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3 years ago

Running for his life, Riley travels 100m in 9.83s. What was his average velocity in

1 answer:

5 0

100m / 9.83 sec = 10.17 m/s (rounded)

There are 3,600 seconds in an hour.

(100 m / 9.83 sec) x ( 3600 sec/hr) = 36,622.6 m/hr = 36.62 km/hr

That's about 22.8 miles per hour.

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Scientific _____ must be supported by observations and results many investigations and are not absolute

n200080 [17]

The correct answer you're looking for would be Theories.

7 0

3 years ago

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

$\Delta E_{system} = E_{in} - E_{out}$

$\Delta U = W_{in} - Q_{out}$

Hence, $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

$\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

$T_{2} = \frac{20 psia}{14.7}(638 R)$

= 868.03 R

Now, we will calculate the mass as follows.

m = $\frac{P_{1}V}{RT_{1}}$

= $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

= 1.031 lbm

Hence,

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

$W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

$W_{in}$ = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0

3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

3 years ago

Depending on circumstances, there are times when divers may wish to be positively, negatively or neutrally buoyant. True. False.

ser-zykov [4K]

Answer:

True

Explanation:

Buoyancy is the most important factors for divers. All they do underwater is to observe the life down there but they also have some other work. However, divers may want to be negatively buoyant when they want to go on deep exploration. When they reach a destination, they may want to observe and neutral buoyancy then will be useful. When they want to go back on surface, they’ll utilize positive buoyancy.

7 0

4 years ago

The dark bands on the wall are from ______________ interference. and i will give brainlyest

Fittoniya [83]

Light can be seen as an electromagnetic wave.
What happens when two waves, with the same frequency, superpose is called interference.

If at a certain point two waves arrive both with a crest, we have constructive interference and the amplitudes sum up, reaching the maximum value, resulting in bright spots.

If at a certain point one of the waves arrives with a crest and the other wave arrives with a trough, we have destructive interference, and the two amplitudes cancel out, resulting in dark spots.

Therefore, the dark bands on the wall are from destructive interference.

4 0

4 years ago