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Blababa [14]
3 years ago
9

A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete

rmine that the Slinky has an inductance 130 µH, when it has been stretched to a length of 3 m. The permeability of free space is 4π × 10−7 N/A 2 . If a slinky has a radius of 4 cm, what is the total number of turns in the Slinky?
Physics
1 answer:
anzhelika [568]3 years ago
8 0

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

L = \frac{\mu _{o}N^{2}A}{l}

130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}

N = 248

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Is an electron cloud bigger than an atom?
Rashid [163]

Answer:

no

Explanation:

An electron is one of the components of an atom, so it cannot be larger than that.

5 0
2 years ago
A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse
NISA [10]

Answer:

The time interval is t  =  5.48 *10^{-3} \ s

Explanation:

From the question we are told that

   The length of the string is  l  =  3.00 \ m

    The  mass of the string is m  =  5.00 \ g  =  5.0 *10^{-3}\ kg

     The  tension on the string is  T  =  500 \ N

   

The  velocity of the pulse is mathematically represented as

      v  = \sqrt{ \frac{T}{\mu } }

Where \mu is the linear density which is mathematically evaluated as

       \mu  =  \frac{m}{l}

substituting values

     \mu  =  \frac{5.0 *10^{-3}}{3}

     \mu  = 1.67 *10^{-3} \  kg /m

Thus  

     v = \sqrt{\frac{500}{1.67 *10^{-3}} }

    v = 547.7 m/s

The time taken is evaluated as

    t  =  \frac{d}{v}

substituting values

      t  =  \frac{3}{547.7}

      t  =  5.48 *10^{-3} \ s

5 0
4 years ago
The electric current leaves the battery through the --- complete this blank
aev [14]
The bulb

Hope this helps!
5 0
4 years ago
If the same amount of force was applied to an object that was twice the mass,​
Olegator [25]

Answer:

the object with greater mass will experience a smaller acceleration and the object with less mass will experience a greater acceleration.

Explanation:

According to Newton's second law, when the same force is applied to two objects of different masses, the two objects will change mass.

4 0
3 years ago
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