The students in a class had the following scores on a test:

allochka39001 [22]

The atmosphere is made up of 21% oxygen, 78% nitrogen, 0.9% argon, and 0.03% carbon dioxide with very small %s of other elements.

3 0

3 years ago

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Learning goal: to practice problem-solving strategy 20.1 electric forces and coulomb's law. two charged particles, with charges

attashe74 [19]

1) Let's put $q_1$ and $q_2$ both on the x-axis such that $q_2$ is on the right side of $q_1$ . The distance between the two charges is $d=2~cm$ .

2) Let's put $q_3$ on the x-axis and let's call $r$ the distance between charge 1 and charge 3. As a consequence, the distance between charge 2 and 3 will be $d+r$ : if r is positive, then charge 3 will be located on the left of charge 1, if r is negative, charge 3 will be located between charge 1 and charge 2.

3) The electrostatic force between charge 1 and charge 3 is
$F_{13} = \frac{q_1 q_3}{4 \pi \epsilon_0 r^2}$
while the force between charge 2 and charge 3 is
$F_{23} = \frac{q_2 q_3}{4 \pi \epsilon_0 (r+d)^2}$

4) The problem says the two forces are equal in intensity. Therefore we can write $F_{13} = F_{23}$ . Using the equations written at step 3), and substituting $q_1=q$ and $q_2=4q$ as mentioned in the problem, $F_{13} = F_{23}$ becomes
$\frac{q}{r^2}= \frac{4q}{(r+d)^2}$
Using $d=2~cm$ , we can solve the equation, and we get two solutions:
$r=2~cm$
$r=-0.67~cm$
Both solutions are correct. With the former, the charge 3 is located 2 cm on the left of charge 1, while with the latter, charge 3 is located 0.67 cm on the right of charge 1, between charge 1 and 2.

4 0

3 years ago

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A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i

slavikrds [6]

Answer:

The inside Pressure of the tank is $4499.12 lb/ft^{2}$

Solution:

As per the question:

Volume of tank, $V = 0.25 ft^{3}$

The capacity of tank, $V' = 50ft^{3}$

Temperature, T' = $80^{\circ}F$ = 299.8 K

Temperature, T = $59^{\circ}F$ = 288.2 K

Now, from the eqn:

PV = nRT (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT' (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

$\]frac{n'}{n} = 2$

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

$\frac{PV}{P'V'} = \frac{nRT}{n'RT'}$

$P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}$

7 0

4 years ago

A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the

xxMikexx [17]

Answer:

Time of race = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

s = ut + 0.5at²

17 = 0 x t + 0.5 x 3.89 x t²

t = 2.96 s

Velocity after 2.96 seconds

v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

$t=\frac{83}{11.50}=7.22s$

Time of race = 2.96 + 7.22 = 10.18 s

3 0

3 years ago

Consider the possibility of using two rotating cylinders to replace the conventional wings on an airplane for lift. Consider an

EleoNora [17]

Answer:

27.35m

Explanation:

For the calculation of the Support Force we rely on the formula for obtaining the force in a cylinder of a certain length l,

$F_y = - \rho Ul\Gamma$

Here each term is,

$F_y$ = Lift force

$\rho$ = density of air

$\Gamma$ = vortex strength

For this last equation, its mathematical representation is given by,

$\Gamma = 2\pi av_{\theta}$

Here each term is,

a= 1m, radios of cylinder

$v_{\theta}= 20 Km/hr=5.5m/s$ , the velocity of cylinder surface.

$\Gamma = 2\pi (1)(5.5) = 34.90m^2/s$

In order to find the density of the area at 2000m we will refer to the table of Standard Atmosphere of the United States, that is $1.007kg/m^3,$

$U= 150Km/hr = 41.6m/s, F_y = 40000N, \Gamma = 34.90m^2/s$

Replacing the values,

$40000 = -(1.007)(41.6)l(34.90)$

Clearing l and solving for it we have,

$l=-27.35m$

<em>In this way we can conclude that the length of the cylinder must be 27.35m</em>

7 0

3 years ago