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3 years ago

Assume it takes 5.00 minutes to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3).

1 answer:

7 0

   (50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)

   = (50 · 0.0037854 · 1) / (5 · 60) m³/sec

   = 0.000631 m³/sec

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A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240

andrew11 [14]

Answer:

2,25 g/cm3

Explanation:

Hi, you have to know one thing for this.. Density = mass/Volume,

When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get.. mass= density x volume

Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density = 2,25 g/cm3

8 0

3 years ago

A light wave has a wavelength of 450 nanometers. What is the frequency of this light?

Zolol [24]

Answer:

Frequency, $f=6.67\times 10^{14}\ Hz$

Explanation:

Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,

$c=f\lambda$

f is the frequency of this light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{450\times 10^{-9}}\\\\f=6.67\times 10^{14}\ Hz$

So, the frequency of this light is $6.67\times 10^{14}\ Hz$ .

3 0

3 years ago

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in

Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0

3 years ago

If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre

creativ13 [48]

Answer:

$-112.876J$

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

$16KNO_3(s) + 24C(s) + S_8(s) \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)$

Let us define $P - V$ work as;

$w_{pv} = - P_{external} \triangle Volume$

where $\triangle (Volume) = (V_{final} - V_{initial})$

External pressure is given as $1.00$ $atm$ , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence, $V_i = 0.$

To find the volume of the products, we need to first find the amount of moles of the product made from $2.40_gKNO_3,$ using the molar mass of $KNO_3$ which is 101.1032 g/mol