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Olenka [21]
3 years ago
15

In water, acids form _____.

Physics
2 answers:
Amiraneli [1.4K]3 years ago
8 0
The answer is hydrogen ions
NeTakaya3 years ago
7 0

<em>Answer:</em>

<em>D.)hydrogen ions</em>

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A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0
3 years ago
How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se
Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats


3 0
3 years ago
Read 2 more answers
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
A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a
prohojiy [21]

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

6 0
3 years ago
Read 2 more answers
The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele
Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

P=P_0e^{-\frac{Mg}{RT}h}

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

R_{H_2}=\frac{R}{M}

Therefore yo can write the barometric formula as:

P=P_0e^{-\frac{g}{R_{H_2}T}h}

at the surface of the planet (h =0) the pressure is P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}

applying to the previuos equation:

\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}

solving for h:

h=17357.9m

3 0
4 years ago
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