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Igoryamba
3 years ago
14

Does friction always oppose relative motion?

Physics
1 answer:
Ilya [14]3 years ago
3 0

Answer:

Yes

Explanation:

Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other.

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An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,
maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is

43.8 N = <em>k</em> (0.155 m)   ==>   <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

5 0
3 years ago
Can you help me please the question says
bearhunter [10]

Answer:

p= 4 m/v

Explanation:

v=l*w*h

v=(25)(2)(3)

v=150

p=600/150

p=4

7 0
3 years ago
Which is an example of someone responding from a personal perspective?
Anika [276]

Answer:

C

Explanation:

If Ami is saying she likes it then it it personal. If you are speaking from statistics and studies it is impersonal and technically not from there perspective. All of these do this except C.

5 0
3 years ago
Read 2 more answers
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr
lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

4 0
3 years ago
Approximately how many kilograms (kg) of carnivore (secondary consumer) biomass can be supported by an area of a field containin
Anastasy [175]

Only 10 Kg of secondary consumers can be supported by the ecosystem.

There exists such a thing as the 10% rule in the transfer of energy in biomass from one trophic level to another. This rule ultimately controls the mass of biomass that is available at the next higher trophic level.

This rule state that only 10% of the energy in biomass is transferred to the next trophic level. By this rule, only 10% of the energy in the plant material is transferred to the carnivore population. Therefore, only 10 Kg of secondary consumers can be supported by the ecosystem.

Learn more about trophic level: brainly.com/question/13267087

7 0
2 years ago
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