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3 years ago

Akia is balancing the equation Na + H2O NaOH + H2. He tries to find the coefficients that will balance the equation.

Physics

2 answers:

Pepsi [2]3 years ago

5 0

<h3>Answer;</h3>

A: by counting up each individual atom and make sure the atom numbers are the same in the reactants and the products

<h3>Explanation;</h3>

According to the law of conservation of mass, the mass of reactants should always be the same as the mass of the products in a chemical equation. Therefore, the number of atoms of each element in a chemical equation should always be the same on both sides of the equation, that is the side of reactants and side of products.

Thus, any chemical equation requires balancing to ensure that the number of atoms of each element is equal in both sides of the equation. Balancing is a try and error process that ensures that the law of conservation of mass holds.

In this case, the balanced equation would be; 2Na +2H₂O → 2NaOH +H₂

ycow [4]3 years ago

3 0

Answer : The correct option is, (A) by counting up each individual atom and make sure the atom numbers are the same in the reactants and the products

Explanation :

According to the law of conservation of mass, the mass remains conserved that means the total number of atoms present on the reactant side should always be equal to the number of atoms present on the product side.

The given unbalanced reaction is,

$Na+H_2O\rightarrow NaOH+H_2$

The number of atoms of hydrogen are 2 on reactant side and 3 atoms of hydrogen on product side. Thus equation is not balanced.

The balanced reaction will be,

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Hence, the correct option is, (A)

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Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

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(a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
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Yes, The intensity of the electric force is still
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(d) does a magnetic field do so?
Yes. As we said in point b, the magnetic force is
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And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

(e) does an electric field exert a force on a straight current-carrying wire?
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3 years ago

In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and t

vfiekz [6]

Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.

The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:

v = (vx, vy)

where vx is the component of v in the horizontal and vy is the component of v in the vertical.

In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0 then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = $\sqrt{(x0 + v0* t* cos angle)^{2} }$

65 m = x0 + v0 t cos angle ( x0 = 0)

65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 = y0 + v0 t sin angle + 1/2 g t²

replacing t = 65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)

using g = -9.8 m/s²

-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

(using trigonometric identity: sin x cos x = sin (2x) / 2

sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

angle = 14.72°

3 0

3 years ago

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