All categories

4 years ago

How many cubic feet are in a 55-gallon drum?

1 answer:

5 0

1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

   (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

   = (55 x 231 / 1728) ft³

   = 7.352 cubic feet (rounded)

You might be interested in

Can I PLEASE get some help? I REALLY need it!

soldi70 [24.7K]

The answer is C. Hope this helps.

7 0

3 years ago

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

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

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

6 0

3 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

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

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

7 0

3 years ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

gizmo_the_mogwai [7]

<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>

5 0

3 years ago

how do u calculate the kinetic energy of a ball of mass 0.25kg being kicked vertically upwards with a speed of 5m/s

vfiekz [6]

Answer:

3.125J

Explanation:

K.E.= 1/2(mass)(velocity)^2

K.E.=1/2(0.25)(5)^2=3.125

6 0

3 years ago