PZZZZZZZ just Answer this question 20 POINT What happens to the total energy as the skater moves down and up the U-shaped ramp?

The intersection angle of a 3 degree curve is 45.2 degrees. What is the length of the curve? of Select one: O a. 455 m O b.573 m

ValentinkaMS [17]

Explanation:

Relation between length of a curve and angle is as follows.

l = $R \times \Theta$

where, R = radius of curve

$\Theta$ = angle in radians

Also, l = $R \times \Theta \times \frac{\pi}{180}$ .......... (1)

If curve has a degree of curvature $D_{a}$ for standard length s, then

R = $\frac{s}{D_{a}} \times \frac{180}{\pi}$ ........... (2)

Now, substitute the value of R from equation (2) into equation (1) as follows.

l = $\frac{s \times \Theta}{D_{a}}$

If s = 30 m, then calculate the value of l as follows.

l = $\frac{s \times \Theta}{D_{a}}$

= $30 \times \frac{45.2}{3}$

= 452 m

thus, we can conclude that the length of the curve is 452 m.

7 0

3 years ago

PLZZ HELP!! If light traveling through air is then passed through glass, what is the result?

Radda [10]

Answer : The light rays are bent and refraction will occur - B.

5 0

3 years ago

What family contains the most<br> reactive nonmetals: halogen or alkali?

mart [117]

Halogen are the most reactive due to their electronic configuration

5 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K

vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

5 0

3 years ago