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3 years ago

Which one of the following is a balanced equation?

Physics

1 answer:

yarga [219]3 years ago

5 0

D. 2Na (s) + MgF2 → 2NaF (s) + Mg (s)

Explanation:

From the given choices, 2Na (s) + MgF2 → 2NaF (s) + Mg (s) is balanced

A balanced equation is one that obeys the law of conservation of matter which states that "matter is neither created nor destroyed but transformed from one form to another".

We expect to find the same name of atoms on both sides of the equation.

let us examine the choices;

Zn (s) + HCl (g) → ZnCl2 (s) + H2 (g)

Reactants Products

Zn 1 1

H 1 2

Cl 1 2

Na (s) + O2 (g) → Na2O

Reactants products

Na 1 2

O 2 1

Na (s) + MgF2 → 2NaF (s) + Mg (s)

Reactants products

Na 1 2

Mg 1 1

F 2 2

2Na (s) + MgF2 → 2NaF (s) + Mg (s)

Reactants products

Na 2 2

Mg 1 1

F 2 2

learn more:

Balanced equation brainly.com/question/2947744

#learnwithBrainly

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**HELP ASAP**

Tcecarenko [31]

The answer of this question is B.

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2 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Dovator [93]

It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
All you need is F=(K*Q1*Q2)/r^2 
Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C

3 0

3 years ago

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

2 years ago

A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

larisa [96]

We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m

Substitute these values into...
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N

Therefore, your answer should be 2.6 × 10–10

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3 years ago

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Why is it difficult to walk on muddy road?

Gnesinka [82]

Because of internal friction between layers of mud particles called viscosity. When you walk, your foot exerts a force on the mud; and according to Newton, the mud also (is supposed to) exert an equal opposite force, which leading to an equal net resultant force in your direction, propelling you forward.

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3 years ago