I need the letters after the questions

A 2.5 kg steel gasoline tank can holds 20.0 L of gasoline when full. What is the average density (in Kg/m^3) of the full gas can

yarga [219]

Answer:

$D=792.3\ kg/m^3$

Explanation:

<u>Average Density </u>

The density of an object of mass m and volume V is

$\displaystyle D=\frac{m}{V}$

If we know the density and the volume occupied by the object, the mass can be computed as

$m=D.V$

The tank can hold 20 L of gasoline when full. Converting to cubic meters

$V_g=20*0.001=0.02\ m^3$

That's the volume of the gasoline it contains. Knowing the density of the gasoline, we get the mass of gasoline.

$m_g=680*0.02=13.6\ kg$

To know the total mass of both, we add the 2.5 kg of the tank

$m=m_t+m_g=13.6+2.5=16.1\ kg$

The volume of the tank is computed solving for V

$\displaystyle V=\frac{m}{D}$

$\displaystyle V_t=\frac{2.5}{7800}=3.205\times 10^{-4}\ m^3$

The total volume is

$V=V_g+V_t=0.02+3.205\times 10^{-4}=0.0203\ m^3$

The average density is

$\displaystyle D=\frac{16.1}{0.0203}$

$\boxed{D=792.3\ kg/m^3}$

7 0

3 years ago

Which of the following is not an application of Doppler technology?

allsm [11]

The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

$F \ \alpha \ \frac{1}{D^{2}}$ for a distance D

If we move them so that D is doubled:

$\frac{1}{2^{2}.D^{2} }$ = $\frac{1}{4} \eq \frac{1}{.D^{2} } \eq$

Then the force they experience is one fourth of the original.

5 0

2 years ago

Hằng số phổ biến chất khí

padilas [110]

Answer:

please in english......................................

Explanation:

8 0

2 years ago

A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0

3 years ago