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3 years ago

Name the following ionic bonds.

Physics

2 answers:

Gennadij [26K]3 years ago

7 0

D !.!.!.!.!.!!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!!..!!..!

Svetllana [295]3 years ago

4 0

A.Ionic bond
B.Ionic bond
C.Ionic bond
D.Ionic bond

An ionic bond is between a metal and non metal

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The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

mihalych1998 [28]

Answer:

a) Ra = 0.517 Ω

Rb = 0.032 Ω

Rc = 0.129 Ω

b) Ia = 5.8A

Ib = 93.75A

Ic = 23.2 A

Explanation:

a) The resistance is equal to:

Resistance for case a:

$R_{a} =\frac{pL_{a} }{A_{a} } =\frac{p*4*L_{0} }{2L_{0}*L_{0} } =\frac{2p}{L_{0} }$

Where

p = 1.5x10⁻²Ωm

L0 = 5.8 cm = 0.058 m

$R_{a} =\frac{2*1.5x10^{-2} }{0.058} =0.517ohm$

Resistance for case b:

$R_{b} =\frac{pL_{b} }{A_{b} } =\frac{pL_{0}}{2L_{0}4L_{0} } =\frac{p}{8L_{0}} =\frac{1.5x10^{-2} }{8*0.058} =0.032ohm$

Resistance for case c:

$R_{c} =\frac{pL_{c}}{A_{c} } =\frac{p2L_{0}}{L_{0}4L_{0}} =\frac{p}{2L_{0}} =\frac{1.5x10^{-2} }{2*0.058} =0.129ohm$

b) The current is equal to:

Current for case a:

$I_{a} =\frac{V}{R_{a} } =\frac{3}{0.517} =5.8A$

Current for case b:

$I_{b} =\frac{V}{R_{b} } =\frac{3}{0.032} =93.75A$

Current for case c:

$I_{c} =\frac{V}{R_{c} } =\frac{3}{0.129} =23.2A$

4 0

3 years ago

A car is traveling to the right with a speed of 2.0 m/s on an icy road when the brakes are applied. The car begins sliding with

quester [9]

Answer:

t = 3.0s

Explanation:

U = 2.0m/s , V = 0 (stop) , S = 3m , t =?

From V^2 - U^2 = 2aS

=) a = -4/6 = -0.667m/s^2

Now again by V-U = at

We have t = -U/a = 2/0.667 = 3s

Required time is t = 3.0s

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3 years ago

Calculate the (absolute) pressure at the bottom of a neighborhood swimming pool 30.0 m by 8.0 m whose uniform depth is 2.0 m. Th

Alika [10]

Answer:

Total pressure= 120945[Pa]

Force exerted = 29026800 [N] or 29.02*10^6 [N]

Explanation:

We know that the total pressure is the result of the sum of the atmospheric pressure plus the manometric pressure. The equation is:

$Ptotal=Patm + Pman$

In this problem we know the atmospheric pressure 101.325x10^3 [Pa], therefore we need to find the manometric pressure.

The manometric pressure in the bottom of the swimming pool depends only on the water column of water generated (depth of the swimming pool)

$Pman = density*g*h$

where:

density = density of the water 1000 [kg/m^3]

g= gravity [m/s^2]

h= column of water (meters)

replacing the values:

$Pman= 1000 *9.81* 2 = 19620 [Pa]\\\\$

The total pressure will be:

$Ptotal= 101325+19620 = 120945 [Pa]\\\\$

The force exerte on the bottom is defined by the following expression:

$Pressure=Force/area\\\\Force= Pressure*Area\\\\Area = 30m*8m= 240 m^2Force= 120945*240\\Force= 29026800N or 2958 Ton$

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4 years ago

A cyclical motion occurs because of density differences in the mantle. Heated, less dense lower regions of the fluid mantle rise

Marta_Voda [28]

The answer for this is B

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2.15 Un móvil en trayectoria rectilinea ocupa las posiciones indicadas en la tabla para cada minuto:

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To make sure I don’t has 35 012345 bishops

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