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3 years ago

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A janitor comes across a spilled substance on the floor of a highschool chemistry lab. he wants to make sure that the liquid is

neutral before cleaning it up. what would be safest to add to the puddle?
a. pour on a base
b. pour on a buffer
c. pour on a acid

1 answer:

MAVERICK [17]3 years ago

5 0

A is the answer I think

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A car is moving with speed 60 m/s and acceleration 4 m/s2 at a given instant. (a) using a second-degree taylor polynomial, estim

pantera1 [17]

Answer:

Explanation:

Using second degree taylor polynomials

let $S(t)$ be position function and set $S(0)=0$

where S(0) is the initial position

Then $v(t) = S^i(t)$ and $a(t)=S^{ii}(t)$

we have $v(0) = 60m/s$ , $a(0) = 4m/s^2$

so $T_{2}(t) =S(0)+v(0)t+\frac{a(0)}{2} t^2\\\\\\=0+60t+\frac{4}{2} t^2\\\\T_{2}(t)=60t+2t^2\\\\S(1)=T_{2}(1)=60+2=62m$

b.) yes

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3 years ago

If the pressure of a gas is increased, then the volume will decre

pogonyaev

Answer:

The answer is b

Explanation:

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3 years ago

Which of the following are transferred or shared when two atoms react chemically? *

xxTIMURxx [149]

Answer:

One of the basic principles of chemistry is the electrostatic attraction between atoms or compounds. Electrons are on the outside of an atoms and that's where the charges come from and the interaction between those charges is what happens during a chemical bond. Therefore the answer would be electrons.

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2 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

3 years ago