All categories

3 years ago

2NaOH + H2SO4 → Na2SO4 + H2O

Physics

2 answers:

irakobra [83]3 years ago

7 0

Its actually oxygen instead of sodium, you got the rest right
Not Balanced - Oxygen - Not Equal

mina [271]3 years ago

6 0

The equation is unbalanced because the number of hydrogen atoms and sodium atoms is not equal in the reactants and the products

You might be interested in

Please help!!30 points and will give the brainliest answer for a clear explanation.I need it really fast. Thank you!

kompoz [17]

I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway

4 0

3 years ago

A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq

Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>

The following expression can be used to relate the fundamental frequency to the driving frequency;

f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

f(4) = 4× f (1)

480 = 4× f(1)

f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
For, n = 5,

f(n) = n 120Hz,

f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

#SPJ4

8 0

1 year ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

The sum of the kinetic and potential energies of a system of objects is conserved: Group of answer choices only when no external

uysha [10]

The sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

<h3>Conservation of mechanical energy</h3>

The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system (absence of external force) is always constant.

M.A = P.E + K.E

where;

P.E is potential energy

K.E is kinetic energy

Thus, the sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

Learn more about conservation of mechanical energy here: brainly.com/question/24443465

7 0

2 years ago

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t

inna [77]

Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

Stanley measured the events at a time interval longer than ∆to by γ,

such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

which is the same as in part (a)

7 0

3 years ago