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dolphi86 [110]
3 years ago
15

A metal wire of resistance R is cut into five equal pieces that are then placed together side by side to form a new cable with a

length equal to one-fifth the original length. What is the resistance of this new cable
Physics
1 answer:
Alina [70]3 years ago
8 0

Answer:

R / 25

Explanation:

Let the initial length of the wire is L and its resistance is R.

Now the length is cut into 5 equal parts.

The resistance of a wire is directly proportional to the length of the wire, thus, the resistance of each portion of the wire is R/5.

Now the five pieces are connected in parallel combination, the equivalent resistance is

\frac{1}{R_{eq}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}

R_{eq}=\frac{R}{25}

Thus, the resistance of the new cable is R / 25.

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A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

W=0.31 J

4 0
3 years ago
Read 2 more answers
If an isotope has 8 protons and 10 neutrons in it's nucleus, what is it's atomic mass?
viva [34]
The atomic mass is always equal to the sum of protons and neutrons in the nucleus. If you add the number of protons and neutrons (8 + 10) = 18 you will find that the atomic mass is 18.
8 0
4 years ago
If a car go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/
Taya2010 [7]

Answer:

v = 87.57 m/s

Explanation:

Given,

The initial velocity of the car, u = 0

The final velocity of the car, v = 60 mi/hr

The time period of car, t = 8 s

                                         = 0.00222 hr

The acceleration of the car is given by the formula,

                                       a = (v -u) / t

                                           = 60 / 0.00222

                                            = 27027 mi/hr²

If the car has initial velocity, u = 50 mi/hr

The time period of the car, t = 5.0 s

                                         = 0.00139 hr

Using first equations of motion

                                      <em> v = u + at</em>

                                          = 50 + (0.00139 x 27027)

                                          = 87.57 mi/hr

Hence, the final velocity of the car, v = 87.57 mi/hr

3 0
3 years ago
Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice
OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0
3 years ago
1. What are the two factors affecting friction
bija089 [108]
Answer: a.) Roughness of the surfaces in contact with each other .
Higher the roughness of surfaces in contact with each other, greater is the friction between bodies. Force of friction will be less between smooth surfaces.

b.) Weight of the sliding/rolling body: greater the weight of the moving body on the surface, more is the force of friction on the body by the surface.

I hope this helps

5 0
2 years ago
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