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Grace [21]
3 years ago
12

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.542 and t

he coefficient of kinetic friction is 0.294. How much force is needed to begin to move the book and how much force is needed to keep the book moving at a constant velocity
Physics
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

Force is needed to begin to move the book and  force needed to keep the book moving at a constant velocity is 10.78 N and 5.85 N.

Explanation:

Given :

Mass of book , M = 2.03 kg.

Coefficient of static friction , \mu_s=0.542 \ .

Coefficient of kinetic friction , \mu_k=0.294\ .

Force, required needed to begin to move the book ,

F=\mu_sN=\mu_s(mg)=0.542\times 2.03\times 9.8=10.78\ N.

Now, We know kinetic friction acts when object is in motion .

Therefore , Force, required o keep the book moving at a constant velocity

F=\mu_kN=\mu_k(mg)=0.294\times 2.03\times 9.8=5.85\ N.

Hence, this is the required solution.

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2. If you were trying to determine if cars in your neighborhood were travelling at or below the posted speed limit, how would yo
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4 years ago
While at a construction site, you see a crane lift a 1000kg steel beam up to a height of 10m in a time period of 5.0 seconds. A
storchak [24]

Answer:

Car has more power output than crane      

Explanation:

We have given that mass of the crane m = 1000 kg

Height through which crane lift the steel beam h = 10 m

Acceleration due to gravity g=9.8m/sec^2

So work done by crane W=mgh=1000\times 9.8\times 10=98000j

Time period is given as t = 5 sec

We know that power P=\frac{W}{t}=\frac{98000}{5}=19600watt

Now mass of the car = 1000 kg

Initial velocity u = 0 m /sec

Final velocity v = 10 m/sec

We know that work done is equal to the change in kinetic energy

So work done =\frac{1}{2}mv^2-\frac{1}{2}mu^2

=\frac{1}{2}\times 1000\times 10^2-\frac{1}{2}\times 1000\times 0^2=50000j

Time ids given as t = 2 sec

So power P=\frac{W}{t}=\frac{50000}{2}=25000watt

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3 years ago
In a certain process, the energy of the system decreases by 250 kJ. The process involves 480 kJ of work done on the system. Find
gregori [183]

Answer:

Q = - 730 KJ

730 KJ is transferred out of the system

Explanation:

According to the first law of thermodynamics, energy can neither be created nor destroyed, but can be transformed from one form to another.

For a particular process/system, the first law is interpreted as

ΔU = Q + W (depending on convention, some textbooks give this relation as ΔU = Q - W)

ΔU is the change in internal energy of the system, in my convention, it is positive if the internal energy increases and negative otherwise.

Q = Heat transferred into or out of the system. Q is positive for heat transferred into the system and negative for heat transferred out of the system.

W = workdone by the system or work done on the system. W is positive for workdone on the system and W is negative for when work is done by the system.

ΔU decreases by 250 KJ, that is, ΔU = - 250 KJ

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Q = - 250 - 480 = - 730 KJ

The heat is transferred out of the system.

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