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ICE Princess25 [194]
3 years ago
10

If the force applied to an object is not greater than the starting friction, what will happen to the object?

Physics
2 answers:
quester [9]3 years ago
6 0

Answer:

D the object will not move.

Explanation:

Friction force: It is that force which is applied in opposite direction of the external applied force.It is oppose the motion.

When external applied force is greater than the friction force then the object will move.

When external applied force is less than the friction force then the object will not move.

We are given that if the force applied to an object is not greater than the starting friction , then the object will not move.

Hence, option D is true.

Answer:D the object will not move.

bonufazy [111]3 years ago
3 0

Answer:

Explanation:

the object will not move as the force exerted is not sufficient enough to overcome its force of friction

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One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

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temperature at the cooler end, T_C=0^{\circ}C

length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

mass of ice melted at zero degree Celsius, m=8.7\times 10^{-3}\,kg

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thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

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dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

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8 0
3 years ago
Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4
Zepler [3.9K]

Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

7 0
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