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2 years ago

15

A compound consisting of Cr3+ ions and OH ions would be named chromium (III) hydroxide

2 answers:

lianna [129]2 years ago

6 0

The answer here is indeed true, I hope this helps

Vedmedyk [2.9K]2 years ago

3 0

The answer to ya question is true

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which statement best describes what would happen if the number of wore coils in a electromagnavneg were increased

jekas [21]

Where are the following choices

4 0

2 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

2 years ago

Why can’t a real machine ever have 100% efficiency

Harman [31]

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

4 0

2 years ago

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

3 0

2 years ago

Cells work ALONE and do not depend on each other to function. True or False? hELPpPP plSSSS

k0ka [10]

Answer:

False

Explanation:

3 0

2 years ago

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