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3 years ago

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On a frozen pond, a 8.54-kg sled is given a kick that imparts to it an initial speed of ð£0=1.87 m/s. The coefficient of kinetic

friction between sled and ice is ðð=0.087. Use work and energy concepts to find the distance the sled moves before coming to rest.

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:2.05 m

Explanation:

Given

Mass of sled $m=8.54 kg$

initial speed $u=0.187 m/s$

coefficient of kinetic Friction $\mu _k=0.087$

According to work-energy theorem work done by all the forces is change in Kinetic energy

$W_{friction}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$\mu _kmg\times d=\frac{1}{2}\times 8.54\times 1.87^2$ ,where d= distance moved

$0.087\times 8.54\times 9.8\times d=14.93$

$d=\frac{14.93}{7.281}$

$d=2.05 m$

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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

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Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l

goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

$\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}$

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

$\Delta t = \frac{0.18\ m}{340\ m/s}$

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

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The apparent backward motion of a planet along the celestial sphere is called ____________ motion.

motikmotik

Answer:

Retrograde

Explanation:

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A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = $\frac{a}{g}$

μ = $\frac{1.25}{9.8}$

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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Select the properties that apply to superconductors.

makkiz [27]

made from pure metals . . . no;
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conduct electricity with zero resistance . . . yes;
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just like any other conductor, the magnetic field depends on the current that's flowing in the conductor.

no loss of energy in the transfer of electricity . . .
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something else, then there would be some loss.

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