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3 years ago

List five (5) compounds h2o can form (10 points)

Physics

1 answer:

Elena-2011 [213]3 years ago

8 0

Oxygen, hydrogen, sodium, chlorine, lead, iron. Hope this helps!!!

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What are the agency that monitors the tectonic phenomena?

matrenka [14]

Answer:

ASDASDSAD

Explanation:

ASDSADASDSADASD

8 0

3 years ago

A 1.0-kg object moving in a certain direction has a kinetic energy of 2.0 J. It hits a wall and comes back with half its origina

saw5 [17]

Lets calculate the initial speed of the block. We know that kinetic energy is given by:

$K_0 =\dfrac{mv_0^2}{2}$

Solving for v₀:

$v_0 = \sqrt{ \dfrac{2K_0}{m}} = \sqrt{ \dfrac{2(2\;J)}{1\;kg}} = 2\;m/s$

If the speed after it hits a wall is half its original speed then:

v = v₀/2 = 2 m/s / 2 = 1 m/s

Then the kinetic energy at this point is:

$K =\dfrac{mv^2}{2} = \dfrac{(1\;kg)(1\;m/s)^2}{2} = \dfrac{1}{2}\;J = 0.5\;J$

The inetic energy of this object at this point is 0.5 J.

5 0

3 years ago

A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a

AysviL [449]

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

$EMF = \frac{\phi_2 - \phi_1}{\Delta t}$

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

$\phi_1 = NBAcos 0 = BA$

$\phi_2 = NBAcos 90 = 0$

now we will have

$EMF = \frac{NBA}{t}$

$10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}$

$B = 0.126 T$

3 0

3 years ago

The cells of collenchyma have at the corners

saw5 [17]

Explanation:

cellulose and pectin are present on the cells of collenchyma...

8 0

3 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago