All categories

3 years ago

How many cubic feet are there in a gallon?

Physics

2 answers:

Kazeer [188]3 years ago

8 0

There are 0.133681 cubic feet in a gallon.

shtirl [24]3 years ago

5 0

1 gallon has 0.133681 cubic feet

You might be interested in

Which of these is most likely a step in the formation of soil? a. erosion of soil b. crystallization of rocks c. animals digging

Lyrx [107]

<span>C. Measure the soil's volume and then the volume of water that the soil can absorb.</span>

7 0

3 years ago

Read 2 more answers

Part A: Explain why x = 5 makes 4x − 1 ≤ 19 true but not 4x − 1 < 19. (5 points) Part B: What value from the set {6, 7, 8, 9,

Elis [28]

Answer:

A: ≤ means less than OR equal to. < only means less than

B: 9

Explanation:

A: Because it would equal 19, and 19 is EQUAL than 19. 4(5) - 1 would equal 19, which is equal to 19, and not less than. ≤ means less than or equal to. < means less than. So its not true.

B: 47 - 2, 45. Then 5 x 9 equals 45. So 5 x 9 equals 45, then add 2 would equal 47.

Hope this helps <3

5 0

3 years ago

Heat moves from hot to cold either through air or liquid <br>

KonstantinChe [14]

Answer:

both

Explanation:

because when it is hot in summer 5hat is the air and the sund u can warm things up and then it get hot

5 0

2 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a force of 12.344 N

at distance r

and we know Electrostatic force is given

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{kQ\cdot Q}{r^2}$

$F=\frac{kQ^2}{r^2}$

Now the magnitude of charge is 2Q and is at a distance of $\frac{r}{2}$

$F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}$

F'=16F

F'=197.504 N

4 0

3 years ago

Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can

svetlana [45]

Answer:

3.6 m

Explanation:

$\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \ 0.25 \ mm\\\\= 0.25 *10^{-3} m$

$\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\$

Also

$\beta = 1 \ mm \ fringe \ width$

$D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} = 3.6 m$

Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m

4 0

3 years ago