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Nat2105 [25]
4 years ago
13

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

yringe tightly. 3. Pull the plunger back to double the volume of gas in the syringe. Which best describes the purpose and outcome of the demonstration? This is a demonstration of Charles’s law. As the volume increases, the temperature decreases, and the marshmallow will freeze. This is a demonstration of Charles’s law. As the volume increases, the temperature increases, and the marshmallow will melt. This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. This is a demonstration of Boyle’s law. As the volume increases, the pressure increases, and the marshmallow will shrink.
Physics
2 answers:
lana [24]4 years ago
7 0
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. 
</span><span>
 Keisha follows the instructions for a demonstration on gas laws.
 1. Place a small marshmallow in a large plastic syringe.
 2. Cap the syringe tightly.
 3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change.  Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
MrRissso [65]4 years ago
5 0

Answer:

C) This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

Explanation:Took the quiz on edg

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If we compare the force of gravity too strong nuclear force we would conclude that
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The force of gravity is much weaker than the strong nuclear force. But the strong nuclear force only acts over short distances, such as within the nuclues. The gravitational force can act over infinite distance.

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A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outst
Contact [7]

Answer:

v_{i}=10.10 m/s

Explanation:

The equation of the position is:

y=y_{i}+v_{i}t-0.5gt^{2}

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

v_{i}=\frac{y+0.5gt^{2}}{t}

v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}

v_{i}=10.10 m/s

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!

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3 years ago
Which one of the following types of electromagnetic radiation causes certain substances to fluoresce?
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The answer is D. <span>Uv causes more stuff to fouresce </span>
8 0
3 years ago
A 10.0g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New Y
GalinKa [24]

Answer:

1. The final velocity of the 30.0 g marble is 0.100 m/s to the left.

2. The final velocity of the 10.0 g marble is 0.500 m/s to the right.

3. The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

4. The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

5. The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J  

6. The change in kinetic energy for the 10.0 g marble is 4.5 × 10⁻⁴ J

Explanation:

Hi there!

Since the collision is elastic both the momentum and kinetic energy of the system comprised by the two marbles is conserved, i.e., it remains constant after the collision.

momentum before the collision = momentum after the collision

mA · vA + mB · vB = mA · vA´ + mB · vB´

Where:

mA and vA = mass and velocity of the 10.0 g marble.

mB and vB = mass and velocity of the 30.0 g marble.

vA´ and vB´ = final velocities of marble A and B respectively.

The kinetic energy of the system is also conserved:

kinetic energy before the collision = kinetic energy after the collision

1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

Then, replacing with the available data:

mA · vA + mB · vB = mA · vA´ + mB · vB´

0.010 kg · (-0.400 m/s) + 0.030 kg · 0.200 m/s = 0.010 kg · vA´ + 0.030 kg · vB´

2 × 10⁻³ kg · m/s =  0.010 kg · vA´ + 0.030 kg · vB´

Solving for vA´

0.2 kg · m/s - 3 kg · vB´ = vA´

Now, using conservation of the kinetic energy:

1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

0.010 kg · (-0.400 m/s)² + 0.030 kg · (0.200 m/s)² = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

Replacing vA´:

2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

2.8 × 10⁻³ kg · m/s = 0.010 kg · (0.2 kg · m/s - 3 kg · vB´)² + 0.030 kg · (vB´)²

(I will omit units from this point for more clarity in the calculations)

2.8 × 10⁻³  = 0.010  (0.2 - 3 · vB´)² + 0.03 · (vB´)²

2.8 × 10⁻³ = 0.010(0.04 - 1.2 vB´ + 9(vB´)²) + 0.03(vB´)²

divide by 0.01 both sides of the equation:

0.28 = 0.04 - 1.2 vB´ + 9(vB´)² + 3(vB´)²

0 = -0.28 + 0.04 - 1.2 vB´ + 12(vB)²

0 = -0.24 - 1.2 vB´ + 12(vB)²

Solving the quadratic equation:

vB´= 0.200  m/s

vB´ = -0.100  m/s

The first value is discarded because it is the initial velocity. Then, the final velocity of the 30.0 g marble is 0.100 m/s to the left.

The velocity of the 10.0 g marble will be:

0.2 kg · m/s - 3 kg · vB´ = vA´

0.2 kg · m/s - 3 kg · (-0.100 m/s) = vA´

vA´ = 0.500 m/s

The final velocity of the 10.0 g marble is 0.500 m/s to the right.

The change in momentum of the 30.0 g marble is calculated as follows:

Δp = final momentum - initial momentum

Δp = 0.030 kg · (-0.100 m/s) -(0.030 kg · 0.200 m/s) = -9.00 × 10⁻³ kg · m/s

The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

The change in momentum of the 10.0 g marble is calculated in the same way:

Δp = final momentum - initial momentum

Δp = 0.010 kg · 0.500 m/s -(-0.010 kg · 0.400 m/s) = 9.00 × 10⁻³ kg · m/s

The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

The change in kinetic energy for the 30.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.030 kg · (-0.100 m/s)² - 1/2 · 0.030 kg · (0.200 m/s)²

ΔKE = -4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J

The change in kinetic energy for the 10.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.010 kg · (0.500 m/s)² - 1/2 · 0.010 kg · (-0.400 m/s)²

ΔKE = 4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is 4.5 × 10⁻⁴ J

8 0
3 years ago
A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around
kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

h = 2.86 m

8 0
3 years ago
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