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ella [17]
3 years ago
9

A weather balloon is filled to a volume of 2 liters with helium gas at a pressure of 101 kPa, at a temperature of 22oC. The ball

oon is released and floats to a height where the pressure is 90 kPa and the air temperature is now 16oC. What is the new volume of the weather balloon?
Physics
2 answers:
kvv77 [185]3 years ago
7 0

Answer: The new volume of  weather balloon will be 2.2 Liters

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas =  101 kPa

P_2 = final pressure of gas = 90 kpa

V_1 = initial volume of gas = 2 L

V_2 = final volume of gas = ? L

T_1 = initial temperature of gas = 22^oC=273+22=295K

T_2 = final temperature of gas = 16^oC=273+16=289K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{101\times 2L}{295K}=\frac{90\times V_2}{289K}

V_2=2.2L

Therefore, the volume will be 2.2 Liters.

OLEGan [10]3 years ago
4 0
The answer is B (: Hope it helps

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Citrus2011 [14]

Sally's average speed is <u>35.3 mi/h.</u>

Average speed of a body is the total distance traveled in the given time interval.

Express the distance  d traveled in miles.

d=\frac{97 km}{1.6 km/m\\i} \\ = 60 mi

Express the time t traveled in hours.

t  =\frac{102 min}{60 min/hr}\\ = 1.7 h

Calculate the average speed v.

v =\frac{d}{t} \\ =\frac{60 mi}{1.7h} \\   = 35.3 mi/h

Her average speed is 35.3 mi/h, which is less than the speed limit of 65 mi/h.

However, the average speed of an object is different from its instantaneous speed. It could be possible that at the time when the officer apprehended her, Sally could have been travelling at a speed greater than the prescribed speed limit, which would have prompted the officer to issue a speeding ticket to her.

Thus, the average speed of a person cannot be considered as a bench mark for speeding offences, since her instantaneous speed could have been higher than the speeding limit and yet she could have had an average speed less than the speeding limit.


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4 years ago
How to get a + b on a graph
Nataly [62]

The first positively essential requirement is that
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4 years ago
What traction of the radioisotope<br>remains in the body after one day?​
r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is (\frac{1}{2})^{\frac{1}{\tau}}, with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}

where:

m_0 is the mass of the radioactive sample at t = 0

\tau is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}

where the half-life \tau must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0
3 years ago
A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

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2250 + 147000 = 149250

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3 years ago
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