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just olya [345]
3 years ago
13

2. How does the direction of your shadow change over the course of a day?

Chemistry
1 answer:
Radda [10]3 years ago
6 0
The direction of your shadow is going to change directions due to the moving of the sun
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Is it easy for gallium to change from solid to liquid? Why<br> subject is science
Alekssandra [29.7K]

Answer:

Solid gallium is a blue-gray metal with orthorhombic crystalline structure; very pure gallium has a stunning silvery color. Gallium is solid at normal room temperatures, but as well as mercury, cesium, and rubidium it becomes liquid when heated slightly

Explanation:

gallium is like water it freezes to turn solid!

...or like Bruce lee get it?

5 0
3 years ago
Glucose and oxygen are the products of
Artemon [7]
Photosynthesis by plants.
5 0
3 years ago
Help ASAP! ........;.;
Damm [24]

Answer: D and A

Explanation: just took the test

6 0
3 years ago
A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0
3 years ago
9)
Trava [24]

Answer : The volume of gas will be, 113.5 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 60 atm

P_2 = final pressure of gas = 30 atm

V_1 = initial volume of gas = 29 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 115 K

T_2 = final temperature of gas = 225 K

Now put all the given values in the above equation, we get:

\frac{60atm\times 29L}{115K}=\frac{30atm\times V_2}{225K}

V_2=113.5L

Therefore, the volume of gas will be, 113.5 L

7 0
3 years ago
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