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a_sh-v [17]
2 years ago
11

The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere ​

Physics
1 answer:
spin [16.1K]2 years ago
5 0

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

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5 0
3 years ago
Suppose you dissolve 12.8 g of one substance in 11 g of another. Is this reasonable? Explain
RUDIKE [14]

Answer:

No, its not reasonable.

Explanation:

The substance that is to be dissolved is known as solute. The substance that is dissolving is known as solvent.

The amount of solvent  in the mixture should be greater than that of solute.

Suppose we are taking a solvent in a beaker and we are continuously adding solute in it. Initially the solute dissolve quickly. At some point the solute stops dissolving in the solvent. This is known as saturation point of the solvent. After saturation point if solute is added further it does not dissolve in the solvent.

So, its not possible to dissolve 12.8 g of one substance in 11 g of another.

5 0
3 years ago
A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is
erica [24]
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b. Newtonian focus
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4 0
3 years ago
A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?
Inga [223]

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

6 0
3 years ago
a car whose initial speed is 30 seconds m / s slows uniformly to a stop in 5.00 seconds what was the cars displacement ​
CaHeK987 [17]

Answer:

60

Explanation:

4 0
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