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3 years ago

What is the velocity of an object that has been in free fall for 1.5s?

Physics

1 answer:

Crank3 years ago

8 0

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

3 years ago

So... if bedbugs live in beds, where do cockroaches live?

Deffense [45]

Answer:

oo.p i wish I could answer that

Explanation:

6 0

3 years ago

Read 2 more answers

What is silviculture?

expeople1 [14]

Sivilculture isis the art and science of managing forests for desired outcomes.

8 0

3 years ago

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

3 0

3 years ago

Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The

Alexus [3.1K]

Answer:

52.5°C

Explanation:

The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state

mh1 + W = mh2

h2 = h1 + W/m

h1 + Wα1/V1

242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg

=287.4 kJ/kg

From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation

i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)

= 50°C + 60 - 50/295.15 - 284.79

(287.4 - 284.79)°C

= 52.5°C

7 0

3 years ago