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3 years ago

8)

2 answers:

7 0

Hey!!

here is your answer >>>

We know that, acids and strong bases are good conductors of electricity!. And gold conducts electricity , while the best is silver and niext comes gold!. And H20 which is water contains less ions , so it is not a good conductors , it is a conductors but not a good conductor!. Salt in it's solid for cannot produce ions, hence, it cannot conduct electricity , but when this salt is mixed with water or when it is in it's molten state it can conduct electricity!. So now,the answer for the question is ,

B) III and IV only!.

Hope my answer helps!

Sergio039 [100]3 years ago

7 0

The correct answer here is B) III IV only

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You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving

Alecsey [184]

<h2>The balloon is moving when it is halfway down the building at 20.78 m/s.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 0.5 x 44 = 22 m

Substituting

v² = u² + 2as

v² = 0² + 2 x 9.81 x 22

v² = 431.64

v = 20.78 m/s

Velocity at 22 m = 20.78 m/s

The balloon is moving when it is halfway down the building at 20.78 m/s.

7 0

3 years ago

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid $E=9\times 10^{-16}J$

We have to find the current in the toroid

Magnetic energy stored is equal to $E=\frac{1}{2}Li^2$

$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

$I=17.32\times 10^{-3}A$

So current in the toroid will be $I=17.32\times 10^{-3}A$

5 0

3 years ago

A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili

lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

* Bimetallic brass / steel tape

* Initial temperature, room temperature T = 20ºC

* Final temperature, boiling water = 100ºC

* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

* perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass α_{brass} = 19 10⁻⁶ ºC⁻¹

Steel α_{steel} = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass L_{f \ brass} =θ r₁

steel L_{f \ steel} = θ r₂

we substitute in our equations

θ r₁ - L₀ = α_{brass} L₀ ΔT

θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

e = r₁ -r₂

θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

tan θ = Δy / L₀

Δy = L₀ tan θ

Δy = 22.2 tan 0.155

Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0

3 years ago

At what height does a 1000-kg mass have a potential energy of 1J relative to the ground? Be sure to show all work to support you

nikdorinn [45]

Potential energy = (mass) x (gravity) x (height)

1 joule = (1,000 kg) x (9.8 m/s²) x (height)

Height = 1 joule / (9,800 newtons)

= 1/9800 meter

= 0.000102 meter

= 0.102 millimeter (rounded)

7 0

3 years ago

An Airplane taxis onto the runway going at 10 m/s. If it can accelerate steady at 3

kirza4 [7]

Answer: 1333 m

the length of runway it will need is S = $\frac{90^{2}-10^{2} }{2.3}=1333$ (m)

Explanation:

4 0

3 years ago