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2 years ago

5

What is the specific heat of silver if the temp ig a 15.4g sample of silver is increased frin 20 degrees celsius to 31.2 degrees

celsius when 40.5 Joules of heat is added?

1 answer:

Genrish500 [490]2 years ago

4 0

heat = mass x spec heat x temp rise

40.5=15.4x10^-3xspec heatx11.2

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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

katrin2010 [14]

Answer:

b

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e

h

Explanation:

Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both

The question show something in common ...acceleration

So let's look at the statements and pick the correct ones

a is false while b is correct as the train is accelerating

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2 years ago

A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

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Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

Explanation:

We have given mass of the crate m = 49.6 kg

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Coefficient of kinetic friction $\mu _k=0.399$

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So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

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2 years ago

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Answer:

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2 years ago

What is the mass of an object if a force of 2N and gives it an acceleration of 2 m/s2

tino4ka555 [31]

Answer:

1kg

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2 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

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$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

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$=0.645m/s^2$

B)

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using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

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2 years ago