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3 years ago

A student increases the temperature of a 300 cm^3 balloon from 30c to 60c. what will the new volume of the ballon be

2 answers:

5 0

<span>This problem uses the ideal gas law, PV = nRT. T for temperature must be in Kelvin, so we add 273.15 to the Celsius temperatures. We are then changing T from 303.15 K to 333.15K. The ratio of the temperatures 333.15/303.15 ~= 1.1. So holding pressure constant, the new volume of the balloon will be approximately 10% larger, or 330cm^3.</span>

Mama L [17]3 years ago

4 0

From Charles law the volume of a fixed mass of gas is directly proportional to the absolute temperature when pressure is kept constant.
Volume α temperature, mathematically; k = V/T
Therefore, comparing two gases V1/T1 = V2/T2
V1 =300 cm³ T1= 30°C (30 +273= 303K) V2= ? T2= 60°C (60 +273= 333K)
Thus, 300/303 = V2/333
V2 = (300 × 333)/ 303
= 329.703 cm³
Hence the new volume will be 329.703 cm³

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Please help me with physics ks3!! Sound waves and hearding (pic incl)

Nina [5.8K]

Your answers for a), b), and c) are correct. Good work !

d). The high noise levels in Technology and PE are more of
a concern to the teachers than to the students because the
students are only in there for 1 or 2 periods a day, but the
Technology and PE teachers are in there ALL day.

e). Mr. Jones can't hear as high frequency as Jenny can because
he is much older than Jenny is. Sadly, even without damage due to
loud noise, the ability to hear high frequencies does decrease with age.

f). The wooden surfaces in the gym cause the gym to be louder
than it would be if it had carpet on the floor. Carpet ... and soft
walls and ceilings ... absorb a lot of the sound that hits them.
But hard surfaces don't absorb much of the sound that hits them,
so it just keeps bouncing around until it finally fades away.

You can see this easily ... just go into the gym at your school, clap
your hands once, and notice how long you keep hearing the sound
after you clap.

7 0

3 years ago

Read 2 more answers

Mike took less time to run the 100 meters than mitchell what does this tell us about his speed

Dahasolnce [82]

Mike was running at a higher speed than Mitchell

8 0

3 years ago

Which one of the following statements concerning kinetic energy is true? a The kinetic energy of an object always has a positive

hodyreva [135]

a The kinetic energy of an object always has a positive value.

Explanation:

The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Let's now analyze each statement:

a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term $v^2$ is always positive as well, so the kinetic energy is always positive.

b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, $K\propto v^2$ .

c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.

d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.

e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

What is a dependent or responding variable? Can someone help me?

Svetllana [295]

They are a variable that changes as a result of the changes in the manipulated variable

8 0

3 years ago

How do I do these? My teacher didn’t show us how.

melisa1 [442]

Explanation:

Displacement is simply the change in position. So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m. So the displacement is:

Δx = -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m. The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m. The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s. It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s. So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet. Unfortunately, this isn't as easy as looking for where the lines cross on the graph. We need to calculate their displacements. We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2. Half of the area is below the x-axis, and half is above. Without doing calculations, we can say the total displacement for this car is 0. This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when. For Car 1, the area under the curve is a triangle. So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time. Since the line has a slope of 1 and y intercept of 0, we know v(t) = t. So:

Δx = ½ t²

Now look at Car 3. The area under the curve is a rectangle. So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for. Where are the cars at this time? Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0

3 years ago