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3 years ago

Which statement is true regarding a chemical reaction?

Physics

1 answer:

liberstina [14]3 years ago

7 0

Answer:

The total number of atoms does not change, so mass is conserved in the reaction.

Explanation:

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A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid

igor_vitrenko [27]

Answer:

a) attractiva, b) dF = $k \frac{Q_1 \ dQ_2}{dx}$ , c) F = $k Q_1 \frac{Q_2}{d \ (d+L)}$ , d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

dF = $k \frac{Q_1 \ dQ_2}{dx}$

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

∫ dF = $k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2$

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

λ = dQ₂ / dx

DQ₂ = λ dx

we substitute

F = $k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}$

F = k Q1 λ ( $-\frac{1}{x}$ )

we evaluate the integral

F = k Q₁ λ $(- \frac{1}{d+L} + \frac{1}{d} )$

F = k Q₁ λ $( \frac{L}{d \ (d+L)})$

we change the linear density by its value

λ = Q2 / L

F = $k Q_1 \frac{Q_2}{d \ (d+L)}$

d) we calculate the magnitude of F

F =9 10⁹ (-4.2 10⁻⁶) $\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}$

F = -1.09 N

the sign indicates that the force is attractive

3 0

3 years ago

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

alexgriva [62]

Answer:

a) I = 1.44 Kg.m²

b) I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.

Explanation:

Given

The skater's mass: M = 64 Kg

Mass of both arms: m = (M/8) = 64 Kg / 2 = 8 Kg

Mass of one arm: = marm = m/2 = 8 Kg / 2 = 4 Kg

Mass of her rest body: Mcyl = M - 2m = 64 Kg - 8 Kg = 56 Kg

rarm = 0.20 m

Rcyl = 0.20 m

Larm = 0.60 m

a) We apply the equation (Using the Steiner Theorem)

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.20 m)²

I = 1.44 Kg.m²

b) Suppose that the center of the mass of her outstretched arm is in the middle, so that the mass of the arm will be at a distance of 50 cm, then

rarm = 0.50 m

We apply the same equation, then

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.50 m)²

I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.

8 0

3 years ago

In a circuit what do lightbulbs and a toaster represent

dolphi86 [110]

Answer:

RESISTANCE

Explanation:

5 0

4 years ago

Read 2 more answers

What are radio isotopes

sweet [91]

Radioisotopes is a atom with unstable nuclei. They also decay to change different from the electrons over time.
Hope it helped you.
-Charlie

6 0

4 years ago

For a given centrifugal pump, if the speed of rotation of the impeller is cut in half, how does the power required to drive the

Andru [333]

Answer:

The power required to drive the pump is reduced in $\frac{1}{8}$ .

Explanation:

From Turbomachinery, we find that power required to drive the pump ( $\dot W$ ) is directly proportional to the cube of the speed of rotation of the impeller ( $\dot n$ ). That is:

$\dot W \propto \dot n^{3}$

$\dot W = k\cdot \dot n ^{3}$ (Eq. 1)

Where $k$ is the proportionality constant.

And now we eliminate the proportionality constant by constructing the following relationship:

$\frac{\dot W_{B}}{\dot W_{A}} = \left(\frac{\dot n_{B}}{\dot n_{A}}\right)^{3}$ (Eq. 2)

If we know that $\frac{\dot n_{B}}{\dot n_{A}} = \frac{1}{2}$ , then:

$\frac{\dot W_{B}}{\dot W_{A}} = \frac{1}{8}$

The power required to drive the pump is reduced in $\frac{1}{8}$ .

4 0

3 years ago