The formula for the specific heat capacity (C) = Q/ m × ΔT

Q is energy added and the value, which is 334J

m is the mass of the lemonade, which is 300g = 0.3 kg

Δ T is the change in temperature, which is 19°C =(273+19)= 292 K

So, C = 334/ 0.3 × 292

         = 3.81 J/kg/k

So the specific heat capacity of the lemonade is 3.81 J/kg/k.

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Answer:

sample B

Explanation:

The samples given all have the same mass which is 100g but their volume differs significantly.

Density is the mass per unit volume of a substance. The mass is the amount of matter contained in a substance. Volume is the space a body occupies.

 Density  = \(\frac{Mass}{Volume}\)  

Density is inversely proportional to volume. The higher the volume the lesser the density provided the mass is constant.

So, the sample with the least volume will have the greatest density value.

This sample is sample B

Answer:

the answer is b

Explanation:

The required, it experiences a downward force due to gravity and a force due to air resistance.

Projectile motion is the movement of an entity projected into space. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.

Here,

When throwing a ball vertically upward, there is a displacement of about 1.0 m from the initial position of the hand to the position where the ball leaves the hand. The mass of the ball is 0.80 kg and it reaches a maximum height of about 20 m above the initial position of the hand. While the ball is in the hand after it leaves, it experiences a downward force due to gravity and a force due to air resistance.

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Answer:

Volume measures liquid

Explanation:

Volume can be measured in meters and centimeters

Complete question is;

One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.

The ratio of their potential energies at highest points of their journey, will be:

Answer:

u² : (u cos θ)²

Explanation:

Maximum potential energy for the first ball will be at a maximum height of;

H = u²/2g

Thus;

PE = mg(u²/2g)

For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g

PE = mg((u cos θ)²/2g)

The ratios of the potential energies are;

mg(u²/2g) : mg((u cos θ)²/2g)

mg will will cancel out since they are of same mass.

Thus;

(u²/2g) : (u cos θ)²/2g

Again 2g will cancel out to give;

u² : (u cos θ)²

Answer:

Explanation:

From the given information:

Let's assume that the missing function is:

s(t) = t³ - 6t², t ≥ 0

From part (b), we are to find the given  required terms when time t = 2

So; from the function s(t) =  t³ - 6t², t ≥ 0

\(velocity \ v(t) \ = \dfrac{d}{dt}s(t)\)

\(velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)\)

\(velocity \ v(t) \ = 3t^2 - 12t\)

\(acceleration a(t) = \dfrac{d}{dt}*v(t)\)

\(acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)\)

\(acceleration\ a(t) = 6t - 12\)

At time t = 2

The position; S(2) = (2)² - 6(2)²

S(2) = 8 - 6(4)

S(2) = 8 - 24

S(2) = - 16 ft

v(2) = 3(2)² - 12 (2)

v(2) = 3(4) - 24

v(2) = 12 - 24

v(2) = - 12 ft/s

speed = |v(2)|

|v(2)|  = |(-12)|

|v(2)| = 12 ft/s

acceleration = 6t - 12

acceleration = 6(2) - 12

acceleration =  12 - 12

acceleration =  0 ft/s²

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

\(E_P=\int dE \cos\)

\(E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}\)

\(\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ\)

\(\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}\)

If we put an electron on point P, then force on point e is :

\(\vec{F}=-|e|\vec{E_P}\)

\(F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}\)

If r >> x , then    \($\frac{x^2}{r^2} \approx 0$\)

Then,  \($\frac{-eKQ}{r^3}x$\)

\($ma =\frac{-eKQ}{r^3}x$\)

\($a =\frac{-eKQ}{mr^3}x$\)

Compare, a = -ω²x

We get,

\($\omega^2 = \frac{eKQ}{R^3m}$\)

\($\omega = \sqrt{\frac{eKQ}{r^3m}}$\)

\($2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$\)

\($f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$\)

To determine the angle through which the wheel rotates in 1.00 second, we can start by finding the angle covered in 0.0710 seconds and then scale it up to 1.00 second.

In 0.0710 seconds, the wheel completes 3.10 revolutions. One revolution corresponds to an angle of 360 degrees or 2π radians. Therefore, in 0.0710 seconds, the wheel rotates through an angle of:

Angle = 3.10 revolutions * 2π radians/revolution = 6.20π radians

To find the angle in 1.00 second, we can use proportional reasoning. Since the time increases by a factor of 1.00/0.0710, the angle covered will also increase by the same factor:

Angle in 1.00 second = 6.20π radians * (1.00/0.0710) = 87.32π radians

Approximately, the angle through which the wheel rotates in 1.00 second is 274.39 radians.

Therefore, the wheel rotates through an angle of approximately 274.39 radians in 1.00 second.

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Answer:

hzdydutsydududfudfufxudyxdtz

In physics, a lower fixed point refers to the minimum temperature at which a particular substance or system can reach and below which it cannot be cooled further.

It is a fundamental concept in the study of thermodynamics, specifically in relation to phase transitions and the behavior of substances at low temperatures.

The lower fixed point is often associated with the concept of absolute zero, which is the lowest possible temperature in the Kelvin scale (-273.15 degrees Celsius or -459.67 degrees Fahrenheit).

At absolute zero, particles in a substance possess the minimum amount of energy and their motion ceases, resulting in the absence of thermal energy.

The lower fixed point serves as a reference point for temperature scales, such as the International Temperature Scale of 1990 (ITS-90), which defines temperature measurements based on fixed points like the melting point of certain substances and the triple point of water.

These fixed points provide reproducible and well-defined temperature values for calibration and measurement purposes.

Understanding the lower fixed point is crucial for various scientific and technological applications, such as cryogenics, superconductivity, and the study of quantum phenomena at extremely low temperatures.

By pushing the boundaries of cooling techniques, researchers aim to approach the lower fixed point and explore the fascinating properties and behaviors of matter at such extreme conditions.

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Answer:

The acceleration would be 10m/s.

Explanation:

To find this out, use the formula:

A = speed/time:

30 divided by 3 = 10m/s.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Given:

The work done on the box from y=0 to 6.0m is 120 J.

To calculate the work done on the box from y=0 to 6.0m, we need to find the area under the force vs. position graph over that interval.

First, we can find the work done from 0 m to 2 m. Since the force is constant at 40 N over this interval, the work done is simply:

W = F * d = 40 N * 2 m = 80 J

From 2 m to 6 m, the force is constant at -20 N, so the work done is:

W = F * d = (-20) N * 4 m = -80 J

Note that the negative sign indicates that the work is done by the box on the force (since the force is in the opposite direction of the displacement).

Therefore, the total work done on the box from y=0 to 6.0m is:

W_total = 80 J - 80 J = 0 J

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The diameter of the smaller end of the pipe is approximately 0.78 meters.

To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.

The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity

Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2

Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2

Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30

Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30

0.36 * 13 = (diameter2/2)^2 * 30

4.68 = (diameter2/2)^2 * 30

Dividing both sides by 30:
0.156 = (diameter2/2)^2

Taking the square root of both sides:
0.39 = diameter2/2

Multiplying both sides by 2:
0.78 = diameter2

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Answer:

Neither the source or wall is moving so the sound takes 3 sec to get to the wall and 3 sec to return

D = 3 * v = 330 m/s * 3 sec = 990 m

There is more to the Earth than what we can see on the surface. In fact, if you were able to hold the Earth in your hand and slice it in half, you'd see that it has multiple layers. But of course, the interior of our world continues to hold some mysteries for us. Even as we intrepidly explore other worlds and deploy satellites into orbit, the inner recesses of our planet remains off limit from us.

However, advances in seismology have allowed us to learn a great deal about the Earth and the many layers that make it up. Each layer has its own properties, composition, and characteristics that affects many of the key processes of our planet. They are, in order from the exterior to the interior – the crust, the mantle, the outer core, and the inner core. Let's take a look at them and see what they have going on.

Like all terrestrial planets, the Earth's interior is differentiated. This means that its internal structure consists of layers, arranged like the skin of an onion. Peel back one, and you find another, distinguished from the last by its chemical and geological properties, as well as vast differences in temperature and pressure.

Explanation:

The work can be seen as a transfer of energy, and can be calculated with the formula below:

Where W is the work in Joules, F is the force in Newtons and d is the distance in meters.

When we have a positive work, that means the body or system is using its own energy and force to move a certain distance, for example.

A negative work means the system is receiving energy from an external source.

Therefore the statement is TRUE.

Answer:

A) Arrangement A

Explanation:

The rate of heat conduction is given by Fourier's Law of Heat Conduction. It is given as follows:

Q = KAΔT/L

where,

Q = Rate of Heat Transfer or Conduction

K = Thermal Conductivity

A = Cross-Sectional Area

ΔT = Difference in Temperature

L = Thickness

So, it is clear from the formula that for a constant temperature difference and value of thermal conductivity, the rate of heat transfer is directly proportional to the cross-sectional area and it is inversely proportional to the thickness.

Therefore, the arrangement with larger cross-sectional area and smaller thickness will be the one with the greatest heat transfer rate and as a result greatest heat shall be conducted through that arrangement.

It is clear that the parallel arrangement that is arrangement A, has higher cross-sectional area and smaller thickness. Therefore, the correct option is:

A) Arrangement A

The resistance that must be placed in parallel to achieve a combined resistance of 15 ohms is also 15 ohms.

To determine the resistance that must be placed in parallel to achieve a combined resistance of 15 ohms, we can use the formula for calculating the total resistance of resistors in parallel.

In a parallel combination, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. Mathematically, this can be expressed as:

1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn

In this case, we are given that the combined resistance should be 15 ohms, and we need to find the resistance that should be placed in parallel.

Let's denote the unknown resistance as R_unknown.

Using the formula for resistors in parallel, we can write:

1/15 = 1/R_unknown

To solve for R_unknown, we take the reciprocal of both sides:

15/1 = R_unknown/1

Therefore, R_unknown = 15 ohms.

So, the resistance that must be placed in parallel to achieve a combined resistance of 15 ohms is also 15 ohms.

By adding a 15-ohm resistor in parallel to the existing circuit, the combined resistance will decrease, as resistors in parallel offer a smaller equivalent resistance compared to individual resistors. The new circuit will have a combined resistance of 15 ohms, allowing for the desired total resistance.

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Answer:

Explanation:ans:0 . Because when we connect this conductive sphare with earth then all charge will be gone in earth