Wheel A has a radius of 0.5m and Wheel B has a radius of 0.25m. Both wheels are turning at 200 rpm
and 100 rpm (revolutions per minute) each, when we simultaneously apply a brake to them, remaining at
rest in 10s and 20s respectively. Find the frequency of each wheel. Find the initial angular velocity of
each wheel. The angular acceleration of each wheel. Find the time that both wheels will be turning with
the same angular velocity. Find the angular velocity of both wheels at the time found above. Find
for the previous time, the value of the tangential acceleration at a point located on the periphery of both wheels.
Find for the previous time the value of the centripetal acceleration at a point located on the periphery of both
wheels. When the wheels are stopped, what total angle will it have turned from the moment we apply
the breaks?​

Resistance of snorkeler, R = 605 Ω.

Current, I = V/R 500/605  = 0.83 A

⇒ I = 0.83 A.

The flow of charge across the cell membrane creates an electric current. Cells control the flow of specific charged elements across membranes using proteins that reside on the cell surface and create openings for specific ions to pass through. These proteins are called ion channels.

For example, if there is a voltage difference between one arm and the other a current will flow through the body. The human body has a higher impedance to DC current than to its AC current. That is, a person can withstand the effects of an electric shock caused by a DC voltage more than he can by an AC voltage. It interferes with the normal functioning of the nervous and muscular systems, causing severe muscle contractions.

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Answer: 4.7× 10-8 gigameter.

Exgiga meter. Hope this helps :)

Answer:

quizlet

Explanation:

they help

Answer:

230° C

Explanation:

A substance's specific heat tells you how much heat much either be added or removed from 1 g of that substance in order to cause a 1∘C

C) a bicyclist riding up a steep hill

The metaphor for a system with rising gravitational potential energy is "a bicyclist riding up a steep hill." Let's get into greater detail:

A cyclist faces resistance from gravity as they ride up a steep slope. The cyclist's elevation, or height above the ground, rises as they cycle and climb uphill. Gravity is pulling the cyclist down the hill by exerting downward force. The cyclist must apply force to the pedals in order to move forward and overcome the pull of gravity. In order to do this, the bicyclist must transform chemical energy from their body into mechanical energy. The distance of the cyclist from the centre of the Earth grows as they ride up the hill. The height and mass of an object affect its gravitational potential energy. In this scenario, as the bicyclist's height rises, their gravitational potential energy also rises.

     Due to the higher elevation, the energy input from the biker is stored as increased potential energy. When the bicycle descends the hill or does work, this potential energy can be transformed back into kinetic energy or other types of energy.

A student sets up four cups with 40 mL of water in each and adds different amounts of ice to each cup.

To determine the connection between the temperature change and the transfer of kinetic energy, the student can measure the initial and final temperatures of the water in each cup and the mass of the ice added to each cup. Then, the student can use the following equation to calculate the amount of heat transferred from the ice to the water

Q = m × c × ΔT

Where Q is the amount of heat transferred, m is the mass of the ice added, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature of the water.

The student can then compare the amount of heat transferred from the ice to the water in each cup to the change in temperature of the water. If the temperature change is greater in a cup where more heat was transferred, this suggests a direct connection between the transfer of kinetic energy (as heat) from the ice to the water and the temperature change of the water.

Hence, This can be further supported by calculating the temperature change per unit of heat transferred, which should be approximately the same for each cup if there is a direct connection between the transfer of heat and the temperature change.

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(a). The chair's dynamic friction factor with the floor is = 0.602.

(b). The chair's as well as the floor's kinetic friction factor is = 0.475.

The metre per second (m/s) is the accepted unit of velocity magnitude (also known as speed).

a)Let's practice using Newton's second law. In the reference system, the x axis is perpendicular to the axis and parallel to the floor.

Y Axis

        N-W = 0

        N = W

X axis

         F-fr = ma

The formula for the friction force is

  fr = μN

The acceleration is zero where the movement starts.

Let's replace

F = μ (mg) = 0

F = μ mg

μ = F / mg

μ = 160/30 9.81

μ = 0.602

b) The formula is the same in this instance, but the friction is kinetic.

    μ = F2 / mg

    μ = 125/30 9.81

    μ = 0.475

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Answer:

a reaction between an acid and a base

Explanation:

Answer:

The answer is A. A reaction between an acid and a base

Explanation:

The acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity with time.

The  acceleration of the crate is calculated from the net force acting on the crate as shown below;

F(net) = applied force - frictional force

F(net) = 10 N - 5N

F(net) = 5 N

Apply Newton's second law of motion to calculate the acceleration of the crate.

F(net) = ma

where;

a = F(net) / m

a = 5/5

a = 1 m/s²

Thus, the acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

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The amount of force required by the farmer to move the pig is 1176N. A force in physics is an effect that has the power to alter an object's motion.

A mass-containing object's velocity can be altered by a force (e.g. moving from a state of rest). Motion can be characterised as a shift in an object's location with regard to time. Motion is demonstrated by things like a book falling off a table, water running from a faucet, rattling windows, etc. There is motion even in the air we breathe! The universe is a moving thing. The universe in which we exist is constantly in force.

Fmax = 0.8*150*9.8 Fmax = 1176N is the maximum force needed to move the pig.

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The complete question is -

Problem 5.29 7 of 15> A stubborn 150 kg pig sits down and refuses to move. To drag the pig to the barn, the exasperated farmer ties a rope around the pig and pulls with his maximum force of 800 N. The coefficients of friction between the pig and the ground are p, = 0.80 and μ'= 0.50. Part A Calculate the force which farmer needs to apply to budge the pig. Express your answer to two significant figures and include the appropriate units. HA F Value Units Submit Request Answer Part B This question will be shown after you complete previous question(s).

The expression for the initial velocity at point A is:

0 = (velocity at point A - 0) / time

Simplifying the equation, we find:

Velocity at point A = 0

The initial velocity at point A is zero, indicating that the object starts from rest before sliding on the horizontal plane AB.

To write an expression for the initial velocity at point A, we need to analyze the forces acting on the object and apply the principles of motion.

Given:

Mass of the object, m

Radius of the semi circle, R

Coefficient of kinetic friction, μ\(_k\) = 0.5

Force applied at point C, F = 3mg

The object is initially at rest.

Let's break down the motion into two parts: the motion on the horizontal plane AB and the motion along the semi circle BCD.

1. Motion on the horizontal plane AB:

The only force acting on the object on the horizontal plane is the force of kinetic friction. The frictional force can be calculated using:

Frictional force, f = μ\(_k\)* Normal force

The normal force is equal to the weight of the object, which is mg.

Normal force, N = mg

Frictional force, f = μ\(_k\) * mg

The frictional force acts in the opposite direction to the motion, so its magnitude is negative. Thus, the net force on the object on the horizontal plane is:

Net force = -f = -μ\(_k\)* mg

Using Newton's second law, we can relate the net force to the acceleration:

Net force = mass * acceleration

-μ\(_k\) * mg = m * acceleration

The acceleration can be expressed as the rate of change of velocity:

Acceleration = (final velocity - initial velocity) / time

Since the object is initially at rest, the initial velocity is zero.

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The coefficient of thermal conductivity (k) is related to the rate of heat flow (Q), the cross-sectional area (A), the length (L), the temperature difference (ΔT), and the thermal resistance (Rth) by the following equation:

k = Q / (A * ΔT * L) = Rth * (A * ΔT)

Reorganizing this equation gives:

Rth = k / (A * ΔT)

The given information in the problem is:

Rate of heat flow (Q) = 14.0 watts

Thermal resistance (Rth) = (350 mm × 350 mm × 15 mm) / (14.0 watts) = 31.5 mm⁴/C

Temperature difference (ΔT) = 28°C

Substituting these values into the equation, we have:

k = Q / (A * ΔT) = 14.0 W / (0.35 m² * 28°C) = 1.94 W/mK

So the coefficient of thermal conductivity (k) for this wood is approximately 1.94 W/mK.

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

\(F=\dfrac{Gm_em_m}{r^2}\)

It means \(F\propto \dfrac{1}{r^2}\)

So,

\(\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}\)

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

\(\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}\)

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

Answer:

1.  Energy

2. All of the above

3. Store and use

4.  Kinetic

5. mass and velocity

6. Potential

7. mass, height, and gravity

8. Chemical Potential

9. Elastic Potential

10.  (`⌒*)O-(`⌒´Q)

Explanation:

The work done by the force on the block is approximately 311.2 Joules.

To calculate the work done by the constant force of magnitude F = 45 N over a distance of 8 m at an angle of 30 degrees to the horizontal, we need to find the component of the force that acts parallel to the displacement.

The horizontal component of the force can be calculated using trigonometry:

F_horizontal = F * cos(angle)

            = 45 N * cos(30 degrees)

            = 45 N * (√3 / 2)

            ≈ 38.9 N

Now, we can calculate the work done by the force using the equation:

Work = Force * Distance * cos(theta)

where theta is the angle between the force and the displacement.

Work = F_horizontal * Distance * cos(0)

     = 38.9 N * 8 m * cos(0)

     = 38.9 N * 8 m

     = 311.2 Joules

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Answer:

D

Explanation:

The CV system does all of the above

The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.

The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;

The total voltage in loop 1 is calculated as;

2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

The current flowing in loop 2 is calculated as;

I = V/R

I₂ = ( 6 V - 4 V ) / (3 + 4)

I₂ = 0.286 A

The value of the current flowing in loop 1 is calculated as;

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0

6 - 3I₁ - 3₁ + 0.858 = 0

-6I₁ = -6.858

I₁ = 6.858 / 6

I₁ = 1.14 A

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It's not 65 s, I can tell you that for sure

The cone emits a single-frequency sound of 100 Hz and moves in a straightforward harmonic manner. The cone moves a maximum of 2.0 millimetres when it is making a loud sound.

Simple harmonic motion is a particular type of periodic motion of a body that arises from a dynamic equilibrium between an inertial force that is proportional to the acceleration of the body away from the static equilibrium position and a restoring force on the moving object that is directly proportional to the magnitude of the object's displacement and acts towards the object's equilibrium position.

In mechanics and physics, SHM is sometimes used to refer to this motion. If friction or any other energy dissipation is not present, it leads to an oscillation that is represented by a sinusoid and that lasts indefinitely.

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When the masses of the blocks are doubled and the force applied is constant, the tension in the string will be the same.

The given parameters:

The acceleration of the blocks is calculated as follows;

\(a = \frac{F}{m + 2m} \\\\a = \frac{F}{3m}\)

The tension in the string is calculated as follows;

\(T = ma\\\\T = m(\frac{F}{3m)}\\\\T= \frac{F}{3}\)

When the masses of the blocks are doubled and the force applied is constant;

\(a = \frac{F}{2m + 4m} \\\\a = \frac{F}{6m}\)

The new tension in the string is calculated as follows;

\(T_2 = ma\\\\T_2 = (2m) (\frac{F}{6m} )\\\\T_2 = \frac{F}{3} = T\)

Thus, when the masses of the blocks are doubled and the force applied is constant, the tension in the string will be the same.

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We have that for the Question, it can be said that

From the question we are told

From,

\(tan\theta = \frac{h}{2}\)

differentiate with respect to h

\(sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min\)

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Mark's work decreases when he climbs the same flight of stairs again in the same amount of time without the cello.

The correct answer is option D.

The amount of work Mark does depends on the weight of the cello, as well as the distance he climbs and the time it takes. Work is calculated using the formula :

Work = Force × Distance.
In the given scenario, Mark is carrying a full-sized cello while climbing the stairs. The weight of the cello adds to the force he exerts. So, the total force Mark exerts is the weight of the cello plus his own weight (375 N).

When Mark climbs the stairs with the cello, he is doing work against the force of gravity.

The work done is equal to the force exerted multiplied by the distance climbed (375 N + weight of cello) × 4 m.

Now, if Mark were to climb the same flight of stairs again in the same amount of time (3 seconds), but this time without the cello, the amount of work he does would decrease. This is because without the cello, the force exerted would only be Mark's weight (375 N), which is less than the total force exerted with the cello.

Therefore, mark's work decreases.

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Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

Answer:

C. Mass

Explanation:

If something has the same weight than it has the same mass

Answer: C

Explanation: the mass is the same as weight so if they have weight they also must have mass.

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

A simplified version of Rutherford’s calculation of the size of the gold nucleus. A piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all particles go through undeflected. The density of gold is 19,300 kg/m3.

Rutherford's experiment involved firing alpha particles (helium nuclei) at a thin sheet of gold foil to study the structure of atoms. Based on the results of this experiment, Rutherford was able to deduce that atoms have a small, dense nucleus at their center.

In this problem, we will go through a simplified version of Rutherford's calculation of the size of the gold nucleus.

First, we need to calculate the total number of gold atoms in the foil. We know that the foil is 0.010 cm thick and has an area of 1 cm x 1 cm, so its volume is

V = thickness x area = 0.010 cm x (1 cm x 1 cm) = 0.010 \(cm^{3}\)

The density of gold is 19,300 kg/\(m^{3}\), which is equivalent to 19.3 g/\(cm^{3}\)Therefore, the mass of the gold foil is

m = density x volume =  19.3 g/\(cm^{3}\) x 0.010 \(cm^{3}\) = 0.193 g.

The molar mass of gold is 197 g/mol, so the number of gold atoms in the foil is

N = (0.193 g) / (197 g/mol) x (6.022 x \(10^{23}\) atoms/mol) = 1.86 x \(10^{21}\) atoms

Next, we need to determine the fraction of alpha particles that are deflected by the gold nucleus. We are told that 99.93% of all alpha particles go through undeflected, which means that only 0.07% of the alpha particles are deflected. This is a very small fraction, which suggests that the size of the gold nucleus must be very small compared to the size of the atom.

Assuming that the alpha particles are deflected only by the gold nucleus and not by the electrons, we can use the principle of conservation of momentum to estimate the size of the gold nucleus. When an alpha particle approaches the gold nucleus, it experiences a repulsive electrostatic force that causes it to change direction. The magnitude of this force is given by Coulomb's law

F = k\(q_{1}\)\(q_{2\) / \(r^{2}\)

Where k is Coulomb's constant, \(q_{1}\) and \(q_{2}\) are the charges of the alpha particle and gold nucleus, respectively, and r is the distance between them. Since the alpha particle has a positive charge and the gold nucleus has a positive charge, the force is repulsive.

If we assume that the alpha particle is initially moving directly toward the center of the gold nucleus, then at the point of closest approach, the alpha particle will have a velocity v that is perpendicular to the direction from the alpha particle to the gold nucleus. At this point, the force on the alpha particle will be perpendicular to its velocity, which means that it will change only the direction of the alpha particle's velocity, not its magnitude.

Using conservation of momentum, we can relate the angle of deflection θ to the distance of closest approach r.

m\(v^{2}\) / r = k\(q_{1}\)\(q_{2\) / \(r^{2}\)

Where m is the mass of the alpha particle. Solving for r, we get

r = k\(q_{1}\)\(q_{2\) / m\(v^{2}\)

To estimate the size of the gold nucleus, we assume that the alpha particles are deflected by a single, stationary gold nucleus at the center of the atom. In reality, the gold nucleus is not stationary, but this assumption gives us a rough estimate of its size.

Hence, the alpha particles are undeflected with a probability of 0.9993, we can assume that they do not interact with the gold nucleus and that their path is a straight line through the foil.

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Answer:

D. Gas

Explanation:

There are only three states of matter: solid, liquid and gas. This makes choice c (Mixture) incorrect because it is not part of the group.

The particles in gas are widely separated, which means there's a lot of free space between the particles. This makes it easily compressible. The particles then move past each other, which causes no regular arrangement.