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4 years ago

How do I find initial velocity and time?

Physics

1 answer:

forsale [732]4 years ago

3 0

Explanation:

Given:

Δx = -100

v = 10

a = 20

To find v₀, use a kinematic equation that's independent of time.

v² = v₀² + 2aΔx

(10)² = v₀² + 2(20)(-100)

100 = v₀² − 4000

v₀² = 4100

v₀ = ±64.0

As your teacher said, v₀ can't be +64.0. So v₀ = -64.0.

Next, to find time, use a kinematic equation that's independent of initial velocity.

Δx = vt − ½ at²

-100 = (10) t − ½ (20) t²

-100 = 10 t − 10 t²

-10 = t − t²

t² − t − 10 = 0

Solve with quadratic formula:

t = [ -(-1) ± √((-1)² − 4(1)(-10)) ] / 2(1)

t = (1 ± √41) / 2

t > 0, so:

t = (1 + √41) / 2

t ≈ 3.70

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which of the following statements best justifies the inclusion of test tube v as a control in the experiment? responses it will

prohojiy [21]

The following statement which best justifies the inclusion of test tube v as a control in the experiment is that it will show the color change that occurs in the absence of enzyme activity and is denoted as option C.

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Enzymes are known as proteinous in nature and are denatured by heat and other substances is included in the test tube v as a control in the experiment is that it will show the color change that occurs in the absence of enzyme activity.

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The options are:

It will provide a measurement of amylase activity at an acidic pH.
it will provide a measurement of amylase activity at a basic pH.
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it will show the color change that occurs in the absence of the amylase protein.

It will show the color change that occurs in the absence of enzyme activity.

5 0

1 year ago

What is astrometry? What is astrometry? measuring the positions of stars on the sky using metric units for distance (e.g. meters

Fofino [41]

Explanation:

It is a branch of astronomy wherein measurements of the positions and movements of celestial bodies are precisely are made.

For the measurement, minute wobbling of the position of the body is detected. When the shift is detected periodically, the astronomers come to know that the star is orbited by a planet.

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3 years ago

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

6 0

3 years ago

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german

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8 0

3 years ago

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-Dominant- [34]

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3 years ago