Do you think all elements are pure substances? Why or why not?

Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the

Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} =$ mass flow rate

$\rho =$ Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3$

$V_1 = 20m/s$

$A_1 = 60*10^{-4}m^2$

$\rho_2 = 0.762kg/m^3$

$V_2 = 160m/s$

PART A) Applying the flow equation we have to

$\dot{m} = \rho_1 A_1 V_1$

$\dot{m} = (2.21)(60*10^{-4})(20)$

$\dot{m} = 0.2652kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}$

$A_2 = \frac{0.2652}{(0.762)(160)}$

$A_2 = 2.175*10^{-3}m^2$

Therefore the Area at the end is $21.75cm^2$

3 0

3 years ago

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5

svlad2 [7]

The magnetic force experienced by the proton is given by
$F=qvB \sin \theta$
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and $\theta$ the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so $\sin \theta=1$ and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
$F=ma$

So we have
$ma=qvB$
from which we can find the magnitude of the field:
$B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T$

4 0

2 years ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

$Q=cm\Delta T$ (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and $\Delta T$ the change in temperature of the sample. So, if we solve (1) for

Sample A:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}$

$\Delta T=-\frac{Q}{16744}$

Sample B:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}$

$\Delta T=-\frac{Q}{5400}$

Sample C:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}$

$\Delta T=-\frac{Q}{9450}$

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression $-\frac{Q}{cm}$ that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of $\Delta T$ implies the temperature is dropping)

8 0

2 years ago

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

2 years ago

Which of the following is an example of a properly written testable hypothesis? *

sveta [45]

Answer:

D. if it is dark, then an owl will find a mouse by the sound the mouse makes

3 0

3 years ago