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3 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

1 answer:

Anastaziya [24]3 years ago

6 0

Answer:

As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.

This will mean that the denser objects will always go to the bottom.

This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.

There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.

The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.

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Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

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velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

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c )

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Water flows over a section of niagara falls at the rate of 1.2x10^6 and falls 50.0 m. how much power is generated by the falling

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The solution for this is:

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4 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

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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