A car travels a distance of 540km in 6 hours. What speed did it travel at?

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    

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.

That would be a conduction.

conduction

hope this helped

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

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)

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)

The kinetic energy of the object is 53775 Joules.

Given the following data:

To find the kinetic energy of the object:

Kinetic energy is an energy that is possessed by a physical object or body due to its motion.

Mathematically, kinetic energy is calculated by using the formula;

[tex]K.E = \frac{1}{2} MV^2[/tex]

Where:

Substituting the values, we have;

[tex]K.E = \frac{1}{2}[/tex] × [tex]478[/tex] × [tex]15^2[/tex]

[tex]K.E = 239[/tex] × [tex]225[/tex]

Kinetic energy = 53775 Joules

Therefore, the kinetic energy of the object is 53775 Joules.

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Answer:

I think it would be B

Explanation:

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.

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

[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]

Final answer:

Density is a physical property of a substance, which can be observed and measured without changing the substance's chemical identity.

Explanation:

Density is a characteristic of a substance that refers to how much mass is contained in a given volume. It is considered a physical property because it can be observed and measured without changing the substance's chemical identity. Density can be particularly helpful in identifying substances and is measured in units such as grams per cubic centimeter (g/cm³) for solids and liquids or grams per liter (g/L) for gases.

Contrasting with physical properties are chemical properties, which describe how a substance can change or react to form new substances. These require a chemical change to be observed, such as rusting iron or burning wood, which both result in new chemical substances being formed.

Therefore, to answer the student's question, density is not a physical change, a chemical change, nor a chemical property; it is a physical property.

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.

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.

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.

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.

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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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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.

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.

If a person does aerobic exercise but does not work to build muscle, they may improve cardiovascular fitness and endurance, but may not see significant changes in muscle size or strength.

Doing aerobic exercise without working to build muscle can have several effects. Aerobic exercise primarily targets and improves cardiovascular fitness, which includes strengthening the heart and improving lung capacity. However, without resistance training, muscle mass and strength may not significantly increase. This means that while a person may improve their endurance and cardiovascular health, they may not see significant changes in muscle size or strength.

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It sounds like the ball's velocity along the [tex]y[/tex]-axis should be taken to be constant, in which case the ball has velocity vector

[tex]\mathbf v(t)=\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]

We note that at [tex]t=t_0[/tex], we're told the ball is at the origin, so its position at this point is [tex]\mathbf r(t_0)=\mathbf 0[/tex]. We then integrate the velocity vector to find the position vector:

[tex]\mathbf r(t)=\mathbf r(t_0)+\displaystyle\int_{\tau=t_0}^{\tau=t}\mathbf v(\tau)\,\mathrm d\tau[/tex]

[tex]\mathbf r(t)=\displaystyle\int_{\tau=t_0}^{\tau=t}\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}\,\mathrm d\tau[/tex]

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\tau\,\hat{\jmath}\bigg|_{\tau=t_0}^{\tau=t}[/tex]

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)(t^2-{t_0}^2)\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)(t-t_0)\,\hat{\jmath}[/tex]

For the sake of simplicity, I'll assume [tex]t_0=0[/tex], so that

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\hat{\jmath}[/tex]

(a) Then at [tex]t=2.50\,\mathrm s[/tex], the ball's position in the plane is given by the vector

[tex]\mathbf r(2.50\,\mathrm m)=(6.56\,\mathrm m)\,\hat{\imath}+(3.25\,\mathrm m)\,\hat{\jmath}[/tex]

which is to say its position in the plane is at the point (6.56, 3.25).

(b) The ball's displacement is the same as its position, but in vector form (as above).

(c) The ball has velocity

[tex]\mathbf v(2.50\,\mathrm s)=\left(5.25\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\imath}+\left(1.3\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]

That is, the ball has a speed of [tex]\|\mathbf v(2.50\,\mathrm s)\|=5.41\,\dfrac{\mathrm m}{\mathrm s}[/tex] and is traveling in a direction of about [tex]13.9^\circ[/tex] relative to the positive [tex]x[/tex]-axis.

At t=2.5s, the ball's position will be at coordinates (6.5625, 3.25), have a displacement of approximately 7.23 metres, and have a velocity of 5.25 m/s in the x direction and 1.3 m/s in the y direction.

The subject of this question is physics, specifically kinematics, and the topic is motion in two dimensions. We will calculate the x and y coordinates, the displacement, and the velocity for the given conditions. First we will calculate the x-coordinate using the equation of motion x = ut + 0.5at^2, where 'u' is initial velocity, 'a' is acceleration and 't' is time. Here, the initial velocity along x-axis is zero, acceleration is 2.10 m/s^2, and time is 2.5 s. Substituting these values, we get x = 0 + 0.5 * 2.1 * (2.5)^2 = 6.5625 m.

Next, for the y-coordinate, since there is no acceleration along y-axis, its position at any time is given by y = ut, where 'u' is initial velocity along y-axis, and 't' is time. So, y = 1.30 * 2.50 = 3.25 m.

Next, the displacement, which is the shortest distance from the initial to the final position, can be found using the Pythagorean theorem, d = sqrt(x^2 + y^2). Thus, d = sqrt((6.5625)^2 + (3.25)^2) = 7.23 m approximately.

Lastly, the velocity of the ball at any point is given by v = u + at, where 'a' is the acceleration along the x-axis, 't' is time and 'u' is the initial velocity (which is zero in this case along the x-axis). So, v = 0 + 2.1 * 2.5 = 5.25 m/s along the + x-direction. And, the velocity along the y-axis remains constant as 1.30 m/s because there's no acceleration in the y direction.

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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.

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.

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.

All the bats living in a cave form a __colony__of bats in the cave ecosystem.

The bats living in a cave form a population of bats within that cave ecosystem. The term 'population' in biology refers to all individuals of a particular species inhabiting a specific area. Factors such as diseases and resource exploitation can though impact a population.

All the bats living in a cave form a population of bats in that cave ecosystem. A population refers to all individuals of a species living within a specific area. For instance, the bats in a cave represent the population of bats in that particular cave. It's important to note that populations aren't static - they can be affected by various factors such as disease. For instance, the tragedy of the commons principle suggests that resources held in common, such as a cave, could potentially be overexploited, leading to population decline. A specific example is white-nose syndrome, a disease impacting bat populations that inhabit caves in the eastern United States and Canada.

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