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Artemon [7]
3 years ago
5

Oxygen gas is used to fill a balloon and a short time later the balloon is observed to only be half full. Explain this observati

on. A) The balloon leaks and the temperature went down. B) More oxygen was added to the balloon and the temperature went up. C) The pressure outside the balloon went down and the pressure inside went down. D) The pressure outside the balloon went down and the temperature inside went up.
Physics
2 answers:
Margaret [11]3 years ago
7 0

Answer: A The balloon leaks and the temperature went down

Explanation: Cause I did the USAtestprep

lesya692 [45]3 years ago
3 0

Answer:

It would be A.

Explanation:

When oxygen is released, that oxygen rises and the ballons temperature will go down. Heat rises so if the ballon is dropping, then the temperature is dropping.

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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

F_{drag}=12v+4v^2

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

F_{drag}=mg

12v+4v^2=80\times 9.8

4v^2+12v=784

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0
3 years ago
Bro i need help omg.
sergey [27]

Answer:

25.

Explanation:

8 0
3 years ago
Read 2 more answers
How much heat is needed to boil 5.30 kg of water at its boiling point?
vivado [14]

Answer:

Required heat Q = 11,978 KJ

Explanation:

Given:

Mass = 5.3 kg

Latent heat of vaporization of water = 2,260 KJ / KG

Find:

Required heat Q

Computation:

Required heat Q = Mass x Latent heat of vaporization of water

Required heat Q = 5.3 x 2260

Required heat Q = 11,978 KJ

Required heat Q = 12,000 KJ (Approx.)

4 0
3 years ago
Read 2 more answers
Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

#SPJ9

7 0
2 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
4 years ago
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