If you use a horizontal force of 30n to slide a 12kg wooden crate across a floor at a constant velocity. what is the coefficient of kinetic friction?

Answer:

F_{resultant} = 1 [m/s]

Explanation:

These types of problems can be solved by means of relative velocities, where vectors (forces or velocities) are necessarily handled, these velocities depending on the velocity are added or subtracted.

Forces to the right are taken as positive and negative to the left.

\(F_{velocity}-F_{air}=F_{resultant}\\2-1 = F_{resultant}\\\\F_{resultant} = 1 [m/s]\)

The magnitude of the acceleration due to gravity, denoted with a lower case g, is 9.8 m/s2. g = 9.8 m/s2. This means that every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

The maximum height of the ball is approximately 6.096 m and the work done by gravity is -4.95 J.

Given information,

m = 0.071 kg

V₀ = 11 m/s

(a) Finding the maximum height:

The initial kinetic energy of the ball is converted into potential energy at its maximum height.

Initial kinetic energy = Potential energy at maximum height

(1/2)mv₀² = mgh

(1/2)(0.071 kg)(11 m/s)² = (0.071 kg)(9.8 m/s)h

h = (0.5)(11²)/(9.8) ≈ 6.096 m

Part (b) Calculating the work done by gravity:

The work done by gravity during the ball's flight to its maximum height is equal to the change in potential energy. Since the gravitational potential energy is defined as zero at the initial height, the work done by gravity is equal to the negative of the potential energy at the maximum height.

Wg = -mgh

Wg = -0.071 × 9.8 × 6.096

Wg ≈ -4.95 J

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Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

mass = 15 kg

acceleration = 10 m/s²

Force = 15 × 10 = 150

We have the final answer as

Hope this helps you

This is due to the fact that occluded fronts combine two cold air masses, which causes one of the cold air masses to go downward.

When a warm air mass is sandwiched between two cold air masses, an occluded front occurs. In an occlusion, the warm front passes over the cold front, which dives beneath it.

In a front is obscured, the warm front is fully supplanted by the cold front, in which the warm air masses have completely disappeared. Furthermore, there are frequent shifts in the various weather producing circumstances because of the cold front's relatively low temperature.

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Answer:

junk food, alcohol, soda, microwave food, candy, some sugary drinks

Answer:

The intensity of the net electric field will:

\(E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C\)

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

\(|E_{1}|=k\frac{q_{1}}{d^{2}}\)

Where:

So, we will have:

\(|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}\)

\(|E_{1}|=1.8*10^{7}\: N/C\)

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

\(|E_{2}|=k\frac{q_{2}}{d^{2}}\)

\(|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}\)

\(|E_{2}|=7.2*10^{6}\: N/C\)

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

\(E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C\)

I hope it helps you!

The net electric field midway between these two charges is 2.52 x 10⁷ N/C.

The electric field strength between two point charges is the force per unit charge.

Electric field strength due to +20.0 μC charge

\(E _1 = \frac{kq_1}{r^2} \\\\E_ 1 = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{(0.1)^2} \\\\E _1 = 18 \times 10^6 \ N/C\)

Electric field strength due to -8μC charge

\(E _2 = \frac{kq_2}{r^2} \\\\E_ 2 = \frac{9 \times 10^9 \times 8 \times 10^{-6}}{(0.1)^2} \\\\E _2 = 7.2 \times 10^6 \ N/C\)

The net electric field midway between these two charges is calculated as follows;

\(E_{net} = 18 \times 10^6 \ + \ 7.2 \times 10^6\\\\E_{net} = 25.2 \times 10^6 \ N/C\\\\E_{net} = 2.52 \times 10^7 \ N/C\)

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Fire detection pull stations are referred to as "manual fire alarm boxes" in NFPA 72.The code stipulates that a specific floor's closest pull station shall be no and over 200 feet away.

NFPA 1, Fire Code, makes numerous references to NFPA 72.The principal section of the Code governing detection, alarm, & communication systems is 13.7.

Pathway resiliency In accordance to NFPA 13, Level 1 shall include any connecting wires, cables, or even other physical paths that are properly covered by an automated sprinkler system, as well as any such pathways that are covered by metal wire harnesses or metal-armored cables.

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Radii is a vital feature of the elements, and it can be useful in determining the characteristics of elements in various chemical and physical processes. The radii of atoms and ions of the same element differ due to their various charge and mass characteristics.

Atomic and ionic radii increase as you move down a group on the periodic table, and decrease as you move across a period from left to right due to increased nuclear charge, making the electrons closer to the nucleus. The size of an atom and ion also changes due to the number of electrons charge, and electronic configuration.In order of increasing radius, the arrangement of \(Ne, F-, O2-, Br-, Rb\) is given as follows:

\(Ne < F- < O2- < Br- < Rb+\)

Rb+ has the smallest radius due to its large nuclear charge and fewer electrons in the valence shell.

As a result, they are larger than Rb+. O2- has more electrons than Ne and is the largest among the given ions and atoms. It is important to note that in certain conditions, the trends in radii may not be valid because of hybridization and other factors. Nonetheless, this arrangement is valid for the given ions and atoms.

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Answer:

______________________________

use tall safety and position hoist arms

Answer: \(CBr_4\) : carbon tetrabromide

Explanation:

\(CBr_4\) is a covalent compound because in this compound the sharing of electrons takes place between carbon and bromine. Both the elements are non-metals. Hence, it will form covalent bond.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

Hence, the correct name for \(CBr_4\) is carbon tetrabromide.

Answer:

Carbon tetrabromide ~ CBr4

Chlorine monofluoride~ ClF

Explanation:

EDGU 2021

ANSWER:

(a) 0.100 m^3

STEP-BY-STEP EXPLANATION:

We have that the force is equal to mass times gravity, therefore, we can calculate the mass, because the difference in force would be the volume of the stone.

Therefore

Now, we know that the density is equal to the quotient between the mass and the volume, therefore we can calculate the volume like this:

Therefore the volume of the piece is 0.100 cubic meters.

(a)The pressure at B is 0.1248 atm.

(b)The temperature at C is 727.1 K.

(c)The total work done by the gas in one cycle is -1979J

General calculation:

We can use the First Law of Thermodynamics to analyze the heat engine cycle:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. For a complete cycle, ΔU = 0, so:

Q = W

We can also use the ideal gas law to relate the pressure, volume, and temperature of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.

(a)How to find the pressure at B segment?

To find the pressure at B, we can use the fact that the segment AB is an isothermal expansion. This means that the temperature remains constant, so:

PV = nRT

PB = (nRT)/(2V) = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/(2V) = (0.0821 L·atm/mol)(600 K)/V

Since the pressure at A is 5.00 atm, we can use the fact that the temperature is constant to find the volume at A:

PV = nRT

VA = (nRT)/P = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/5.00 atm = 197.76 L

Since the volume at B is twice the volume at A, we have:

VB = 2VA = 395.52 L

Substituting into the expression for PB, we get:

PB = (0.0821 L·atm/mol)(600 K)/395.52 L = 0.1248 atm

Therefore, the pressure at B is 0.1248 atm.

To find the temperature at C, we can use the fact that the segment BC is an adiabatic expansion. This means that no heat is added or removed from the system, so:

\(PV^\gamma\)= constant

where γ is the ratio of specific heats (for a monoatomic gas, γ = 5/3). We can use the fact that the volume at C is equal to the volume at A to find the pressure at C:

\(PAV^\gamma = PCV^\gamma\)

PC =  \(PA(V/A)^\gamma\) = 5.00 atm\((1/2)^(^5^/^3^)\) = 1.556 atm

Since the segment BC is adiabatic, the temperature changes but no heat is added or removed from the system. Using the ideal gas law, we can relate the pressure, volume, and temperature:

PV = nRT

TC = (PCVC)/(nR) = (1.556 atm)(197.76 L)/(2.00 mol)(0.0821 L·atm/mol·K) = 727.1 K

Therefore, the temperature at C is 727.1 K.

The total work done by the gas in one cycle is the sum of the work done in each segment of the cycle:

W = WAB + WBC + WCD + WDA

For segment AB, the work done is:

WAB = -QAB = -∫PdV = -nRT∫(1/V)dV = -nRT ln(VB/VA) = -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2) = -602 J

For segment BC, the work done is:

WBC = -QBC = -∫PdV = -nγRT∫(1/V)dV = -nγRT

We know that VB = 2VA and VC = 2VD, so we can express the ratio VB/VC in terms of VA/VD:

VB/VC = (2VA)/(2VD) = VA/VD

Substituting into the expression for WBC, we get:

WBC = -nγRT ln(VA/VD)

For segment CD, the work done is:

WCD = -QCD + PCDΔV = -nCpΔT + PCDΔV

where Cp is the specific heat at constant pressure, ΔT is the change in temperature, and ΔV is the change in volume. We know that the segment CD is isobaric, so ΔV = VB - VA = (2VA) - VA = VA. We can also use the ideal gas law to relate the pressure, volume, and temperature:

Substituting into the expression for WCD, we get:

WCD = -nCpΔT + (nRT/VD)VA = -nCp(TC - TD) + (nRT/VD)VA

For segment DA, the work done is:

WDA = -QDA + ΔU = -nCvΔT

where Cv is the specific heat at constant volume. We know that the segment DA is isovolumetric, so ΔV = 0. Using the First Law of Thermodynamics, we know that ΔU = 0 for a complete cycle, so:

QDA = -WDA = nCvΔT

Substituting into the expression for WDA, we get:

WDA = -nCvΔT

Adding up the work done in each segment, we get:

W = WAB + WBC + WCD + WDA

= -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(Cp)(TC - TD) + (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(Cv)(TC - TA)

We know that Cp and Cv for a monoatomic gas are related by Cp = Cv + R, so we can express Cp in terms of Cv:

Cp = Cv + R = (3/2)R + R = (5/2)R

Substituting and simplifying, we get:

W = (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(727.1 K)+ (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(600 K)

W = -966.2 J - 4957 J - 7476 J + 5154 J

    = -1979 J

Therefore, the total work done by the gas in one cycle is -1979 J

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Answer:

392 N

Explanation:

Mass of the ball=40kg

g=9.8m/s^2

Weight(w)=40×9.8

=392N

Because of Kinetic Theory Postulate, the gas molecule's collision will affect.

Gases can be compressed because most of the volume of gas is an empty space according to Kinetic Molecular theory. If we compress a gas without changing its temperature, the average K.E of the gas particles remains the same. There is no change in the speed of the gas molecules with which the particles move in a container, but the container is smaller. So particles travel from one side of the container to the other in a very small period of time. So it means that they strike the walls mostly. Any increase in the frequency of collisions with the walls must increase the pressure of the gas. from Boyle's law, the pressure of gas becomes more significant as the volume of the gas becomes smaller have opposite relation.

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Answer:

It is cold at saturn and far away

Explanation:

The Earth moves at a uniform speed of approximately 29.8 kilometers per second (18.5 miles per second) around the Sun in a circular orbit with a radius of 1.50×1011 meters.

According to Kepler's laws of planetary motion, planets move in elliptical orbits around the Sun, but the Earth's orbit is nearly circular. The Earth's average orbital speed is approximately constant due to the conservation of angular momentum. By dividing the circumference of the Earth's orbit (2πr) by the time it takes to complete one orbit (approximately 365.25 days or 31,557,600 seconds), we can calculate the average speed. Thus, the Earth moves at an average speed of about 29.8 kilometers per second (or 18.5 miles per second) in its orbit around the Sun, covering a distance of approximately 940 million kilometers (584 million miles) each year.

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Explanation:

Given mass of the object is 1200kg and it's placed at a height of 45m above the ground. As we know that potential energy is ,

\(\longrightarrow PE = mgh \)

where,

Substituting the respective values ,

\(\longrightarrow P.E. = 1200kg * 10m/s^2* 45m\)

Multiply ,

\(\longrightarrow P.E. = 540000J \)

Hence the potential energy is 540000J .

I hope this helps.

It takes 6.4 seconds for the car to accelerate steadily from rest at a rate of 2.5 m/s² up to a speed of 16 m/s.

The car accelerates steadily from rest at a rate of 2.5 m/s² up to a speed of 16 m/s.To calculate the time it takes for the car to reach a speed of 16 m/s, we can use the following equation: v=u+at. where: v is the final velocity (16 m/s in this case), u is the initial velocity (0 m/s since the car starts from rest), a is the acceleration (2.5 m/s² in this case), t is the time.  Substituting the given values: Therefore, it takes 6.4 seconds for the car to accelerate steadily from rest at a rate of 2.5 m/s² up to a speed of 16 m/s.

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Answer:

The rocket hit the top half of the building.

Explanation:

Here is the formula for displacement

\(x(t)=\frac{at^{2} }{2}vt\)

Because the projectile is being launched at zero degrees, all of its velocity is in the horizontal direction and so there is no initial velocity for the fall, therefore

\(h=\frac{at^{2} }{2}\)

\(2h=at^{2}\)

\(\frac{2h}{a}=t^{2}\)

\(\sqrt{|\frac{2h}{a}| } =t\)

Now that we have the time taken for the object to fall, we can use it to find d by applying the displacement formula again. In this case we will remove the initial term because the acceleration is zero in the horizontal direction. Therefore

\(d=vt\)

\(d=v*\sqrt{|\frac{2h}{a}| }\)

\(d=22*\sqrt{|\frac{2*30}{9.81}| }\)

\(d=54.41\)

If there wasn't a building in the way the displacement in the x direction is 54.41.

\(y=(tan0*40)-[\frac{-9.81}{2(22*cos0)^{2}}]40^{2}\)

\(y=16.21\)

Like you said the building is 30 meters tall. You were looking to find out is it was going to hit the top or bottom half.

\(15 < 16.21\)

So it hit the top half of the other building.

Answer:

D: The Moon's orbit is tilted compared

to Earth's orbit, so Earth is not in the

Moon's shadow most months.

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

In order to rank the speeds of the three balls in the figure, we need to consider the laws of physics that govern their motion. Since all three balls have the same mass and are fired from the same height above the ground, their potential energy is the same. This means that the only factor that determines their speeds is the conversion of potential energy to kinetic energy as they fall towards the ground.

According to the law of conservation of energy, the total amount of energy in the system must remain constant. Therefore, the sum of the potential and kinetic energies of each ball must be equal at all times. This can be expressed mathematically as:

Potential energy + Kinetic energy = Constant

Since the potential energy is the same for all three balls, the only way for their kinetic energies to be different is if they experience different amounts of air resistance or drag during their fall. However, since we are assuming that all three balls are fired with equal speeds, we can assume that they experience the same amount of air resistance and drag.

Therefore, we can conclude that the speeds of the three balls as they hit the ground are equal, and they should be ranked as equivalent. This means that va, vb, and vc should be overlapped in the ranking order

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The instantaneous speed of the point of the disk that makes contact with the surface is zero m/s (Option A).

To determine the instantaneous speed of the point of the disk that makes contact with the surface, we must consider the following terms:

The answer to the question is A) zero m/s. The reason for this is that the point of contact between the disk and the surface is stationary for an instant, as it constantly changes due to the rotation of the disk. At that specific moment, the instantaneous speed of the point of contact is zero.

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the height of the image of the insect will be the same as the height of the insect, but with a negative sign.

To determine the height of the image of the insect formed by a lens with a focal length of f = -125 mm, we need to apply the sign conventions used in lens imaging.

In the sign convention:

Distances measured from the lens towards the object are positive.

Distances measured from the lens towards the image are negative.

Focal length (f) is positive for a converging lens (convex) and negative for a diverging lens (concave).

Since the given focal length is f = -125 mm (negative value), we know that we are dealing with a diverging lens.

When an object is placed in front of a diverging lens, the resulting image is virtual, upright, and located on the same side as the object. The image formed by a diverging lens is always reduced in size compared to the object.

Since the image is virtual and upright, the height of the image will have the same magnitude as the height of the object, but it will be negative to indicate that it is located on the same side as the object.

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Answer:

yes

Explanation:

Because they have logic duh!!! ;D

that is true. most ppl in general cant resist flattery. some can resist tho,  bc they are smart.

Answer:

Away from all ppl

Explanation:

Answer: Energy is released to the environment.

Explanation: