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4 years ago

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

Physics

1 answer:

olganol [36]4 years ago

4 0

Answer:

Explanation:

Case-A

Block weight $W=150\ lb$

Friction Force $f=30\ lb$

and friction force is

$f=\mu _kN$

where N=normal reaction (Usually equals to weight)

$\mu _k=\frac{30}{150}$

$\mu _K=\frac{1}{5}=0.2$

Case-B

Block weight $W=20\ N$

Friction Force $f=5\ N$

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

$\mu _k=\frac{5}{20}=\frac{1}{4}$

$\mu _k=0.25$

Case-C

Block weight $W=1500\ lb$

Friction Force $f=60\ lb$

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

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

$\mu _k=0.04$

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Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 181 ms. How

lara [203]

Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

First, we must find the distance at which the objects are in time t = 360s

We will use the formula for vertical distance in free fall

$h=v_{0}t+\frac{1}{2} gt^2$

$v_{0}$ is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time ( $g=9.81m/s^2$ )

At $t=360s$ the first stone has been falling for the entire 360 seconds, its position h1 is:

$h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m$

And at 360 seconds the second stone has been fallin for $t= 360s -181 s = 179s$ , so its position h2 is:

$h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m$

And finally using the equation for the center of mass:

$CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}$

We know that the mass of the second stone is twice the mass of the first stone so:

$m_{1}=m\\m_{2}=2m$

replacing these values in the equation for the center of mass

$CM=\frac{mh_{1}+2mh_{2}}{m+2m}$

$CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}$

Finally, replacing the values we found fot h1 and h2:

$CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m$

the center of mass is 316,670m bellow the release point.

8 0

4 years ago

1 hr 16 minutes equal how many minutes ?

Len [333]

Answer:

76 minutes.

Explanation:

There are 60 minutes in an hour + 16 minutes= 76 minutes.

4 0

3 years ago

The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 84 MW at a cruising speed of 35.0 kn

frosja888 [35]

Answer:

$F = 4669.201\,kN$

Explanation:

Maximum speed is get at maximum power. Let assume that ship travels at constant speed, the expression for power is equal to:

$\dot W = F\cdot v$

Where $F$ and $v$ are the forward force and speed of ship measured in newtons and meters per second, respectively.

The forward force can be determined by clearing it in the expression described above:

$F = \frac{\dot W}{v}$

$F = \frac{84\times 10^{3}\,kW}{(35\,knots)\cdot \left(\frac{0.514\,\frac{m}{s} }{1\,knot} \right)}$

$F = 4669.201\,kN$

4 0

3 years ago

Read 2 more answers

Two positive point charges of 12uc and Suc

GarryVolchara [31]

Answer:

54 N

Explanation:

We have two positive charges 12uC and 5 uC ,they are kept at a distance of 10cm.

We have a relation for force between two charges q1,q2 as

$F=\frac{kq1q2}{r^{2} }$

Value of k is $9*10^9$

On substituting the values into the equation we get,

$F=\frac{9*10^9*12*5}{10^9*10}\\\\ F=54N$

Hence the force between them is 54 N.

7 0

3 years ago