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2 years ago

14

Hey, guys I kinda need help with this question, 1. The number of atoms of each element are not equal on both sides of the equati

on

2 answers:

Xelga [282]2 years ago

5 0

Answer:

The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.

Explanation:

I hope thats what u needed.

nadya68 [22]2 years ago

3 0

Answer: Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

Explanation:

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A uniform disk with radius 0.650 m

VashaNatasha [74]

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

$a = \sqrt{a_{t}^{2}+a_{n}^{2}}$

$a = 13.758\,\frac{m}{s^{2}}$

3 0

2 years ago

How many meters are in a gigameter?

gladu [14]

Answer:

1 gigameter = 1e+9 meter

7 0

3 years ago

À stone is thrown

masya89 [10]

A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.

3 0

2 years ago

Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two road

Gwar [14]

Explanation:

It is given that,

$\frac{dx}{dt}$ = -40 mi/h, $\frac{dx}{dt}$ = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find $\frac{dz}{dt}$ . Equation for the given variables according to the Pythagoras theorem is as follows.

$z^{2} = x^{2} + y^{2}$

Now, we will differentiate each side w.r.t 't' as follows.

$2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

or, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As, $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

= $\frac{1}{5}(4 \times (-40) + 3 \times (-40))$

= $\frac{-140 - 120}{5}$

= 52

Thus, we can conclude that the cars are approaching at a rate of 52 mi/h.

7 0

3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

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