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son4ous [18]
4 years ago
15

Define work done against friction(helpp)​

Physics
1 answer:
77julia77 [94]4 years ago
7 0

Answer:

The work done against friction is the work done on an object that overcomes this frictional force allowing the object to move - it doesn't include any extra work used to accelerate an object for example - only the work used to beat the frictional force.

Explanation:

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Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of
liubo4ka [24]

Explanation:

hope this will help u

5 0
3 years ago
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Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At lef
Sladkaya [172]

Question:

Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. In which direction should an arrow at position 1 point?

left

right

up

down

Answer:

The correct answer is

Left

Explanation:

Magnetic circuits describe the path of a magnetic flux. In the same way electricity follows a complete closed circuit, the path of a magnetic flux is also a complete and closed circuit which leaves from the N pole, migrates through the air  and reenters the magnet through the S pole through which it passes back into the magnet to come to the N pole again.

As such the magnetic field lines emanate from the N pole which is on he right to the S pole which is on the left. Hence the arrow should point in the left direction.

3 0
3 years ago
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How do I find time?
kari74 [83]

<u>Answer:</u>   3

H = 45 m ,

time required for the stone

  S = u.t+(1/2) g.t²

45  = (1/2) 9.81. t²      since u( initial velocity) = 0 ; S =H.

 t²  = 9.17 s

<em>  t = 3.02 s  </em>

therefore answer is 3 m/s

<em>  if you satisfy with explanation, please give brainiest ; </em>



6 0
3 years ago
A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di
Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

F . t = m(v_f - v_i)

so here we know that object is at rest initially so we have

80 (12) = 80( v_f - 0)

v_f = 12 m/s

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

F = 80 - \frac{40}{3} t

now again by same equation we have

\int F .dt = m(v_f - v_i)

\int (80 - (40/3)t) dt = 80(v_f - 12)

80 t - \frac{40t^2}{6} = 80(v_f - 12)

put t = 6 s

480 - 240 = 80(v_f - 12)

v_f = 12 + 3

v_f = 15 m/s

6 0
3 years ago
An engineer is designing a runway for an airport. Of the planes that will
ziro4ka [17]

Answer:

704m

Explanation:

Given, initial velocity, u=0m/s and final velocity v=65m/s

Acceleration a=3m/s 2

If d be the allowed length for the runway.

Using formula v*2 −u*2 =2aS

(65)*2−0*2 =2(3)d

⟹ d=65*2 /6=704.16∼704m

8 0
3 years ago
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