¿Por qué el origen de ciertos metales se relaciona con la dinámica terrestre?

An example of a point source water pollutant is a sewer pipe draining directly into a river.

According to the United States Environmental Protection Agency, a  point source pollution is defined as;  “any single identifiable source of pollution from which pollutants are discharged, such as a pipe, ditch, ship or factory smokestack.”

Hence, an example of a point source water pollutant is a sewer pipe draining directly into a river. The pollution is coming from an easily identifiable source which is a sewer pipe connected to a river directly.

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Answer:

[tex]12.79 m/s^2[/tex]

Explanation:

1. Apply Newton's 2nd law:

[tex]F=ma[/tex]

2. Use the definition of weight to find the mass of the box:

[tex]W=mg=m=\frac{W}{g}[/tex]

3. Solve for acceleration in Newton's 2nd law and substitute for mass:

[tex]a=\frac{F}{\frac{W}{g} }[/tex]

4. Plug in the given values from the question and assume [tex]g=9.81m/s^2[/tex]:

[tex]a=\frac{777N}{\frac{596N}{9.81m/s^2} } } =12.79m/s^2[/tex]

Answer:

a1 = -w^2 A sin w t        where w is the angular frequency

a2 =  - (2 w)^2 A sin 2 w t       where w2 = 2 w1

a2 / a1 = 4 sin 2 w t / sin  w t

Since sin 2 w t / sin w t = 2     sin w t will change twice as fast

a2 / a1 = 8      the max acceleration must change 8 tims as fast

Explanation:

static electricity draws energy

Answer:

F-F(gr) = ma

a= {F-F(gr)}/m =

=(15-10)/15=0.33 m/s² (upward)

Answer: HOPE THIS HLEP YOU

Explanation:

The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.

Answer:

relating to or supporting democracy or its principles.

Explanation:

Answer:

b

Explanation:

when the wavelength increase it doesnt affect the frequency of a wave.

Answer: The wavelength increases by a factor of 3

Explanation:

Explanation:

I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.

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,

Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.

The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.

This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.

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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:b

Explanation:

Answer:

The lobster let the claw go. so, B.

Explanation:

i hope this helps.

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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Answer: Terrestrial biomes are based on land, while aquatic biomes include both ocean and freshwater biomes.

Explanation:

The patient eating a candy bar instead of fasting for the test should be told

that the test results will be wrong and he may receive a wrong diagnosis.

The medical practitioners told him to fast when coming for a reason and he

forgetting and eating something means the objective for the test may have

been defeated.

This is why it's best to explain to him about the consequences of his actions

which is likely getting a wrong result and diagnosis.

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Answer:

2 times more work

Explanation:

Work is a force times distance

W = Fd

The barbell does not change mass so its weight (force) is constant.

That means work is directly proportional to the distance traveled.

twice the distance means twice the work.

As water cools the volume expands.

Answer: B. The volume of water will increase if cooled from 3°C to 2°C.

Answer:

The two main factors that affect kinetic energy are mass and speed.

Explanation:

Kinetic energy is the energy that is caused by the motion. The kinetic energy of an object is the energy or force that the object has due to its motion. Your moving vehicle has kinetic energy; as you increase your vehicle's speed, your vehicle's kinetic energy increases.

Have a great day! :D

Answer:

3h/4

Explanation:

At speed v/2 height will be 3/4 h

Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics

When block is at top of first ramp

u=0 ( block was at rest )

a = g ( acceleration due to gravity

using equation of motion                                                                                                                                                                                                                                                                    

2as = v^2 - u^2

2gh = v^2

Then the block continued and reached a speed of v1 = v/2 on second ramp

now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2

u= [tex]\sqrt{2gh\\}[/tex]

s= h1

using equation of motion , we get

2as = v^2 - u^2

2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]

2(-g)h1 = (g h - 4 g h) / 2

h1 = 3/4 h

At speed v/2 height will be 3/4 h

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Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So

m₁ = 30.0 g = 0.0300 kg

m₂ = 13.0 g = 0.0130 kg

v₁ = + 20.5 cm/s = 0.205 m/s

v₂ = + 15.0 cm/s = 0.150 m/s

and we want to find v₁' and v₂', the final velocities of either object after their collision.

Momentum is conserved throughout the objects' collision, so that

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

where v₁' and v₂' are the first and second object's velocities after the collision.

Kinetic energy is also conserved, so that

1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²

or

m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²

From the first equation (omitting units), we have

0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'

0.0810 = 0.0300 v₁' + 0.0130 v₂'

81 = 30 v₁' + 13 v₂'

From the second equation,

0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²

0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²

1.55 ≈ 30 (v₁')² + 13 (v₂')²

Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields

v₁' ≈ + 0.172 m/s

and

v₂' ≈ + 0.227 m/s

Answer:

True

Explanation:

Answer:

(d) is your answer for your question

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

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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From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees

Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,

Initial velocity U = 0 ( since it starts from rest)

Final velocity V = 5.3 m/s

distance S = 6.4 m

Let us first calculate its acceleration by using  third equation of motion.

[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS

[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a

28.09 = 12.8a

a = 28.09 / 12.8

a = 2.2 m / [tex]s^{2}[/tex]

To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.

acceleration a = gsin∅

substitute all the relevant parameters

2.2 = 9.8 sin∅

sin∅ = 2.2/9.8

sin∅ = 0.224

∅ = [tex]Sin^{-1}[/tex](0.224)

∅ = 12.97 degrees

∅ = 13 degrees (approximately)

Therefore, the angle of the slope is 13 degrees

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