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3 years ago

Elements with positive valences usually ______ electrons

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

The answer is donate, therefore elements with positive valences usually donate electrons

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3 years ago

a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a

Whitepunk [10]

Answer:

$W = -2.76\times 10^{-9}~J$

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

$U = \frac{1}{2}CV^2$

where C can be calculated using the plate separation and area.

$C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon$

Therefore, the potential energy in the first case is

$U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon$

In the second case:

$C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon$

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

$W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J$

6 0

3 years ago

Please help me people

saveliy_v [14]

Answer :

(1) The density of asphalt is, $1200kg/m^3$

(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.

(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

Explanation :

Part 1 :

As we are given:

Mass of block = 90 kg

Volume of block = $0.075m^3$

Formula used :

$\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}$

Now put all the given values in this formula, we get:

$\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3$

Thus, the density of asphalt is, $1200kg/m^3$

Part 2(a) :

As we are given that:

Length of aluminium sheet = 35 cm

Width of aluminium sheet = 11 dm

Thickness of aluminium sheet = 15 mm

Now we have to convert these dimensions into meters.

Conversions used:

1 cm = 0.01 m

1 dm = 0.1 m

1 mm = 0.001 m

Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m

Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m

Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m

Part 2(b) :

First we have to calculate the volume of aluminium sheet.

Volume of aluminum sheet (cuboid) = Length × Width × Thickness

Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m

Volume of aluminum sheet (cuboid) = 0.005775 m³

Now we have to calculate the mass of aluminium sheet.

$\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}$

$2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}$

$\text{Mass of aluminium}=15.59kg$

Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

7 0

3 years ago

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