When capacitors are connected in parallel, they have the same:.

What are the relative ages of the features in order of oldest to youngest?

yuradex [85]

Answer:

Layer 1, Rock 2, Rock 1, Fault

8 0

3 years ago

Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

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3 0

2 years ago

150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit

zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

$I=\frac{150}{30}=5 A$

Learn more about current:

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6 0

3 years ago

The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he

VashaNatasha [74]

Answer:

dV/dt = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

$V = \pi r^{2}h$

put r = h/2 so,

$V = \pi \frac{h^{3}}{4}$

Differentiate both sides with respect to t.

$\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}$

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

$\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3$

dV/dt = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0

3 years ago

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