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Bad White [126]
3 years ago
10

If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling

Physics
1 answer:
ryzh [129]3 years ago
5 0

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after  falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter.  After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

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What is the name given to the initial 150 counts in 2 minutes?
Whitepunk [10]

Divide CFU of Dilution. Divide the CFU of the dilution (the number of colonies you counted) by the result from step 4. For this example, you work out 46 ÷ 1/1000, which is the same as 46 x 1,000. The result is 46,000 CFU in the original sample.

7 0
3 years ago
A mass of 2.1 kg the ball leaves her hand with the speed of 30 mass find the energy of the ball
snow_lady [41]
Ke= 1/2 x m x v^2
Ke= 1/2 x 2.1 x 30^2
Energy = 945 J
8 0
3 years ago
PLSLPSLLPSLPSLPS HELP I ONLY HAVE 5 MINS PLEASE
UkoKoshka [18]

Answer:

0.19m/s²

Explanation:

Initial velocity(u) = 50×1000/60×60

=13.88 m/s

Final velocity(v) = 36.5×1000/60×60

=10.13 m/s

Acceleration(a) = v-u/t

=10.13-13.88/19.5

a= -0.19m/s²

-a = 0.19m/s²

The magnitude of retar dation is 0.19m/s²

5 0
2 years ago
Read 2 more answers
Every winter I fly to Michigan. It's a total distance of 3900km. It takes 5 hours. What is my average speed?
olga_2 [115]
\frac{3900}{5} = 780\sf ~kilometer~per~hour

Your average speed is 780 kilometer per hour.
4 0
3 years ago
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For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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