All categories

3 years ago

How many grams does 15.58 L of gasoline weigh? Gasoline has a density of 0.74 g/mL

Physics

1 answer:

Korolek [52]3 years ago

5 0

We are given volume of gasoline = 15.58 L and

Density = 0.74 g/mL.

Density is given in grams(g) per milliliter (mL). So, we need to convert given volume in mL also.

We know, 1 litre = 1000 milliliters

Therefore, 15.58 L = 15.58 * 1000 mL = 15580 mL

We could rewrite,

Volume of gasoline = 15580 mL.

Formula for density is..

Density (ρ)= $\frac{Mass}{ Volume}$

Plugging values in formula,

0.74 g/mL = $\frac{Mass}{15580 \ mL}$

Muliplying both sides by 15580.

$0.74 g/mL * 15580mL =\frac{Mass}{15580 \ mL} * 15580mL$

11529.2g = mass.

Therefore, 11529.2g grams does 15.58 L of gasoline weigh.

You might be interested in

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

4 years ago

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of

Soloha48 [4]

Answer:

The length of her shadow is changing at the rate -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 = (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

7 0

3 years ago

Read 2 more answers

Two trains are traveling side-by-side along parallel, straight tracks at the same speed. In a time t, train A doubles its speed.

sergij07 [2.7K]

Answer:

derdddrdickdd

Explanation:

3 0

3 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

CAN ANYONE GIVE ME SOME INFO ON THE THERMOSPHERE

Pachacha [2.7K]

The thermosphere is the layer of the Earth's atmosphere directly above the mesosphere and below the exosphere. Within this layer of the atmosphere, ultraviolet radiation causes photoionization/photodissociation of molecules, creating ions in the ionosphere.

6 0

3 years ago

Read 2 more answers