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3 years ago

13

What gas law applies to the situation you have described above. Why?

1 answer:

Allushta [10]3 years ago

6 0

If you give us the situation described then I'll be able to help.

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Make the following conversion: 34.9 cL = _____ hL

aliina [53]

We know that: 1 L = 100 cL. Or 1 cL = 0.01 L. Then we will make the conversion: 34.9 cL = 34.9 / 100 L = 0.349 L. Also: 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL. This can be also written as: 3.49 * 10^(-3) hL ( in the scientific notation ). Answer: 3.49 cL = 0.00349 <span>hL </span>

8 0

3 years ago

What is the frequency of a wave with a period of 8s?

PSYCHO15rus [73]

I don’t know the answer

3 0

3 years ago

A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move

Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity = 21.7m/s

Time = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

Acceleration = $\frac{v - u}{t}$

v is final velocity

u is initial velocity

t is the time

Acceleration = $\frac{21.7 - 10.5}{4.34}$ = 2.58m/s²

Distance traveled;

V² = U² + 2aS

21.7² = 10.5² + 2 x 2.58 x S

360.64 = 2 x 2.58 x S

S = 69.9m

3 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

The engine of a jet airplane pushes exhaust gases from burning fuel backward

olga nikolaevna [1]

For more boost and to stop chases of fire

6 0

3 years ago