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brilliants [131]

3 years ago

13

Solve without using a calculator. on a cold day near the ocean, it was found that 8 g of an unknown gas occupies a volume that i

s a little less than 6 l. based on this information, what is the most likely identity of the unknown gas?

1 answer:

liberstina [14]3 years ago

7 0

Assuming that the gas acts like an ideal gas so that the molar volume is Vm = 22.4 L / moles, we can calculate for the number of moles.

number of moles = 6 L / (22.4 L / mol) = 0.2679 moles

So the molar mass of this gas is:

molar mass = 8 g / 0.2679 moles = 29.9 g/mol

The value is around 32 g/mol which is the molar mass of oxygen. Therefore the answer is:

<span>O2</span>

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Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it

Vlad [161]

Answer:

r = 4.62 cm

Explanation:

Mass of a coffee, m = 370 g

Height of the mug, h = 5.5 cm

We need to find the inside radius of the mug. Density of an object is equal to the mass per unit volume. It can be given by :

$d=\dfrac{m}{V}\\\\d=\dfrac{m}{\pi r^2 h}\\\\r=\sqrt{\dfrac{m}{\pi dh}}\\\\r=\sqrt{\dfrac{370\ g}{\pi \times 1\ g/cm^3\times 5.5\ cm}}\\\\r=4.62\ cm$

So, the inside radius is 4.62 cm.

8 0

3 years ago

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB

aliina [53]

The moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

<h3>What is moment of inertia?</h3>

The moment of inertia is the amount of rotation obtained by an object when it is in state of motion or rest.

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB = 2.83 kg, mC = 3.39 kg.

The coordinates of Ball A :(2,1) Ball B :(8,2) and Ball C: (5,8) and axis is at (4,3)

Here, for the moment of inertia of each ball

For ball A

IA = 1.09 x ((4 - 2)^2 + (3-1)^2)

IA = 8.72 Kg.m^2

for the ball B

IB = 2.83 * ((4 - 8)^2 + (3 - 2)^2)

IB = 48.11 Kg.m^2

for the ball C

IC = 3.39 * ((4 - 5)^2 + (3 - 8)^2)

IC = 88.14 Kg.m^2

Total moment of inertia = 8.72 + 48.11 + 88.14

Total moment of inertia = 144.97 Kg.m^2

Thus, the moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

Learn more about moment of inertia.

brainly.com/question/15246709

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3 0

2 years ago

What the exercise or play in the p.e in the philippines?<br> please!!!!, please!!!!

Debora [2.8K]

Answer:

They could be jumping off structures, running or climbing on rocks. They also could be participating in sports like soccer.

Explanation:

sources:

https://www.philippinesbasiceducation.us/2013/06/physical-activity-and-physical.html - about there play schedule

http://sportphil.com/category/articles/ - concil of philippines

Sorry that's all I could find.

5 0

3 years ago

Read 2 more answers

A sheet of glass measures one metre by half metre at 0°C. Calculate its area at 30°C, a for

Yuki888 [10]

Answer:

0.50027 $m^{2}$

Explanation:

ΔA= $2\alpha A_{0} T$

Where ΔA is the change in area

There is a 2 since the coefficient for area = 2 x that for the linear coefficient

$A_{0}$ is the original area

$\alpha$ is the coefficient of thermal expansion

And T is the change in temperature

ΔA=2(0.000009)(0.5)(30-0)=0.00027 $m^{2}$

New area= ΔA+ $A_{0}$ = 0.5+0.00027=0.50027 $m^{2}$

6 0

3 years ago