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victus00 [196]
3 years ago
10

The three pairs of metal same-size spheres have different charges on their surfaces, as indicated. Each pair is brought together

, allowed to touch, and then separated. Rank from greatest to least the total amount of charge on the pairs of spheres after separation.
A) +6 +2
B) +6 -2
C) +6 0
Physics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

A≥C>B

Explanation:

The total charge on the spheres depends on the pair attached. considering pair (A), the pair of +6 and +2 will produce a net charge +8 on pair A. Pair (B) has one positive and negative sphere of + 6 charge and other having -2 charge when brought near the net charge becomes +4. Pair C has a sphere having +6 charge and other is neutral i.e 0 charge when brought near the net charge on the pair becomes +6. In terms of ranking, pair A is the highest while pair B is the lowest.

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Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
Questions 15 out of 20
Olin [163]

Answer:

D

Explanation:

7 0
3 years ago
A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

3 0
3 years ago
1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.
Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] =  11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

7 0
3 years ago
Three long, straight wires are carrying currents that have the same magnitude. In C the current is opposite to that in A and B.
Nadusha1986 [10]

Answer:

(b) B

Explanation:

The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire

On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.

on wires A and C the forces (due to B and C and A and B

respectively) act in opposite directions and therefore tend to cancel out.

5 0
4 years ago
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