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3 years ago

What affects the amount of electricity that can be generated through water power?

Physics

1 answer:

Andre45 [30]3 years ago

8 0

Water is composed of hydrogen and oxygen molecules the hydrogen and oxygen ions in water can be separated by passing an electric current through the water, which in turn will give it a temporary negative charge.

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What is defect of vision

Anuta_ua [19.1K]

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

4 0

3 years ago

A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is

labwork [276]

Answer:

$\mu_s=0.60$

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, $F=639\ N$

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

$F=\mu_smg$

$\mu_s=\dfrac{F}{mg}$

$\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}$

$\mu_s=0.60$

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.

5 0

3 years ago

Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc

Ira Lisetskai [31]

Answer:

0 < r < r_exterior B_total = $\frac{\mu_o I}{2\pi r}$

r > r_exterior B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

outside the inner cable

B₁ (2π r) = μ₀ I

B₁ = $\frac{\mu_o I}{2\pi r}$

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case r> r_exterior

B₂ = \frac{\mu_o I}{2\pi r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

B_total = B₁

B_total = $\frac{\mu_o I}{2\pi r}$

r > r_exterior

B_total = B₁ -B₂

B_total = 0

6 0

3 years ago

An ocean wave is an example of what type of wave

Tanzania [10]

Answer:

It is called a surface wave (rayleigh wave) that transmits its energy with the wind blowing onto its surface.Hope this helps

7 0

3 years ago

A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks

ipn [44]

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

3 0

3 years ago