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3 years ago

8

A glass slides across a bar and slows down due to a kinetic friction of 0. 175N. If the glass weighs 0. 500N, what is the coeffi

cient of kinetic friction between the glass and the bar?

1 answer:

Juli2301 [7.4K]3 years ago

6 0

Answer:

F = M a = Ff force of friction

Ff = μ M g = μ W where M g = weight of glass

The problem gives the kinetic friction as a force (Ff) in Newtons

μ = Ff / W = .175 / .500 = .35

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Answer:

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b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

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$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

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$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

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c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

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4 years ago

An ion has 8 protons and 10 electrons. What is the total charge? *

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The answer is -2 sjs

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3 years ago

Your bedroom has a rectangular shape and you want to measure its size. You use a tape that is precise to 0.001 mm and find that

avanturin [10]

This question is in two parts. This is not the correct multiple choice options for this part a.

The second part had the option

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c)31.37 m^2

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Answer:

A. 17.0 m²

B. 31.4 m²

Explanation:

The formula for the calculation of the area of a rectangle is given as

Area = length x width

The length = 3.547 m

The width = 4.79 m

Then area = 3.547 x 4.79

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This is the most precise measurement for the area of the bedroom.

B.

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= 3.14(6.32/2)²

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3 years ago

The force required to drag several objects across a floor was measured. Calculate the µK for these objects.

olganol [36]

Answer:

Explanation:

Case-A

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$\mu _k=\frac{f}{N}$

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$\mu _k=0.25$

Case-C

Block weight $W=1500\ lb$

Friction Force $f=60\ lb$

$\mu _k=\frac{f}{N}$

$\mu _k=\frac{60}{1500}$

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Answer:

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<em>b) 3344 N</em>

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

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Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = <em>3344 N</em>

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is <em>3344 N </em>

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3 years ago