A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements

Answer:

d. 0.5 m/s along -x direction

Explanation:

Wave: A wave is a disturbance, that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself.

The general equation of a traveling wave can be expressed as

y = Acos(2πft-2πx/λ).................................. Equation 1

Where A = amplitude of the wave, f = frequency of the wave, λ = wavelength of the wave, x = linear distance, t = time, π = pie.

From the question,

the equation of the moving wave is

y = 0.12cos(4x+2t) ................................... equation 2

Comparing equation 1 and 2

-2πx/λ = 4x

λ  = -2π/4

λ  = -2(3.14)/4

λ  = -1.57 m.

Also,

2πft = 2t

f = 2t/2πft

f = 1/π

f = 1/3.14

f = 0.3185 Hz.

Recall that

v = λf.......................... Equation 3

Substitute the value of f and λ  into equation 3

v = -1.57(0.3185)

v = - 0.5 m/s.

Note: v is negative because - x direction

Hence the right option is d. 0.5 m/s along -x direction

They travel at a speed of 0.68 m/s south ([tex]90^{\circ}[/tex] south of west)

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two players must be conserved before and after they collide.

Mathematically:

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]  

where, taking north as positive direction:

[tex]m_1 = 92 kg[/tex] is the mass of the first player

[tex]u_1 = -5.8 m/s[/tex] is the initial velocity of the first player (south, so negative)

[tex]m_2 = 110 kg[/tex] is the mass of the second player

[tex]u_2 = 3.6 m/s[/tex] is the initial velocity of the second player (north, so positive)

[tex]v[/tex] is the final combined velocity of the two players

Solving for v, we find the final velocity of the two players combined:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(92)(-5.8)+(110)(3.6)}{92+110}=-0.68 m/s[/tex]

where the negative sign indicates their final direction is south.

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To find the final speed and direction after the tackle, we use the conservation of momentum. The fullback and lineman's initial momenta are calculated and combined to find the total initial momentum, which equals the final momentum because momentum is conserved. The direction is either south (90 degrees) or north (270 degrees) depending upon the sign of the final velocity.

To solve this physics problem, we will use the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant. The formula for momentum (p) is the product of mass (m) and velocity (v), or p = mv.

Before the tackle, the fullback's momentum is southward and the lineman's is northward. We treat south as the positive direction. Thus, the initial momentums of the fullback and lineman are:

The total initial momentum (pinitial) is the sum of these:

pinitial = pFB + pL

After the tackle, the players stick together and move as one mass (M = 92 kg + 110 kg), with a final velocity (V), so:

pfinal = MV

Since momentum is conserved, pinitial = pfinal, and we can solve for V:

V = pinitial / M

Once we calculate the final velocity, the direction is south if V > 0, north if V < 0. To convert the direction into an angle south of west, note that if they move southward, the angle is simply 90 degrees (directly south), while any northward movement would result in a 270-degree angle.

Answer:

The answer to your question is angle = 18.46°

Explanation:

Data

d = 75 m

v₀ = 35 m/s

α = ?

Formula

                     [tex]d = \frac{vo^{2}sin2\alpha}{g}[/tex]

solve for sin2α

                  sin 2α = [tex]\frac{dg}{vo^{2}}[/tex]

Substitution

                 sin 2α = [tex]\frac{75(9.81)}{35^{2}}[/tex]

Simplify

                sin 2α = [tex]\frac{735.75}{1225}[/tex]

Divide

                sin 2α = 0.600

Get sin⁻¹

                      2α = 36.9°

Divide by 2

                        α = 18.5°                

Answer:

Explanation:

Archimedes principle states that the upward buoyant foce exrted on a body is equal to th wight o the liquid displaced.

Now, the buoyant force on the boat is given by:

[tex](m+m_b)g=V\rho g[/tex]

[tex]V[/tex] is the volum [tex]\rho[/tex] is the density [tex]m_b[/tex] is the mass of the boat and [tex]m[/tex] is the mass added to the boat.

[tex](m+m_b)g=(Sd)\rho[/tex]

[tex]S[/tex] is the surface area and [tex]d[/tex] is the depth.

[tex]m=Sd\rho - m_b...(1)[/tex]

The equation for thebest fit linis,

[tex]d=(0.374m/kg)m+0.11m[/tex]

Re-arrangethis equati for [tex]m[/tex]

[tex]m=\frac{d}{(0.374m/kg)}-\frac{0.11m}{0.374m/kg}...(2)[/tex]

From equations(1) and (2),

[tex]Sd\rho=\frac{d}{0.374m/kg}[/tex]

the density is,

[tex]\rho=\frac{1}{S(0.0374m/kg)}=\frac{1}{(25cm^2)(\frac{1m^2}{10^4cm^2})(0.374m/kg)}=1.069\times 10^3 kg/m^3[/tex]

Therefore, the density of the liquid is

[tex]\rho=1.07\times 10^3 kg/m^3[/tex]

To determine the density of the liquid, one needs to get the slope of the graph from the data given. Using this slope in the formula ρ = k/g, where 'g' is the gravity, gives the density of the liquid.

Let's first understand this concept with the help of Archimedes' Principle, which states that, the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. To determine the density of the liquid, we need the slope of the line from the graph.

Let's assume that the slope of the line is 'k' (which you will obtain from your graph). The slope of the line of best fit in the graph of m versus d will be m/d = ρVg/(Ag), where 'm' is the mass added, 'd' is the depth, 'ρ' is the density of the fluid, 'V' is the volume of the fluid displaced, 'A' is the cross-sectional area of the boat, and 'g' is the acceleration due to gravity.

We can write Volume 'V' = Ad, so the equation simplifies to k = ρg and hence the density ρ = k/g, where 'k' is the obtained slope and 'g' (assuming you are on the earth) is 9.81 m/s². Therefore, once you obtain the value of 'k', you can easily calculate the density of the liquid.

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Explanation:

The Natural Greenhouse Effect However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.

Answer:

E2/E1 =99.2

Explanation:

time after release of E1 (t) = 27.3 s

time after release of E2 (t') = 3.03 s

acceleration (a) = [tex]\frac{QE1}{M}[/tex]

where

after 27.3 s

velocity (V) = a x t = [tex]\frac{QE1}{M}[/tex] x 27.3  = [tex]\frac{27.3QE1}{M}[/tex]

distance to turning point (s) = 0.5a[tex]t^{2}[/tex] = 0.5 x [tex]\frac{QE1}{M}[/tex]x [tex]27.3^{2}[/tex] = [tex]\frac{372.65QE1}{M}[/tex]

now for its return back to its starting point

acceleration (a') = [tex]-\frac{QE2}{M}[/tex]

total distance S' = distance to turning point + distance from turning point to starting point

S' = S + vt' + 0.5 a'[tex]t'^{2}[/tex]

S' = [tex]\frac{372.65QE1}{M}[/tex] + ([tex]\frac{27.3QE1}{M}[/tex] x 3.03) + (0.5 x [tex]-\frac{QE2}{M}[/tex]x [tex]3.03^{2}[/tex])

S' is the distance at the starting point and = 0

0 = [tex]\frac{372.65QE1}{M} + \frac{82.72QE1}{M}-\frac{4.59QE2}{M}[/tex]

[tex]\frac{4.59QE2}{M}=\frac{372.65QE1}{M} + \frac{82.72QE1}{M}[/tex]

multiplying both side by M/Q we have

4.59.E2 = 372.65E1 + 82.72E1

4.59.E2 = 455.37E1

E2/E1 = 455.37 / 4.59

E2/E1 =99.2

The ratio of the magnitude of electric field E2 to E1, when a proton returns to its initial point after the direction of a uniform electric field changes, is approximately 0.0123.

A proton is initially at rest in an electric field that points due north. After 27.3 seconds, the electric field changes direction and points due south, and after 3.03 seconds the proton returns to its starting point. To find the ratio of the magnitudes of E2 to E1, we must consider the distances covered by the proton under the influence of both fields given that it starts and ends at the same position.

The distance covered under E1 can be calculated using the formula s = 0.5 × a × t^2, where 'a' is the acceleration and 't' is the time. Since the proton moves for 27.3 s under E1, the distance s1 is s1 = 0.5 × a1 × (27.3)^2. The proton then moves in the opposite direction under E2 for 3.03 s, covering the same distance in opposing direction, thus s2 = 0.5 × a2 × (3.03)^2. Since s1 = s2, we can equate them: 0.5 × a1 × (27.3)^2 = 0.5 × a2 × (3.03)^2.

Furthermore, the acceleration of the proton is directly proportional to the electric field (since a = F/m and F = qE), giving us a1 ∝ E1 and a2 ∝ E2. By simplifying the above equation, we find that a1/a2 = (3.03/27.3)^2, and therefore the ratio of the magnitudes of electric fields is also E2/E1 = (3.03/27.3)^2. Calculating this, we get the ratio E2/E1 approximately equal to 0.0123.

Answer:

do not agree

Explanation:

For student A

[tex]B=4.2\pm 0.8\ T\\\Rightarrow B=4.2+0.8\ or\ B=4.2-0.8\\\Rightarrow B=5\ T\ or\ 3.4\ T[/tex]

The magnetic field measured by student A = 5 T or 3.4 T

For student B

[tex]B=5.6\pm 0.5\ T\\\Rightarrow B=5.6+0.5\ or\ B=5.6-0.5\\\Rightarrow B=6.1\ T\ or\ 5.1\ T[/tex]

The magnetic field measured by student B = 6.1 T or 5.1 T

The magnetic fields measured are not equal hence they do not agree.

Answer:

The correct answer is: justice

Explanation:

The Belmont Report refers to a report that was published 25 year ago, focusing on the ethical treatment and protection of participants in medical and behavioral research. This report centers around 3 principles:

1. Beneficence- striving to maximize benefits for participants of the research study and minimizing any harms/ risks that might occur.

2. Justice- The fair selection of potential participants for a study. This ensures equitable and fair distribution of risks/ benefits to all potential participants of a research study. Subjects of a study must not be chosen merely out of convenience or easy access. The inclusion/ exclusion criteria should be chosen according to the nature of the study and steps/ treatments that it will involve.

3. Respect for persons- Each participant of a research study should be  able to provide informed consent prior to their participation, protected from controllable harm and treated with respect.

Therefore, moral requirement that there be fair outcomes in the selection of research subjects, expresses the principle of justice.

Answer: work Melvin did=9000J

Explanation:

Given to complete the question: If the sled moved 33.9m,how much work did Melvin do? Answer in unit of J and round to the nearest thousandth.

W = F ×S

W = 317 × cos 33°×33.9

W=9012.6055J

W=9000J to the nearest thousandth

Answer:

To calculate the period of satellite orbiting around a planet, we use Kepler's third law;

Square of T = [(4π)/(G*m)] * R^3.

Therefore,

T = sqrt{[(4π)/(G*m)]*R^3}.

T is the period, m is mass orbiting satellite, G is gravitational constant, R is the radius of of the planet, r is the radius of the orbiting satellite.

For Satellite 1, r is one-half of the planet, that is r = (3/2) * R

For satellite 2, r = R

Explanation:

The speed is 2.5 m/s

Explanation:

The total (mechanical) energy of the mass-spring system at any point during the motion is the sum of the kinetic energy (KE) and the potential energy (PE):

[tex]E=KE+PE=3.24 J[/tex]

and it is constant.

The kinetic energy can be written as

[tex]KE=\frac{1}{2}mv^2[/tex]

where

m = 0.500 kg is the mass

v is the speed

While the potential energy is

[tex]PE=\frac{1}{2}kx^2[/tex]

where

k = 330 N/m is the spring constant

x is the elongation

So the first equation becomes

[tex]E=\frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]

Therefore, if we substitute

x = 0.100 m

We can find the speed when the elongation is x = 0.100 m:

[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(3.24)-(330)(0.100)^2}{0.500}}=2.5 m/s[/tex]

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Correct Answer:

2.5 m/s

Answer:

-M6

-G2

-A5

-O9

Explanation:

Answer:

A. Weight

Explanation:

Mass And Weight

Mass is a fundamental property of objects and is defined as a measure of the amount of matter in the object. The symbol for mass is m and its SI unit is the kilogram (kg). At normal speeds, i.e. much smaller than the speed of light, the mass is considered as a constant property.

The weight W is defined as the force of gravity on the object, that is, the gravitational attraction between the object and the Earth. Its SI unit is the Newton (Nw or N).

Explanation:

Nucleus of every atom contains both protons and neutrons. Protons are positively charged species whereas neutrons are neutral species, that is, neutrons do not contain any charge.

And, when an electron cloud contains a negatively charged ion then it is able to participate in a chemical reaction as it needs to gain stability.

Therefore, we can conclude that the nucleus contains positively charged particles called protons and neutral particles called neutrons, which are bound together by the strong nuclear force. The electron cloud contains negatively charged particles, which participate in chemical reactions.

Answer:

Protons, neutrons,and chemicals.

Explanation:

on edge

Answer:

a. fruit juice

Answer:

a. fruit juice

Explanation:

have a great day everyone! also merry christmas!

There is one mistake in the question as unit of acceleration is not written correctly.The correct question is here

A car traveling at 7 m/s accelerates uniformly at 2.5 m/s² to reach a speed of 12 m/s. How long does it take for this acceleration to occur?

Answer:

time taken =2 seconds

Explanation:

Given Data

Initial Speed Vi= 7 m/s

Final Speed Vf=12 m/s

Acceleration a= 2.5 m/s²

To find

Time taken for this acceleration

Solution

As we know that

Final velocity=Initial velocity + acceleration×time

[tex]V_{f}=V_{i}+at\\  t=\frac{V_{f}-V_{t}}{a}\\ t=\frac{12m/s-7m/s}{2.5m/s^{2} }\\ t=2 seconds[/tex]

So car takes 2 seconds for this acceleration to occur

Answer:

Following are the correct options.

1. In a uniform electric field, the field lines are straight, parallel and uniformly spaced. ( As the field strength does not change in the uniform electric field that is why the field lines are parallel, equally spaced etc).

2. Electric field lines near positive point charges radiate outward. (Due to repulsive forces field lines radiate in outward direction)

3.The electric force acting on a point charge is proportional to the magnitude of the point charge. ( As columb law states that the electrostatic forces between the point charges is proportional to the product of their magnitude and inversely proportional to square of distance between them).

Explanation:

In a uniform electric field, electric field lines are straight, parallel, and uniformly spaced while near positive point charge it emits outwards. Also, the electric force on a point charge is proportional to its magnitude. But, lines near negative point charges radiate inward, not circle clockwise. The electric field created by a point charge isn't constant across space, but diminishes with distance.

Of the five statements you provided:

Therefore, statements 1, 2 and 3 are true, while 4 and 5 are false.

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Answer:

a1 = 3.68m/s²

Explanation:

Given values:

Mass of the block placed on the table, m1 = 12.25 kg

Mass of the block hanging vertically, m2 = 7.5 kg

Acceleration due to gravity, g = 9.8 m/s2

Tension in the string is T

Let the acceleration of mass 1 and mass 2 be a1 and a2

a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,

a = m2*g/(m1 + m2)

a = 7.5 x 9.8 / (12.5 + 7.5)

a = 3.68 m/s²

a1 = a2 = 3.68m/s²

a1 is directed to the right and a2 is directed downwards

Below is a diamonds to show the geometrical arrangements of both masses

Answer:

18 J

Explanation:

Work done: This can be defined as the product of force and distance acting on a body. The S. I unit of work is Joules (J)

From the question, the work that must be done in bringing the block to rest is equal to the kinetic energy of the block.

Ek = 1/2mv².................... Equation 1

Where Ek = kinetic eenrgy m = mass of the block, v = velocity of the block.

Given: m = 4 kg, v = 3 m/s.

Substituting into equation 1

Ek = 1/2(4)(3²)

Ek = 2(9)

Ek = 18 J.

Thus the work that must me done on the block to bring it to rest = 18 J.

Answer:

option (b)

Explanation:

The density of cool air is more so always falls downwards.

Compared to the cool air, the density of air is less and it tends to rise.

Thus, option (b) is correct.

Answer:

218.25 W.

Explanation:

Power: This is defined as the rate at which work is done. The S.I unit is Watt (W). Mathematically, it can be expressed as

P = W/t ............................... Equation 1.

Where P = useful power output, W = useful work, t = time taken to do the work.

Given: W = 6.6×10⁶ J, t = 8.4 h = 8.4×60×60 = 30240 s.

Substitute into equation 1

P = 6.6×10⁶/30240

P = 218.25 W.

Hence the useful power output = 218.25 W.

Answer:

work done on compressing spring will be 135 j

Explanation:

We have given spring constant [tex]K=3kN/cm=3\times \frac{1000N}{10^{-2}m}=3\times 10^5N/m[/tex] ( As 1 kN = 1000N and 1 m = 100 cm )

Spring is compressed by 3 cm

As 1 m = 100 cm

So [tex]3cm=3\times 10^{-2}m[/tex]

Work done on compressing spring is given by [tex]W=\frac{1}{2}kx^2[/tex]

So [tex]W=\frac{1}{2}\times 3\times 10^{5}\times (3\times 10^{-2})^2=13.5\times 10=135J[/tex]

So work done on compressing spring will be 135 j

The work a spring can produce after being compressed from its unloaded length can be calculated using the formula for potential energy stored in a spring. In this case, given the spring constant of 4 kN/cm and a compression of 3 cm, the spring can produce 18 kJ of work.

The work done by a spring, in this case, can be calculated using the formula for the potential energy stored in a compressed or stretched spring, which is given by PE = 1/2 * k * x². Here, k is the spring constant and x is the displacement or compression of the spring from its original length.

Given that the spring constant k = 4 kN/cm = 4000 N/cm and displacement x = 3 cm, we can substitute these values into the formula: PE = 0.5 * 4000 * (3)² = 18,000 N.cm = 18 kJ. The spring can therefore produce 18 kJ of work when it has been compressed 3 cm from its unloaded length.

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The net force on the shopping cart is 12 N to the right.

This is a question related to Newton's Laws. It can be solved by using a Free body diagram, which shows all the forces acting on the object. There are 4 forces acting on the object:

Forces (1) and (2) cancel each other because they have the same magnitude, and the overall force is given by the addition of forces (3) and (4)

[tex]F=Fp+Ff=20N+ (-8N)=12N[/tex]

The positive value indicates that the shopping cart is moving to the right.

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Answer: a. 198.6J b. - 198.6J

Explanation: Parameters given:

m = 15kg

g = 9.8m/s²

∅ = 12°

a. Work done by the force Fp on the cart if the ramp is 6.5m long.

Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N

Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J

b. The work done by the force mg on the cart.

Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance

= 15kg x -9.8m/s² x Sin12° x 6.5m

= - 198.6J

Answer:

(a)The work done by the force Fp is 198.6J

(b) The work done by the weight mg is -198.6J

The net force acting on the cart is zero. This is because the cart is moving with a constant velocity and by newton's first law the the net force on the cart is equal to zero.

Fp was calculated to be equal to 34.94 N.

Explanation:

In order to solve this kind of problems successfully, the best approach is to resolve all forces acting on the cart parallel and perpendicular to the ramp surface. So that the x-axis is parallel to the ramp surface and the y axis is perpendicular to the ramp surface.

Fpx = FpCos 29°

Fpy = FpSin 29°

Wx = mg sin12°

Wy = mg cos 12°

Summation Fx = 0

And Summation Fy = 0

The full solution can be found below in the attachment.

Thank you for reading and I hope this is helpful to you.

The final resistance is [tex]679\Omega[/tex]

Explanation:

The relationship between the resistance of a metal and the temperature is

[tex]R(T) = R_0(1+\alpha (T-T_0))[/tex]

where

[tex]R_0[/tex] is the resistance at a temperature of [tex]T_0[/tex]

R is the resistance at temperature T

[tex]\alpha[/tex] is the temperature coefficient of resistance

In this problem, we have:

[tex]R_0 = 525 \Omega[/tex]

[tex]T_0 = 20^{\circ}C[/tex]

[tex]\alpha = 0.005866 \Omega/^{\circ}C[/tex]

Therefore, the resistance when [tex]T=70^{\circ}C[/tex] is

[tex]R=(525)(1+0.005866(70-20))=679\Omega[/tex]

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Answer:

Therefore the new resistance would be 679 Ω

Explanation:

Resistance is the opposition to the flow of electric current. The resistance of an object given its coefficient of resistance can be obtained with the expression bellow;

R =R_ref [1+ α (T - T_ref)]

Where R is the new resistance

R_ref is the base resistance = 525 Ω

α is the coefficient of resistance at  20°C = 5

T is the new temperature = 70°C

T_ref is the base temperature =  20°C

Substituting the values into the equation we have;

R = 525 x  [ 1 + 0.005866 (70-20)]

R =  525 x  [ 1 + 0.005866 (50)]

R = 525 x 1.2933

R = 678.98

R≈ 679 Ω

Therefore the new resistance is 679 Ω

Answer: The boiling point of  solution is [tex]112.16^0C[/tex]

Explanation:

Elevation in boiling point:

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b[/tex] = boiling point of solution = ?

[tex]T^o_b[/tex] = boiling point of toluene = [tex]110.63^oC[/tex]

[tex]k_b[/tex] = boiling point constant of toluene =[tex]3.40^oC/m[/tex]

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

[tex]w_2[/tex] = mass of solute [tex](I_2)[/tex] = 7.94 g

[tex]w_1[/tex] = mass of solvent (toluene) = 69.2 g

[tex]M_2[/tex] = molar mass of solute [tex](I_2)[/tex]= 254g/mol

Now put all the given values in the above formula, we get:

[tex](T_b-110.63)^oC=1\times (3.40^oC/m)\times \frac{(7.94g)\times 1000}{254\times (69.2g)}[/tex]

[tex]T_b=112.16^0C[/tex]

Therefore, the boiling point (in °C) of a solution is 112.16

Answer:C) BLADDER

D) EPIDERMIS (E) BLOOD VESSEL LINING

Explanation: Autoimmune diseases are diseases caused by the body fighting against its self. In autoimmune diseases the body's defense system tends to attack the body cells, tissues, Organs and the entire system mistaking it for a foreign body such as Viruses or Bacteria,Fungi etc.

Epithelial tissues are tissues which form the surface covering of most of the hollow Organs,all body Organs and are present inside the glands. The body parts that will be involved are the BLADDER, EPIDERMIS AND BLOOD VESSEL LINING.

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s

Explanation:

given information:

moment inertia of arm and leg when in, I₁ = 0.9 kgm²

moment inertia of arm and leg when extended, I₂ = 2.9 kgm²

angular speed when in, ω₁ = 4.5 rev/s

so, her angular speed (in rev/s) when her arms and one leg open outward is

L₁ = L₂

I₁ω₁ = I₂ω₂

ω₂ = I₁ω₁/I₂

     = 0.9 x 4.5/2,9

     = 1.4 rev/s

Answer:

a short-to-ground

Explanation:

Many failures that occur in vehicles can be diagnosed as electrical problems. The reason for these usually relate mainly to the electricity generated in the battery or alternator, in the ignition system as well as those produced by blown wires or fuses.

Cables or fuses: a blown fuse or a damaged electrical wire or one that is making ground or earth effect can be the cause of an electrical problem. Keep in mind that everyone is connected to each other through the wires of the electrical system. Fuses protect car components from power surges. If you have a problem with one, check it to see if you will have to replace it.

A failure where an insulated wire rubs through and touches the steel body of a vehicle is known as a short circuit.

If an insulated (power side) wire rubs through a part of the insulation and the wire conductor touches the steel body of a vehicle, this type of failure is known as a short circuit. This happens when a wire carrying current touches the vehicle's body, which is grounded. As the body acts like a path of least resistance, the current prefers flowing through it rather than its intended path, creating a short circuit. This can result in various problems like damage to components, blowing a fuse, or even triggering a fire.

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Answer:

The net force acting on the otter along the incline is 13.96 N.

Explanation:

It is given that,

Mass of the otter, m = 2 kg

Distance covered by otter, d = 85 cm = 0.85 m

It takes 0.5 seconds.

We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Here, u = 0 (at rest)

[tex]d=\dfrac{1}{2}at^2[/tex]

[tex]a=\dfrac{2d}{t^2}[/tex]

[tex]a=\dfrac{2\times 0.85}{0.5^2}[/tex]

[tex]a=6.8\ m/s^2[/tex]

The net force acting on the otter along the incline is given by :

F = ma

[tex]F=2\ kg\times 6.8\ m/s^2[/tex]

F = 13.6 N

So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.

Answer:

[tex]F=13.6\rm N[/tex] Force acts on the otter along the incline

Explanation:

Given information:

Incline length [tex]s=\rm 85cm=0.85m[/tex]

Time [tex]t=0.5\rm sec[/tex]

Mass of otter [tex]m=2\rm kg[/tex],

Initial velocity [tex]u=0[/tex] as otter is in rest

Muddy incline so we can assume surface as friction less

By use equation of motion,

[tex]s=ut+\frac{1}{2} a t^2[/tex]

[tex]s=ut+\frac{1}{2} a t^2=0\times t+\frac{1}{2}\times a\times t^2[/tex]

[tex]0.85=0+\frac{1}{2} a (0.5)^2\\\\a=\frac{2\times0.85}{0.5^2}=6.8\rm m/s^2[/tex]

The net force act on the otter is

[tex]F=ma[/tex]

[tex]F=2\rm kg\times6.8\rm m/s^2[/tex]

[tex]F=13.6\rm N[/tex]

Hence [tex]F=13.6\rm N[/tex] force acts on the otter along the incline

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