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3 years ago

What do you know about water

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

It is vital for all known forms of life.
It provides no calories nor organic nutrients.
It forms precipitation in the form of rain and aerosols in the form of fog.
It's chemical symbol is H₂O

Explanation:

HOPE THAT HELPS

PLEASE MARK AS BRAINLIEST

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A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha

PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

$F_{e} = mg$

where

$F_{e} =$ the force between the electric field

m = mass

g = gravitational force

$F_{e} =qE$

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

$F_{e} =qE$

mg = qσ/2ε₀

σ = (2mg ε₀)/q

= 2 (0.016) (9.8) (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

= - 3.23 x 10⁻⁷ C/m²

8 0

4 years ago

Two large metal plates of area 0.7 m2 face each other. They are 5.1 cm apart and have equal but opposite charges on their inner

german

Explanation:

The given data is as follows.

Area (A) = 0.7 $m^{2}$ .

Electric field between the plates, (E) = 55 N/C

Since, electric field is related to surface charge density as follows.

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

Also, E = $\frac{\sigma}{\epsilon_{o}}$

or, $\sigma = \epsilon_{o} E$

Therefore, charge will be calculated as follows.

Q = $\sigma \times A = \epsilon_{o} E \times A$

= $8.854 \times 10^{-12} \times 55 \times 0.7$

= $3.41 \times 10^{-10} C$

= 341 pC

Thus, we can conclude that magnitude of the charge on each plate is 341 pC.

8 0

4 years ago

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump.

Effectus [21]

Answer: $2.32\times 10^{-3} kg$

Explanation:

Given

mass of first and second bullet $m_1=m_2=3.90\times 10^{-3} kg$

Velocity of two bullets $v_1=v_2=368 m/s$

velocity of third bullet $v_3=618 m/s$

angles between guns is $120^{\circ}$

Suppose First gun is at $0^{\circ}$ and second is at $120^{\circ}$ and third is at $240^{\circ}$

therefore

conserving momentum in x-direction

$m_1v_1\cos 0+m_2v_2\cos 120+m_3v_3\cos 240=0$

as three bullets club together to become lump

$3.90\times 10^{-3}\times 368+3.90\times 10^{-3}\times 368\times \cos (120)+m_3\times 618\times \cos (240)=0$

$3.90\times 10^{-3}\times 368+3.90\times 10^{-3}\times 368\times (-0.5)+m_3\times 618\times (-0.5)=0$

$0.5\times 3.90\times 10^{-3}\times 368=m_3\times 618\times 0.5$

$m_3=3.90\times 10^{-3}\times \frac{368}{618} kg$

$m_3=2.32\times 10^{-3} kg$

5 0

3 years ago

When you get into a car with hot black leather in the middle of the summer and it hurts your skin whe you sit this is an example

NeTakaya

C.the heat is conducted to your skin

3 0

3 years ago

Read 2 more answers

Two bulbs marked 200 W-250 V and 100 W-250 V are joined in

lozanna [386]

Answer:

Explanation:

<u>Joule Heating</u>

It's the process by which the electric current passing through a conductor produces heat.

Also known as Joule's first law or the Joule–Lenz law, states that the power of heating generated by an electrical conductor (P) is proportional to the product of its resistance (R) and the square of the current (I).

It can be described by the equation that follows:

$P = I^2.R$

Also, we can calculate the voltage V with the formula of Ohm's law:

$V = I.R$

Combining both equations, power can be related to the voltage:

$\displaystyle P=\frac{V^2}{R}$

Given the power and the voltage, the resistance can be calculated by solving for R:

$\displaystyle R=\frac{V^2}{P}$

There are two bulbs marked P=200W V=250V and P=100 W V=250.

The first bulb has a resistance of:

$\displaystyle R_1=\frac{250^2}{200}$

$\displaystyle R_1=312.5\Omega$

The first bulb has a resistance of:

$\displaystyle R_2=\frac{250^2}{100}$

$\displaystyle R_1=625\Omega$

When connected in series, the total resistance is

$R = R_1 + R_2=312.5\Omega+625\Omega$

$R=937.5\Omega$

The total power consumed when connecting them to a V=250 V supply is:

$\displaystyle P=\frac{250^2}{937.5}$

P = 66.67 W

4 0

3 years ago