The reciprocal of resistance or measure of a devices ability to conduct is called

Answer:

I think it would be B

Explanation:

her velocity is 0.82012

Answer:

Dina's velocity relative to the ground is 7.5 m/s.

Explanation:

Given that,

Velocity of Dina v = 0.5 m/s

Velocity of train v' = 8.0 m/s

We need to calculate the Dina's relative velocity

Using formula of relative velocity

Relative velocity=velocity of train - velocity of Dina

[tex]v_{r}=v'-v[/tex]

Where, v' = Velocity of train

v = Velocity of Dina

Put the value into the formula

[tex]v_{r}=8.0-0.5[/tex]

[tex]v_{r}=7.5\ m/s[/tex]

Hence, Dina's velocity relative to the ground is 7.5 m/s.

joule / second is the answer

Answer:

Option (A)

Explanation:

Watt is the SI unit of power.

The rate of doing work is defined as power.

Power = work/time

The SI unit of work is joule and the SI unit of time is second, so the SI unit of power is joule per second.

If joule of work is said to be done in second then we can say the power is watt.

An object moving at constant speed in a straight line would move differently if acted on by unbalanced forces, as indicated by Newton's first and second laws. Unbalanced forces would cause changes in the object's velocity, either in speed, direction, or both.

An object moving at constant speed in a straight line would move differently if it was acted on by unbalanced forces. According to Newton's first law, an object will continue in its state of motion at constant speed in a straight line unless it is compelled to change by an external force. This principle refers to the inertia of the object.

For instance, the answer option 'B. It is acted on by unbalanced forces.' would make the object move differently. Unbalanced forces will cause the object to either speed up, slow down, or change its direction of motion. If there are exactly opposing forces (option 'C'), assuming they are equal in magnitude, they would balance each other out and the object will continue to move at the same speed in the same direction.

Newton's second law further explains that the rate of change of momentum of an object is directly proportional to the unbalanced force acting on it and takes place in the direction in which the force acts. As such, any presence of unbalanced forces will result in a change in the object's velocity - either speed, direction, or both - meaning the object's motion would be different.

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I just answered the same question and you can see it in the link https://brainly.com/question/11346625.

This is the same answer.

Answer: The work OUTPUT of a simple machine can never exceed the work INPUT because FRICTION uses some of the work.

Explanation:

This concept is related to ideal machines and efficiency.

Only ideal machines can convert all the work input into useful work (work ouput).

The efficiency is a concept that tells you the proportion of input work that is transformed into output work.

         Efficiency = (output work / input work) × 100

Ideal machines, those where friction does not exist, have a 100% efficiency, and so ideal machines convert all the input work into useful output work.

Real machines have a efficiency less than 100%. Friction lowers the efficiency and the work output is less than the work input, because friction uses work which is converted, mostly, into heat energy or sound energy.

i would say line graph

Answer: a. line graph

Explanation:   A line graph is formed by joining the points given by the data with straight lines.

A line graph is usually used to show the change of information over a period of time using two variables . This means that the horizontal axis is usually a time scale, for example minutes, days, months or years; and the vertical axis shows the value of the information that varies in time.

[tex]Given:\\m=7kg\\v=4\frac{m}{s} \\\\Find:\\E_k=?\\\\Solution:\\\\E_k=\frac{mv^2}{2} \\\\E_k=\frac{7kg\cdot (4\frac{m}{s})^2}{2} =\frac{7kg\cdot 16\frac{m^2}{s^2} }{2} =7kg\cdot 8\frac{m^2}{s^2} =56J[/tex]

Answer:

b

Explanation:

cuz i no

The slope of the position-time graph represents the corresponding velocity - this relationship forms the basis for the creation of a velocity-time graph. Positive slopes on the position-time graph mean forward motion and negative slopes mean backward motion which is graphically represented as positive and negative slopes on the velocity-time graph respectively.

The position-time graph and the velocity-time graph are closely related. The slope of the position-time graph at any given point represents the velocity at that point in time. For example, if you have a straight line on a position-time graph which has a positive slope, it means the object is moving in a positive direction or forward. Therefore, the velocity-time graph for the same object would also have a straight line with positive slope.

A similar concept applies if the slope on the position-time graph is zero, it means the velocity at that time is zero and this would be represented as horizontal line on the velocity-time graph. Another interesting aspect is, if the slope of a position-time graph is negative, it means the object is moving in a negative or backward direction and the corresponding a velocity-time graph would have a negative slope.

It’s notable that straight lines on velocity-time graphs indicate constant velocity, whereas lines with positive or negative slope indicate acceleration or deceleration respectively.

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At the same time, however, you get less detail or less precision in a chart or graph than you do in the table. Imagine the difference between a table of sales figures for a ten-year period and a line graph for that same data. You get a better sense of the overall trend in the graph but not the precise dollar amount.

A data tables presents information and data in a table, usually in rows. A graph presents statistics and data in graph, or coordinate grid or plane.

B) are compressible.  

gasses are compressible

Answer: Option (B) is the correct answer.

Explanation:

Noble gases group 18 elements. Members of this group are He, Ne, Ar, Kr, Xe, and Rn.

All these elements have completely filled sub-shells and hence they are stable in nature. All the elements of group 18 are gases, this means that their particles are loosely held due to weak intermolecular interaction.

As a result, their particles can move freely from one place to another. These gases can be compressed in order to bring their particles together. It is also known that all gases are compressible in nature.

Thus, we can conclude that noble gases have one property in common is that they are compressible.

The ball's horizontal position [tex]x[/tex] at time [tex]t[/tex] is

[tex]x=v_{0x}t[/tex]

The ball has initial velocity in the horizontal direction only, so [tex]v_{0x}=15\,\frac{\mathrm m}{\mathrm s}[/tex]. Then the time it takes for the ball to travel 83 m horizontally [tex]t[/tex] is given by

[tex]83\,\mathrm m=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)t\implies t=5.5\,\mathrm s[/tex]

Meanwhile, the ball's vertical position [tex]y[/tex] at time [tex]t[/tex], starting at the height of the cliff [tex]h[/tex], is given by

[tex]y=h-\dfrac12gt^2[/tex]

where [tex]g=9.8\,\frac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. After 5.5 seconds, the ball hits the ground, so that [tex]y=0[/tex] when [tex]t=5.5\,\mathrm s[/tex], and we use this to solve for [tex]h[/tex]:

[tex]0=h-\dfrac12g(5.5\,\mathrm s)^2\implies h\approx150\,\mathrm m[/tex]

Answer: 3.63 s

The car would accelerate due to force of friction. Initially the car would have to overcome static force of friction.

Static force of friction:

[tex]F_s=\mu_smg[/tex]

Where[tex]\mu_s[/tex] is the coefficient of static friction, mg is the weight.

Let m be the mass of the car and a be the acceleration, then

[tex]ma=\mu_smg \Rightarrow a=\mu_sg[/tex]

It is given that, [tex]\mu_s=1.00[/tex]

[tex]a=1.00\times 9.8 m/s^2= 9.8 m/s^2[/tex]

Now, using the equation of motion:

[tex] v-u=at[/tex] we can find the shortest time in which the car would accelerate from[tex]u= 0 \hspace{1mm} to \hspace{1mm}  v=80 mph=80\frac{miles}{h} \times 1.6 \frac{km}{mile} \times \frac{5 m/s}{18km/h}=35.56 m/s[/tex]

[tex] \Rightarrow t=\frac{v-u}{a}=\frac{35.56 m/s-0}{9.8m/s^2}=3.63 s[/tex]

For a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.

Given the data in the question;

Since the car accelerated from 0 to 80.

Initial velocity; [tex]u = 0m/s[/tex]

Final velocity; [tex]v = 80mph = 35.7632 m/s[/tex]

To determine the shortest time

We use the expression from the Force of Static Friction:

[tex]f_s = u_sN[/tex]

[tex]f_s = u_smg[/tex]

Also, From Newton's Second Law of Motion:

[tex]F = ma[/tex]

Now, from the diagram ( free body )

[tex]F = f_s[/tex]

Hence,

[tex]ma = u_smg\\a = u_sg[/tex]

We know that acceleration due to gravity;[tex]g = 9.8m/s^2[/tex] and in the question, [tex]u_s = 1.00[/tex]

So we substitute into the equation

[tex]a = 1.00 \ *\ 9.8m/s^2\\a = 9.8m/s^2[/tex]

Next, from the First Equation of Motion;

[tex]v = u + at[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration.

We make time "t", the subject of the formula

[tex]t = \frac{v-u}{a}[/tex]

We substitute in our value

[tex]t = \frac{35.7632m/s- 0m/s}{9.8m/s^2} \\\\t = \frac{35.7632m/s}{9.8m/s^2} \\\\t = 3.649s\\\\t = 3.65s[/tex]

Therefore, for a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.

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The formula is sum of the forces = Mass* Acceleration

If there is no friction (which i'm assuming not)

500N = 100 * A

A = 5 M/S/S (Or M/S^2)

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

[tex]Q = ms \Delta T[/tex]

[tex]Q = 1*s*(250 - 20)[/tex]

[tex]Q = 1*s*230[/tex]

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

[tex]Q = ms'\Delta T[/tex]

[tex]1*s*230 = 1* s' * \Delta T[/tex]

now we have

[tex]\Delta T = (\frac{s}{s'})*230[/tex]

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

A) The second pan would reach a lower temperature.

As the specific heat increases, the temperature change decrease. The second pan would reach a lower temperature.

mercury is the closest planet from the sun

The terrestrial planets formed near the Sun due to the high temperatures that allowed only heavy materials like silicates and metals to condense into solid forms. Beyond the asteroid belt, colder temperatures in the outer solar system permitted gases and ices to condense, leading to the formation of gas giants like Jupiter and Saturn.

The planets that formed near the Sun where the solar system temperatures were very high are known as the terrestrial planets. Close to the Sun, temperatures were too great for light gases like hydrogen and helium to condense; only heavier materials such as silicates and metals could form solid clumps. This is why the inner planets, including Mercury, Venus, Earth, and Mars, are composed mostly of rock and metal.

In contrast, in the outer solar system, colder temperatures allowed gases to clump together and form the gas giants like Jupiter and Saturn. These giant planets attracted and retained hydrogen and helium, becoming much larger than their terrestrial counterparts. With greater volume and mass, the gas giants were able to grow very large, very quickly, and generate significant internal heat through contraction, even after their formation.

The difference in planet composition and formation is a result of the thermal gradients in the solar nebula, being hotter closer to the Sun and cooler further away. This led to terrestrial planets forming close to the Sun and gas giants beyond the asteroid belt, where it was cold enough for ice and gases to condense into large massive planets.

6.10 x 10^9 / 3 x 10^5 = 2.033 x 10^4 seconds = 20.33 / 3.6 = nearly 6 hours .

Light travels from the sun to Pluto in approximately 5.5 hours. This is calculated using the speed of light in vacuum and the average distance from the sun to Pluto.

To calculate the time it takes for light from the sun to reach Pluto, we first need to understand the concept of light speed. Light speed is the fastest speed at which light can travel in vacuum, and it is broadly accepted as being about 299,792 kilometers per second.

Given that the average distance from the Sun to Pluto is approximately 6.1 x 10^9 kilometers, we can utilize the equation

speed = distance / time where time = distance / speed.

In this case, we divide the distance 6.1 x 10^9 kilometers by the speed of light 299,792 kilometers per second, resulting in approximately 5.5 hours.

Therefore, the time it takes for light from the Sun to reach Pluto is around 5.5 hours.

Learn more about Light speed here:

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The value of using the International System of Units when describing motion is that these units are universally accepted - so an observer that uses these units to describe a motion in a place can compare his results to the results of the same experiment made by another obserber in a different place.

For example, if I measure the speed of a car, and I measure it is 20 m/s, and then another observer measures the speed of another car, and he finds it is 25 m/s, we are able to compare the two speeds and tell which one is greater. On the contrary, if we use two different units (e.g. km/h and miles/h), it would be difficult to compare the two results.

according to newton's third law, every action has equal and opposite reaction. in this scenario of making a jump from the boat onto a dock, as i jump my feet in contact with the boat push the boat in backward direction. hence the action force is the push by my feet on the boat. the boat reacts by applying a reaction force on my feet pushing the feet in forward direction. hence reaction force here is the force by the boat on the feet. due to the reaction force of the boat on feet, i am pushed in forward direction to reach the the dock.

When jumping from a boat to a dock, Newton's Third Law of Motion explains that the force you use to push off the boat also pushes the boat in the opposite direction.

When you jump from a small boat onto a dock, Newton's Third Law of Motion is in full effect. According to this law, for every action, there is an equal and opposite reaction. When you push yourself off the boat to jump onto the dock, you exert a force on the boat. The boat, in turn, reacts to your force by moving in the opposite direction.

The same principle is evidenced if a man fixed on the shore pulls a boat by a rope; the man pulls on the rope, exerting a force on the boat. At the same time, the boat exerts an equal and opposite force back on the man through the rope. Neither of these interactions violate the third law of motion.

Applying this concept further, let's consider the swimmer pushing off the side of a pool. The swimmer exerts a force against the pool wall, and the wall exerts an equal and opposite force that propels the swimmer forward. Essentially, this thrust is similar in nature to how rockets and other vehicles generate propulsion.

Given:

u(initial velocity):9.6 m/s

at the maximum height final velocity =0

The acceleration acting on the body is gravity as it is free falling object

=-9.8m/s^2

Now we know that

v^2-u^2 = 2as

Where v is the final velocity measured in m/s

u is the initial velocity which is measured in m/s

a is the acceleration measured in m/s^2

s is the distance traveled by the rock in this case it is considered as the height

Substituting these values we get

0-9.6= 2 x -9.8 x s

s= 0.49 m

Consider t as the time taken for the rock to travel

v=u+at

0=9.6 -9.8t

t=0.98sec

When object reached the terminal speed then its acceleration is zero

So as per Newton's II law we can say

[tex]F_{net} = 0[/tex]

now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed

so we can write

[tex]mg - F_d = 0[/tex]

so here we are given that

[tex]F_d = 30[/tex]

[tex]6*9.8 - 30*v = 0[/tex]

[tex]58.8 - 30 *v = 0[/tex]

[tex]v = \frac{58.8}{30} [/tex]

[tex]v = 1.96 m/s[/tex]

so terminal speed will be nearly 2 m/s

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = [tex]12.0v+4.00v^2[/tex]

Equating both, we have

    [tex]784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0[/tex]

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

The terminal speed of the object is approximately [tex]\( 12.59 \, \text{m/s} \).[/tex]

To find the terminal speed of the object, we need to set the drag force equal to the gravitational force acting on the object. At terminal speed, the net force on the object is zero, meaning the drag force and the gravitational force are balanced.

The gravitational force [tex]\( F_g \)[/tex] acting on the object is given by the mass of the object  m times the acceleration due to gravity g , which is approximately[tex]\( 9.81 \, \text{m/s}^2 \):[/tex]

[tex]\[ F_g = m \cdot g = 80.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \][/tex]

[tex]\[ F_g = 784.8 \, \text{N} \][/tex]

The drag force [tex]\( F_{\text{drag}} \)[/tex] is given by the equation:

[tex]\[ F_{\text{drag}} = (12.0 \, \text{N} \cdot \text{s/m})v + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v^2 \][/tex]

At terminal speed,[tex]\( F_{\text{drag}} = F_g \)[/tex], so we have:

[tex]\[ (12.0 \, \text{N} \cdot \text{s/m})v_{\text{terminal}} + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v_{\text{terminal}}^2 = 784.8 \, \text{N} \][/tex]

Now, we can use the quadratic formula to solve for [tex]\( v_{\text{terminal}} \):[/tex]

[tex]\[ v_{\text{terminal}} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where [tex]\( a = 4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2 \), \( b = 12.0 \, \text{N} \cdot \text{s/m} \)[/tex], and [tex]\( c = -784.8 \, \text{N} \).[/tex]

Plugging in the values:

[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{(12.0)^2 - 4 \cdot 4.00 \cdot (-784.8)}}{2 \cdot 4.00} \][/tex]

[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{144 + 12556.8}}{8} \][/tex]

[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{12700.8}}{8} \][/tex]

[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm 112.7}{8} \][/tex]

Since speed cannot be negative, we take the positive root:

[tex]\[ v_{\text{terminal}} = \frac{-12.0 + 112.7}{8} \][/tex]

[tex]\[ v_{\text{terminal}} = \frac{100.7}{8} \][/tex]

[tex]\[ v_{\text{terminal}} = 12.5875 \, \text{m/s} \][/tex]

Further Explanation:

As the dogs collectively full the sled on the wet snow surface, the sled will experience a net force due to all the dogs acting on it.

Part (a):

The net force applied by the 8 dogs on the sled is:

[tex]\begin{aligned}{F_{{\text{net}}}}&=8\timesF\\&=8\times185\\&=1480\,{\text{N}}\\\end{aligned}[/tex]

The net mass of the dog and the sled combined is:

[tex]\begin{aligned}M&=\left({8\times{m_d}}\right)+{m_s}\\&=\left({8\times19}\right)+210\\&=362\,{\text{kg}}\\\end{aligned}[/tex]

The net acceleration of the dogs and the sled combined is:

[tex]a=\frac{{{F_{net}}}}{M}[/tex]

Substitute the values of net force and mass in above expression.

[tex]\begin{aligned}a&=\frac{{1480}}{{362}}\\&=4.088\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\&\approx4.09\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}[/tex]

Thus, the acceleration of the dogs as they pull the sled on the snow is [tex]\boxed{4.09\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}[/tex] .

Part (b):

Since the sled is moving with an acceleration of [tex]4.09\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}[/tex] , the net force experienced by the sled due to the pull of the dogs is:

[tex]{F_{sled}}={m_s}\times a[/tex]

Substitute the values in above expression:

[tex]\begin{aligned}{F_{sled}}&=210\times4.09\\&=858.9\,{\text{N}}\\&\approx{\text{859}}\,{\text{N}}\\\end{aligned}[/tex]

Thus, the force experienced by the sled due to the dogs is [tex]\boxed{859\,{\text{N}}}[/tex] .

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Newton’s law of Motion

Keywords:

Team of eight dogs, waxed wood runners, loaded sled, with its rider, acceleration, average force 185 N, coupling between dog and sled, average masses of 19 kg each.

The most important characteristics that are exhibited by metals are-

1- Metals are ductile

2-Most metals are conductive in nature.

3-Most metals are malleable.

4- Metals have strong inter molecular force of attraction between the.

5-Metals have luster.

6-Metals are sonorous.

Here we are given Tungsten filament.

Tungsten is  a metal.So it must be conductive and as well as ductile in nature.

The electric filament that we are using in our electric bulb glows due to the heating effect of current.Hence the chosen substances for glowing electric bulb must have high melting point.

The melting point of tungsten is 1650 degree celsius which is very high.That's why it is used in electric bulb.

Hence the correct answer to the question is the third one i.e Tungsten is ductile,has a high melting point, and is electrically conductive.

Answer:

c) Tungsten is ductile, has a high melting point, and is electrically conductive.

Explanation:

i took the test

Velocity = 278m/s

Distance = 7500m

Time = How long it takes?

Time = Distance/Velocity = 7500m ÷ 278m/s = 26.98 seconds (Rounded up to nearest two decimal places)

Answer:

 Position at t= 4 seconds is 144 m

Explanation:

 It is given that acceleration, a = 18 t, where t is the time.

 We know that Velocity, [tex]v = \int { a} \, dt[/tex]

  Substituting value of a,

           Velocity, [tex]v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c[/tex]

 We know that at t = 0, v = -12 m/s

         So, [tex]9*0^2+c=-12\\ \\ c=-12m/s[/tex]

So velocity, [tex]v = (9t^2-12)m/s[/tex]

  We also know that displacement, [tex]x = \int { v} \, dt[/tex]

     Substituting value of v,  

        Displacement, [tex]x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c[/tex]

  We know that at t = 0, particle is at origin, x =0.

               So,  [tex]0=3*0^3-12*0+c\\ \\ c=0[/tex]

   Displacement, [tex]x = 3t^3-12t[/tex]

At t = 4 seconds

   [tex]x = 3*4^3-12*4=192--48=144m[/tex]

Position at t= 4 seconds is 144 m  

]The particle is at 144 meters along the x-axis at t=4.0 seconds.

To find the position, we first need to integrate the acceleration to get the velocity as a function of time, and then integrate again to get the position.

We know that acceleration is the derivative of velocity with respect to time, so:

v(t) = ∫dt = ∫18t dt = 9t2 + C

Where C is the integration constant. Using the initial condition, v(0) = -12 m/s, we find that C = -12.

Now we integrate the velocity to find the position:

x(t) = ∫v(t) dt = ∫(9t2 - 12) dt = 3t3 - 12t + D

Where D is another constant of integration, since the particle is at the origin at time t=0, x(0) = 0, we find D = 0.

Finally, we plug t = 4.0 s into the position function:

x(4) = 3(4)3 - 12(4) = 192 - 48 = 144 meters.

The particle's position at t=4.0 seconds is 144 meters along the x-axis.

The Earth's magnetic field is believed to be generated by electric currents in the conductive material of its core, created by convection currents due to heat escaping from the core.

Answer:

Independent variable: washer detergent

Dependent varaible: cleaing power of effectivenes of the detergent.

Explanation:

When an experiiment is performed there exists one independent variable, one dependent variable, and some other variables that must be controlled.

The goal of the experiment is to test a hypothesis, which is a statement that relates an effect with a cause and that must be proved true or false.

The effect in the dependent variable, and the cause is the independent variable.

Jenny (the scientist) wants to show how an effect or phenomenun, the dependent variable, is explained by the explanatory, independent, variable.

In the stated experiment, Jenny wants to know how the effectiveness of a new brand detergent and an old one compare with each other. Therefore, she designs a test in which she will determine how the brand of detergent may  explain (explanatory or independent variable) the cleaning power or effectiveness, the dependent variable.

Jenny only can manipulate the independent variable and observe the effect on the dependent variable.

To be conclusive, other variables must be controlled, like the fabric and the amount of dirt of the t-shirts.

The answer is: 12.30 degrees.

To solve this problem you must apply the proccedure shown below:

1. You run 22 meters horizontally and then you climb a rope vertically for 4.8 meters.

2. So. let's call the angle [tex]\alpha[/tex], therefore, you must find it using [tex]tan^{-1}[/tex], as following:

[tex]tan^{-1}(\alpha)=\frac{opposite}{adjacent} \\opposite=4.8\\adjacent=22[/tex]

3. Substitute values and solve for the angle:

[tex]tan^{-1}(\alpha)=\frac{4.8}{22}\\\alpha=12.30degrees[/tex]

Your answer would be the;  NET  force on the object.   Refer to Newton's Laws of Forces and Motion.

Hope that helps!!!!!!!!!!!!!!!!                                        : )

Answer:

Net force

Explanation:

When a number of forces acting on a body, and then a non zero value of force is there. It is called net force acting on a body.

There are two types of forces.

1. Balanced forces

2. Unbalanced forces

When a number of forces acting on a body and then net force is zero, Then the forces are called balanced forces.

When a number of forces acting on a body and net force is non zero . Such forces are called unbalanced forces.