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3 years ago

6

If a truck loses 5610 J of energy as it slows down due to an external force of 425 N, what distance will it have moved after the

force is first applied

1 answer:

lara31 [8.8K]3 years ago

3 0

Answer:

13.2m

Explanation:

Step one:

given data

Energy= 5610J

Force F= 425N

Required

The distance traveled

Step two:

We know that work done is given as

WD= force* distance

so

5610=425*d

divide both sides by 425

d= 5610/425

d=13.2m

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Answer:

The tension is $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

The distance between the two poles is $D =14 m$

The mass tied between the two cloth line is $m = 3Kg$

The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium

And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

$= tan^{-1} [\frac{1}{7}]$

$=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

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Rapid gas exchange can be accomplished more easily in air than in water because A. the O2 content of water is higher than that o

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Answer:

C. water is more dense and viscous

Explanation:

Rapid gas exchange can be accomplished more easily in air than in water because water is more dense and viscous.

Gases have the greatest ease of diffusion of their respective particles, as occurs in air, since their molecules have higher speeds and have more distance from each other than liquids.

The molecular diffusion rate in liquids is much less than in gases. The molecules of a liquid are very close (liquids are more dense and viscous) to each other compared to those of a gas, then the gas molecules hits with the molecules of the liquid with more frequency and this causes that the gas moves slower than in other gas (for example in air).

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A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

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Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

b.

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

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