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Nezavi [6.7K]
2 years ago
9

A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does an equally thick wire 6 m long, made

of the same material and under the same tension, stretch?
Physics
2 answers:
shusha [124]2 years ago
8 0

Answer:

2.4 mm

Explanation:

Given that:

Initial Original length of the wire L = 3 mm

The stretch  of the first wire  ΔL= 1. 2 mm

The length of the second wire L'' = 6 mm

The stretch of the second wire ΔL'' = ???

Considering the Tension of the system; the Young modulus and the cross sectional remains constant ; as such:

\frac{Y}{Y''} = \frac{FL}{A \Delta L} *\frac{A \Delta L''}{FL''}

1= \frac{L   \Delta L''}{L'' \Delta L}

\Delta L''= \frac{L'' \Delta L  }{L}

\Delta L''= \frac{6 \ m * 1.2 \ mm  }{3 \ m}

\Delta L''=2.4  \ mm

Thus, the same material under the same tension stretches 2.4 mm

Rainbow [258]2 years ago
6 0

Answer:

2.4 mm

Explanation:

Given that

Length of the wire, L = 3 m

Extensión of the wire, ΔL = 1.2 mm = 1.2*10^-3 m

Tensión of wire, T = 200 N

We use the formula

Y = TL/ΔLA

Since both wires material is same that makes the value of young's modulus the same in both the cases

hence equating

[200 * 3 / 1.2*10^-3 * A] = [200 * 6 / ΔL * A]

ΔL = 2.4*10^-3 m = 2.4 mm

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SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
3 years ago
How many millimeters are 120 meters ? Help ! Please &lt;3
blsea [12.9K]

Answer:

120,000

Explanation:

Millimeters to meters calculation-

Multiply by 1,000.

120 x 1,000 = 120,000.

This is the correct answer and formula.

Hope this helps!

8 0
3 years ago
Read 2 more answers
A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

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\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

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(c)

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e = 1.89 x 10^-7 V

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3 years ago
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joja [24]
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I hoped this was satisfying!:)</span>
3 0
3 years ago
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