Comparing energy resources

The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.

The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.

At standard conditions;

We know that:

Therefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.  

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I think 1980is the answer because you add???

Identify:

In each case the forces are constant and the displacement is along a straight line, so

[tex]$$W=F s \cos \phi \text {. }$$[/tex]

Set-Up:

In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls  [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car  [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.

Execute:

(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].

When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].

(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],

So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].

(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.

Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork

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Answer:

A5 A20 A32 5G A1 G A20 5G T mOBIL

Answer:

d because u now

Explanation:

sana makatulong

Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.

Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:

Bird population:

Increase:

Decrease in predators that increase birds chances to survive (natural factor)

Increase in crops that improve food access for birds (human factor)

Decrease:

Invasive species or increase of predators (natural factor)

Destruction of natural habitats due to an increase in industry and extraction (human factor)

Rodent population:

Increase:

Decrease:

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Machine translators:

Level of accuracy can be very low.

Accuracy is also very inconsistent across different languages.

Machines can't translate context.

Mistakes are sometimes costly.

Sometimes translation simply doesn't work.

Human translators:

Turnaround time is longer.

Translators rarely work for free.

Unless you use a translation agency, with access to thousands of translators, you're limited to the languages any one translator can work with.

Answer:

2

Explanation:

please mark me as brainlist i need it so sorry

Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.

Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.

Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.

Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.

The following is an accurate matching of the ions;

11 protons, 12 neutrons, 10 electrons = +1 - Na^+

31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+

Group 2, period 5 = +2 - Sr^2+

Silver is a lustrous metal.

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Answer:

If your answers are the green check marks all of them are correct

Explanation: I took the test

Answer:

Explanation:

v² = u² + 2as

s = (v² - u²)/2a

s = (0² - 75²) / (2(-14))

s = 200.8928

s = 200 m

Answer:

Identification

Explanation:

:-;..........

The speed of the chord wave allows finding a new speed when the mass is increased and the chord length is:

A wave is a periodic movement of the particles that carries energy, but not matter, the speed of the waves is related to the properties of the medium by the relationship.

           [tex]v = \sqrt{\frac{T}{\mu } }[/tex]  

Where v is the speed of the wave, T the force and μ is the linear density.

Indicates that the applied mass increases by a factor of 18.0 and the length of the head is increased to twice the original.

The linear density of the cable is

          [tex]\mu = \frac{m}{l}[/tex]

Where m is the mass of the cable and l is the length.

Let's use the subscript "o" for the initial conditions.

          M = 18.0 m₀

          l = 2 l₀

We look for the density.

         [tex]\mu = \frac{18.0 m_o}{2.0 l_o}[/tex]mu = 18.0 mo / 2 lo

         [tex]\mu = 9.0 \mu_o[/tex]  

We substitute in the expression for the velocity assuming that the tension is kept constant.

        [tex]v= \sqrt{\frac{T}{9.0 \mu} }[/tex]  

        [tex]v= \frac{v_o}{3}[/tex]  

In conclusion, using the speed of the chord wave we can find a new speed when the mass is increased and the chord length is:

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Answer:

Our atmosphere has five different layers. They are:

1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.

2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.

3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.

4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.

5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.

Answer:

1. C

2. True

3. D

4. True

Explanation:

While distance refers to “how much land an object has traversed throughout its journey,” displacement measures “how far an object is out of place.” In this post, let's look at the difference between distance and displacement. Thus, option C is correct.

Displacement and distance are two distinct ideas. The total distance travelled is larger than the displacement between those two points if an object changes direction while travelling.

Distance is the length of any path connecting any two places. Displacement is the straight-line distance between any two points when calculated along the shortest route.

Therefore, The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.

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Answer:

1

+

1

is

2

Explanation:

because if you add one to a one it would end up being 2 <3

Answer: 2

Explanation: 1+1=2

This was THE hardest question I've ever decided to answer

The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:

       a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:

         y = A sin (kx - wt)

Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.

Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship

           [tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]  

Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.

Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore

               [tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]  

              λ = 4L

              λ = 4 0.20

              λ = 0.80 m

The wave number.

              k = [tex]\frac{2\pi }{\lambda }[/tex]  

              k = [tex]\frac{2 \pi }{0.80 }[/tex]  

              k = 2.5π i m⁻¹

In the associated traveling wave, from the graph we can see that the period of the wave is:

             T = 2.8 s

the angular velocity is related to the period.

             [tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]  

             w = 0.714π  rad/s

indicate the maximum displacement that is the amplitude of the wave.

              A = [tex]y_s[/tex]  

             A = 4.3 cm

Let's write the equation of the traveling wave.

              y = 4.3 sin [π (2.5 x - 0.714 t)]

with this expression we can answer the questions.

a) the displacement of the particle for x = 0.30 m

            y = 4.3 sin (π (2.5 0.30 - 0.714 t))

            y = 4.3 sin π( 0.75 - 0.714 t(

Remember that the angles must be in radians.  For time t = 0 the displacement is

              y = 4.3  0.707

              y = 3.04 cm

b) The displacement for x = 0.4m

              y = 4.3 sin (π 2.5 0.4)

              y = 0 cm

c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s

the speed of the wave is

              [tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]  

              v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)

              v = 9.65 cos π(0.75 - 0.714 t)

For time t = 0.90 s the velocity is:

            v = 9.65 cos π(0.75 - 0.714 0.9)

            v = 9.65 0.9436

            v = 9.11 cm / s

d) The velocity for time t = 1.3 s

           v = 9.65 cos π(0.75 - 0.714 1.3)

           v = 9.65 0.9999

           v = 9.65 cm / s

In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:

      a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

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Answer:

Let ω1 be the initial angular speed and ω2 the final angular speed:

α = (ω2- α1) / t      

corresponding to a = (v2 - v1) / t  

S (distance corresponds to theta)

1 rev = 2 pi      and 3.74 rev = 7.48 pi = 23.5 radians

S = 1/2 a t^2 linear     or S = 1/2 α t^2    angular acceleration

23.5 = 1/2 * 1.08 t^2     and t = 6.60 sec   for first 3.84 rev

b)  ω1 = 1.08 * 6.6 = 7.13 rad/sec    initial speed for second 3.74

23.5 = 7.13 t + .54 t^2        compare to S = v1 t + 1/2 a t^2

.54 t^2 + 7.13 t - 23.5 = 0    

t^2 + 13.2 t - 43.5 = 0

t = 2.7 sec for next 3.74

Check:

7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad

Answer:

138

Explanation:

(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)

1) find the partial derivative of each:

[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]  

 [tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]

2) use partial integrals to make gradient of f:

take whatever you got from partial integral and add them together (if they repeat, just use it once)

[tex]F =[/tex] Vf (V = gradient of)

[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]

3) Evaluate the integrals with given points:

Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)

[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]

= 144 - 6 = 138 units of work

Work done by the force field F in moving an object from A to B = 138 J

Given data :

Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]  

Step 1 : determine the partial derivatives of the vector quantity

Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]

Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]

Equating the partial derivatives :  

[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]  = [tex]6xy^{\frac{3}{2} } + c[/tex]

therefore the gradient of F  i.e. F = vF  = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]

Next step : Determine the work done

Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ]  - [ F(point A) = F( 1,1 ) ]

F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex]  = 144

F( 1,1 )  = 6(1)(1)[tex]^{\frac{3}{2} }[/tex]    = 6

Therefore the work done by the force field = 144 - 6 = 138 J

Hence we can conclude that the work done by the force field F is = 138 J

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Answer:

The remains of producers are broken down by decomposers.

The remains of consumers are broken down by soil decomposers.

Animals break down food molecules to obtain energy.

Explanation:

Answer:

It is A. B. and C.

Explanation:

Hope this helps

Rest in physics generally refers to the state in which an object is stationary.

Hence, in Physics, an object is in a state of rest when it does not move from one point to another.

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A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

Explanation:

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

Answer:



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Join / LoginQuestionIn a lifting machine an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10000 N by a distance of 0.8 m Determine the velocity ratio and mechanical advantage

Determine the velocity ratio and mechanical advantageA

Determine the velocity ratio and mechanical advantageA10 and 35

Determine the velocity ratio and mechanical advantageA10 and 35B

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20Easy

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in App

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by Toppr

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option is

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8m

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEd

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25

Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25the load is 10,000N and effort=500N

MA=effortdloadd

MA=effortdloaddMA=50010000

MA=effortdloaddMA=50010000MA=20

Answer:

statements?

Explanation:

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

[tex]X = \frac{MsXs+McXc}{Mw+Mc} \\\\[/tex]

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

[tex]X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\[/tex]

The center of gravity lies at the center of this boat. Therefore,

[tex]Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5[/tex]

5.00 - 1. 00 = 4 meters

[tex]\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\[/tex]

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

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Answer: An ice hockey player picks up a trophy as he slides past it.

Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.

Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.

The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.

Thus, the correct option is A.

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Your question is incomplete, most probably the complete question is:

Which of the following statements describes a perfectly inelastic collision?

A. Two billiard balls bounce off of each other,

B. A car crashes into a tree and rebounds in the other direction,

C. A wad of chewing gum is thrown and sticks to a moving truck,

D. Two ice skaters hit each other and fall over in opposite directions,

The correct statement is, the system performed 800 J of work on its surroundings.

The given parameters:

Apply first law of thermodynamics;

ΔU = Q + W

where;

In adiabatic process no heat is gained or lost by the system.

Q = 0

ΔU = W  (work is positive when it is done on the system by its surrounding)

If work is done by the system to the surrounding, the new equation becomes;

ΔU = - W

W = - ΔU

W  = -800 J

This implies that the system performed 800 J of work on its surroundings.

"Your question is not complete, it seems to be missing the following information";

In a thermodynamic process, the internal energy of a system in a container with adiabatic walls decreases by 800 J. Which statement is correct?

a. The system lost 800 J by heat transfer to its surroundings.

b. The system gained 800 J by heat transfer from its surroundings.

c. The system performed 800 J of work on its surroundings.

d. The surroundings performed 800 J of work on the system.

e. The 800 J of work done by the system was equal to the 800 J of heat transferred to the system from its surroundings.

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Answer:

Explanation:

The change in force = 9.81(1.95 - 0.300) = 16.2 N

The change in length is 0.750 - 0.200 = 0.550 m

K = ΔF/Δx = 16.2/0.550 = 29.4 N/m

The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;

As we know; the periodic table is an array of elements in order of their atomic number.

In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.

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The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]

The distance between the glass is calculated as follows;

[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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The motion of the rope which is perpendicular to the direction of the

propagation of the wave is a transverse wave motion.

Reasons:

The given function for the wave speed is presented as follows;

[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]

Where;

[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]

Taking the mass of the rope as, m = 2.00 kg

The length of the rope, L = 80.0 m

The mass hanging on the rope, M = 20.0 kg

We have;

T = 20.0 kg × 9.81 m/s² = 196.2 N

[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]

Therefore;

Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;

v = f × λ

Therefore;

v = 7.9 Hz × 7.9 m = 62.41 m/s

Which gives;

[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]

T = 62.41² × 0.025 = 97.3752025

[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]

Where;

g = The acceleration due to gravity which is approximately 9.81 m/s²

[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]

Therefore;

The mass of the box, m ≈ 9.93 kg

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The parameters obtained from a similar question online are;

[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]

Length of the rope, L = 80.0 m

Mass of the rope, m = 2.0 kg

Frequency of a point on the rope, f = 20 Hz