An object with density of 0.8g/ml will float in water

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

3 years ago

Raina is running for student body president and doesn't want to forget her campaign speech. She practices her speech over and ov

mezya [45]

The answer is intentional.

6 0

3 years ago

Read 2 more answers

A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days. How much energy is used? A laptop uses 75 W. The cot of el

zzz [600]

Answer:

See the answers below.

Explanation:

We will solve this problem by calculating each part separately.

A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days.

Energy can be calculated by multiplying the value of the power of the equipment by the amount of time of use.

$500 [W]*[\frac{6min}{1day} ]*[\frac{1day}{24hr} ]*[\frac{1hr}{60min} ]=2.083 [W]$

The cots of electricity is 5.6 cents per kWh. How much would it cost to operate the laptop for 24 hours a day for one week?

We know that the power of the latop is 75 [W], then we can calculate the cost, multiplying the value of the power by the value of the cost by the time of use of the computer.

$0.075[kW]*5.6[\frac{cents}{kw*h}}]*[\frac{24hr}{1day}]*[1week]*[\frac{7days}{1week} ]=70.56 [cents]$

A toaster oven is 85% efficient. It uses 1200 J of energy. How much thermal energy is it producing?

Efficiency is defined as the relationship between the energy obtained on the energy delivered. Almost always the energy delivered is greater than the energy obtained (first law of thermodynamics).

Therefore.

$Effic = E_{obtained}/E_{delivered}\\0.85=E_{obtained}/1200\\E_{obtained}=1020[J]$

4 0

3 years ago

I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s

masha68 [24]

20/9.8 = 2.0 seconds. The ball stops after 2 seconds.

5 0

3 years ago