What materials are difficult to charge because electrons easily flow through them?

From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of

alexandr1967 [171]

Answer:

The value is $h = 11930 \ m$

Explanation:

From the question we are told that

The temperature is $T = 300 \ K$

Generally the root mean square speed of the oxygen molecules is mathematically represented as

$v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }$

Here R is the gas constant with a value $R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}$

M is the molar mass of oxygen molecule with value $M = 0.032 \ kg /mol$

So

$\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h$

=> $h = 11930 \ m$

4 0

3 years ago

If a FM radio station broadcasts at 80. 3 MHz (megahertz), what is its wavelength in m (speed of light 3. 0 x 108 m/s)

hammer [34]

Answer:

Wavelength = 3.74 m

Explanation:

In order to find wavelength in "metres", we must first convert megahertz to hertz.

1 MHz = 1 × 10⁶ Hz

80.3 Mhz = <em>x</em>

<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz

The formula between wave speed, frequency and wavelength is:

v = fλ [where v is wave speed, f is frequency and λ is wavelength]

Reorganise the equation and make λ the subject.

λ = v ÷ f

λ = (3 × 10⁸) ÷ (8.03 × 10⁷)

λ = 3.74 m [rounded to 3 significant figures]

8 0

3 years ago

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, $T_i=4^{\circ} C$

Final temperature of water, $T_f=25^{\circ} C$

The specific heat capacity of liquid water is, $c=4.184\ J/g\ ^oC$

Heat transferred is given by :

$Q=mc(T_f-T_i)$

$Q=710\times 4.184\times (25-4)$

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0

3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0

3 years ago