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aliina [53]
3 years ago
6

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

dard atmospheric pressure?
Physics
1 answer:
PSYCHO15rus [73]3 years ago
8 0

Answer:

volume is 700 mL

Explanation:

pressure = 2 atm

volume = 350 mL = 0.350 L

to find out

volume

solution

we will apply here equation that is

P1×V1 = P2×V2   ..............1

here P1 = 2 and V1 = 0.350 and P2 = 1 for standard atmospheric pressure

so put all value here  in equation 1 and get V2 volume

2 × 0.350 = 1 × V2

V2 = 0.700 L

V2 = 700 mL

so volume is 700 mL

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coldgirl [10]

Answer:

See explanation

Explanation:

Favourite scientific discipline; Chemistry

Definition: Chemistry is the study of the composition, properties and uses of matter as well as the principles governing the changes that matter undergoes.

Source: New School Chemistry by Osei Yaw Ababio (2013)

Least Favourite Scientific Discipline: Botany

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2 years ago
Technician A says that the battery must be in good condition and at least 75% charged to accurately test an alternator. Technici
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Answer:Technician A

Explanation:

Technician A statement is correct as  

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3 years ago
The distance between two electric pole is 800m then express this in terms of mm​
Lana71 [14]

Answer:

Mm stands for milimetres

Explanation:

6 0
2 years ago
Read 2 more answers
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3 0
3 years ago
A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
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