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4 years ago

This element is added to table salt to prevent the development of goiter

Physics

2 answers:

mestny [16]4 years ago

8 0

Iodine is added there

Gemiola [76]4 years ago

5 0

Iodine is added to salt to prevent goiter

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What force is needed to bring a 1.10 × 10^3 kg car moving at 22 m/s to a halt in 20s?

Tems11 [23]

Answer:

A very very big hole and some ramps

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3 years ago

A cyclist and his bicycle have a combined mass of 88 kg and a combined

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4 years ago

swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the

Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

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3 years ago

A proton in the nucleus of an atom has an electrical charge of:<br> neutral<br> -<br> +<br> zero

dlinn [17]

Answer:

proton is positively charged changechar

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3 years ago

Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.

erma4kov [3.2K]

Answer:

The extent to which it would stretch is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

The initial length is $L = 1.00m$

The area is $A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2$

The Young modulus of the steel is $Y = 2.0*10^{11} Pa$

The tension is $T =1500 N$

The Young modulus is mathematically represented as

$Y = \frac{\sigma}{e}$

Where $\sigma$ is the stress which is mathematically represented as

$\sigma = \frac{F}{A}$

Substituting values

$\sigma = \frac{1500}{0.500*10^{-6}}$

$\sigma = 3.0*10^9 N/m^2$

And e is the strain which is mathematically represented as

$e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

$Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values

$2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

$\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

$\Delta L = 0.015 \ m$

7 0

3 years ago