The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.

Show work

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

 Work done:

             Work done  = Weight x height  = 700 x 828

        Work done  = 579600J

Answer:

ummm hehe this is my time to shine

Explanation:

  MERICIA!!!!!!!!!!!!!!!!!!!!!!!

Answer:

Explanation:

It takes 3.1 s for the boat to travel from its highest point to its lowest, so the period of oscillation

T = 2 x 3.1 = 6.2 s

frequency of wave n = 1 / T = .1613 per sec

Amplitude of oscillation = .55/2 = .275 mm

The fisherman sees that the wave crests are spaced 4.2 mm apart. so wavelength of wave λ = 4.2 mm .

A ) velocity of wave v = n λ

.1613 x 4.2 = .677 mm /s

B ) Amplitude of wave = .275 mm

C ) The vertical distance determines only the amplitude which does not affect the velocity , so velocity will remain unchanged .

D ) Amplitude of wave depends only on the vertical displacement .

The amplitude will become .5 / 2 = .25 mm .

Distance and time I think

The question is incomplete, the complete question is;

A student is measuring the mass of 20 paper clips using an electric balance that measures to the thousandths of a gram. The balance displays the following values for the mass of the paper clip: 20.120 g.

Which of the following choices would be acceptable ways to record this mass on a lab report? Select all that apply.

a) 20g

b) 20.1 g

c) 20.12 g

d) 20.120g

e) 20.1200g

Answer:

d) 20.120g

Explanation:

We have been told in the question that the electronic balance measures the mass of the paper clips to thousandths of a gram.

This implies that the value of mass is measured to the third decimal place.

If we look at the options, 20.120 g is the measurement of the mass to thousandths of a gram hence that is the correct answer to the question.

Answer:

A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them

An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances,

Explanation:

For this exercise, the feedback and control mechanisms must be directly related to the kinematic relationships,

It is assumed that the vehicle speed (taken from the speedometer) and the braking capacity (given by the brake manufacturer) are known in the form of negative acceleration,

A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them, for which we know the acceleration that exists. This would be a correct mechanism, in general we can adjust to an error between the sedated distance and the real one, so when they have very different give a maximum acceleration and in decreasing it as the differences between the distances decrease.

An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances, so we would know the amount of acceleration necessary to reach the optimal distance between the two vehicles.

Answer:

76.1N

Explanation:

Given parameters:

Mass of the ball  = 7.77kg

Unknow:

Weight of balloon  = ?

Solution:

Weight is the vertical force applied on a body.

  Weight = mass x acceleration due gravity

So;

  Weight  = mass x acceleration due to gravity

So;

 Weight  = 7.77 x 9.8  = 76.1N

Answer:

[tex]Q = 4.9216 * 10^{-7}C[/tex]

Explanation:

Given

[tex]E = 12596.37 N/C[/tex]

[tex]r = 0.593m[/tex]

Required

Determine the magnitude point charge (Q)

This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula

[tex]E = \frac{kQ}{r^2}[/tex]

Where

[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]

Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]

[tex]E * r^2 = kQ[/tex]

[tex]Q = \frac{E * r^2}{k}[/tex]

Substitute values for E, r and k

[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]

[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]

[tex]Q = \frac{492.16}{10^9}[/tex]

[tex]Q = 492.16 * 10^{-9}[/tex]

Express in standard form

[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]

[tex]Q = 4.9216 * 10^{2-9}[/tex]

[tex]Q = 4.9216 * 10^{-7}C[/tex]

Answer:

force = mass * acceleration

therefore the SI unit is kg*m/s2 or newton's

it's a vector quantity because it has both direction(acceleration) and size (mass)

Answer:

6N

Explanation:

Given parameters:

Mass of object  = 6kg

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time  = 30s

Unknown:

Net force acting on the object  = ?

Solution:

From Newton's second law of motion:

   Force  = mass x acceleration

Acceleration is the rate of change of velocity with time

  Acceleration  = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

  Force  = mass x  [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

So;

 Force  = 6 x [tex]\frac{25 - 5}{30}[/tex]    = 6N

Answer:

change in kinetic energy of the trolley is 53.91 J.

Explanation:

mass of the trolley, m = 5.0 kg

initial velocity of the trolley, u = 1.0 m/s

external force on the trolley, F = 25 N

time of force action, t = 0.75 s

The final velocity of the trolley at the end of 0.75 s is calculated as follows;

[tex]F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force[/tex]

The change in kinetic energy of the trolley is calculated as;

Δ K.E = ¹/₂m(v² - u²)

Δ K.E = ¹/₂ x 5(4.75² - 1²)

Δ K.E = 53.91 J.

Therefore, change in kinetic energy of the trolley is 53.91 J.

Answer:

69.62 minutes

Explanation:

From the information we have here,

1 byte is = 8 bits

So 1 megabyte = 8 megabits

Then

783.216 x 8 megabits = 6265.728 megabits

This player has its reading capacity at 1.5 megabits / second

So 1 minute = 60x1.5

= 90 megabits / min

Then we have the entire reading time of this CD player to be =

6265.728/90

= 69.62 minutes.

This answers the question

Answer: F = 498.04 lbs

Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.

Force of friction is given by the formula

[tex]F_{f}=\mu.F_{N}[/tex]

where

μ is coefficient of friction

[tex]F_{N}[/tex] is normal force

Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.

In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So

[tex]F_{N}=m.g[/tex]

Coefficient of friction is how much friction exists between two surfaces.

Rearraging friction force is

[tex]F_{f}=\mu.m.g[/tex]

Mass for this system is the sum of Julie and the sled, therefore

m = 109 + 12

m = 121 lb

Calculating Friction Force:

[tex]F_{f}=0.42.121.9.8[/tex]

[tex]F_{f}=[/tex] 498.04 lbs

LBS is a unit of measurement referred as pound by weight.

In conclusion, force Mary needs to start moving the sled is 498.04 lbs

Answer:

3kg

Explanation:

Given parameters:

Force  = 9N

Acceleration  = 3m/s²

Unknown:

Mass = ?

Solution:

From Newton's second law of motion:

        Force  = mass x acceleration

So;

             9  = mass x 3

             mass  = 3kg

Answer:

The order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]

Explanation:

To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.

So, let's work it out together

According to Heisenberg's uncertainty principle:

[tex]\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}[/tex]

Also; if we recall from the equation of motion that:

[tex]v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)[/tex]

So, if u = 0 and a = g

Then;

[tex]v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)[/tex]

From (2)

Making (s) the subject, we have:

[tex]s = \dfrac{v^2}{2g}[/tex]

[tex]s = \dfrac{p^2}{2gm^2}[/tex]

By differentiation;

[tex]ds = d (\dfrac{p^2}{2gm^2})[/tex]

[tex]ds = \dfrac{2pdp}{2gm^2}[/tex]

[tex]\Delta \ s = \dfrac{p \Delta p}{gm^2 }[/tex]

where;

[tex]\Delta p = \dfrac{h}{4 \pi \Delta \ s}[/tex] from uncertainty principle

This implies that:

[tex]\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}[/tex]

[tex]\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}[/tex]

[tex](\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }[/tex]

here;

v = 2gH

So;

[tex](\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }[/tex]

[tex]\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }[/tex]

Thus, the order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]

Answer:

it can melt orcan put them past their boiling point

Explanation:

Answer:

400 Ns

Explanation:

Impulse = Change in momentum

i.e I = ΔP

So that,

Impulse experienced by Max = Change in Max's momentum

Change in Max's momentum = m(v - u)

Where m is the mass, v is the velocity after collision, and u is the velocity before collision.

m = 100.0 kg, v = 2.0 m/s, u = 6.0 m/s

Change in Max's momentum = 100 x (2 -6)

                             = -400 kg m/s

The negative sign shows that the change in momentum was against his direction of motion.

Impulse experienced by Max = 400 Ns.

Thus,

Max experienced an impulse of 400 Ns as a result of the collision.

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]   (1)

Where:

x: is the displacement

[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)                    

[tex]v_{0}[/tex]: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:

[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)

Where:

[tex]\Delta s[/tex] - Travelled distance, measured in meters.

[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:

[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]

[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Answer:

Workdone = 0.89 Joules

Explanation:

Given the following data;

Workdone = 6J

Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

6 = ½*k*0.13²

6 = ½*k*0.0169

Cross-multiplying, we have;

12 = 0.0169k

k = 12/0.0169

k = 710.06

a. To find the workdone when e = 30, 35.

Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m

Workdone = ½*710.06*0.05²

Workdone = 355.03*0.0025

Workdone = 0.89 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.

Answer:

Action-at-a-Distance Forces. Frictional Force. Gravitational Force. Tension Force ... The force of gravity on earth is always equal to the weight of the object as ... The friction force is the force exerted by a surface as an object moves across it or ... The force of air resistance is often observed to oppose the motion of an object

Explanation:

Answer:

I believe the answer is gravity.

Explanation:

Because with friction you need to be rubbing multiple object together, in tension two objects must be pulling against each other, and in air resistance the air must be touching the object either pushing or pulling it. While in gravity the mass of an object is pulling another object toward it, the objects don't have to be touching each other making it an at-a-distance force.

Answer:

C) 128 kg*m/s

Explanation:

When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.

Answer:

[tex]m=1.63\times 10^{-27}\ kg[/tex]

Explanation:

The velocity of a particle is 90% of the speed of light.

The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]

We need to find the mass of the particle.

The formula for the wavelength of a particle is given by :

[tex]\lambda=\dfrac{h}{mv}[/tex]

h is Planck's constant

v is 90% of speed of light

m is mass of the particle

[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]

So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].

Answer:

v₁ =√2gh,      v₂ = v₁ /√2

Explanation:

Let's use the concepts of energy and work to analyze each case

hill without rubbing. Energy is conserved

starting point. Highest part

         Em₀ = U = mg h

final point. Lower part

         [tex]Em_{f}[/tex] = K = ½ m v²

         Em₀ = Em_{f}

         m g h = ½ m v²

         v₁ =√2gh

rubbing hill

in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction

        W = Em_{f} - Em₀

they indicate half of the initial mechanical energy is lost due to friction

          W = ½ Em₀

we substitute

          - ½ Em₀ = Em_{f} - Em₀

The negative sign is because the friction work always opposes the movement

         Em_{f} = ½ Em₀

        ½ m v₂² = ½ m g h

         v₂ = √½    √2gh

         v₂ = v₁ /√2

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

[tex]h=\sqrt{18^{2}+6^{2}}[/tex]

[tex]h=\sqrt{360}[/tex]

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

[tex]P_{t}=[/tex] 44 + 44 + 44

[tex]P_{t}=[/tex] 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

Answer:

a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]

b)[tex]dh / dt = 0.2658 mm / min[/tex]

Explanation:

From the question we are told that

Diameter of hole [tex]d_h=4mm=>0.004m[/tex]

Depth of hole [tex]D=0mm=>0.001m[/tex]

Diameter of tank [tex]d_t=2mm=>0.002m[/tex]

Generally the equation for pressure is mathematically given as

[tex]Pressure P= \rho*g*d[/tex]

[tex]P= 1/2*\rho *v^2[/tex]

Where

[tex]v = \sqrt {2gd}[/tex]

[tex]V = Area*v[/tex]

[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]

Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by

[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]

Therefore the level at which the water level will initially drop if the water is not replenished

[tex]dh / dt = 0.2658 mm / min[/tex]

The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

Given data:

The diameter of hole is, d = 4.0 mm = 0.004 m.

The depth of hole is, h = 1.0 m.

The diameter of tank is, d' = 2.0 m.

The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.

Let us first obtain the equation of pressure as,

[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]

Here, v is the velocity of efflux and its value is,

[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]

And the level at which the water level will initially drop if the water is not replenished is mathematically given by,

[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]

Here,

r is the radius of hole.

R is the radius of tank.

Solving as,

[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]

Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

Learn more about the flow rate here:

https://brainly.com/question/11816739

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:

[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)

Where:

[tex]m[/tex] - Mass of the sphere, measured in kilograms.

[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.

[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.

[tex]\Delta t[/tex] - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)

Where:

[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.

[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.

[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]

If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:

[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]

[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Answer: 109.4 mm

Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time:  d = vt

The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:

location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m

To reach that point, both steamrollers will have spent

[tex]v=\frac{\Delta x}{t}[/tex]

[tex]t=\frac{\Delta x}{v}[/tex]

[tex]t=\frac{0.0525}{1.2}[/tex]

t = 0.04375 s

The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:

d = 2.5*0.04375

d = 0.109375 m

Until it is crushed, the fly will have traveled 109.4 mm.

Answer:

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Explanation:

The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:

[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)

Where:

[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.

[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.

The angular frequency and the moment of inertia are represented by the following formulas:

[tex]\omega = \frac{2\pi}{T}[/tex] (2)

[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)

Where:

[tex]T[/tex] - Period, measured in seconds.

[tex]m[/tex] - Mass of the stick, measured in kilograms.

[tex]L[/tex] - Length of the stick, measured in meters.

By (2) and (3), (1) is now expanded:

[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]

[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]

[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]

If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:

[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]

[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Answer:

MINORS,    SYSTEMATIC,   STATISTICAL,  BOTOM LINE, ZERO MATCHES

Explanation:

In general the sources of error or uncertainty can be classified

* Statistics. Which are those that describe the statistical formulas, for example: average, standard deviation, absolute error, etc.

* Systematic. That they occur due to an inappropriate measurement or to an interaction between the system and the instrument that cannot be quantified, in general this error shifts the measurements towards an explicit side

* Random, so errors that sometimes occur in the measurement and sometimes not, for example temperature changes during the medical process

In this case, you are asked to complete the sentences with the appropriate word

the measured diameters were slightly ___ MINORS________ compared to what a non-contact method of measuring would provide

. This represents an ____SYSTEMATIC_______ error in ________________.

This additional source of error ________STATISTICAL________ significant.

When the caliper jaws closed, the zero mark on the sliding Vernier scale, BOTOM LINE AND THE __________ line up with the zero mark on the measuring scale.

This means the caliper ___ZERO MATCHES_____________ calibrated correctly.