Which of the following is a definite indicator of a chemical change?

In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

ANTONII [103]

Is there a picture?

6 0

2 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.

4 0

2 years ago

A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

6 0

2 years ago

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

2 years ago

The moon Phobos orbits Mars

dlinn [17]

Answer:

The moon Phobos orbits Mars

(mass = 6.42 x 1023 kg) at a distance

of 9.38 x 106 m. What is its period of

orbit?

Explanation:

Answer: 27.9816 x 10^3 is the period of orbit

8 0

2 years ago