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Lesechka [4]
3 years ago
9

Three balls are thrown off the top of a building, all with the same speed but with different launch angles (position is given in

meters and time is given in seconds). The components of the initial velocities are given.
The blue ball has an initial velocity of (6 m/s, 8 m/s).
The green ball has an initial velocity of (10 m/s, 0 m/s).
The red ball has an initial velocity of (8 m/s, -6 m/s).

a. Rank the three balls according to which one hits the ground first.
b. Rank the three balls according to which one has the greatest speed the instant before impact with the ground.
c. Calculate the speed of each of the balls the instant before impact with the ground.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

A. Red, Green, Blue

B. None. They all hit the ground with the same speed.

C. Final Kinetic Energy = Initial Gravitational Potential Energy

Explanation:

A. This one is actually just common sense. If you think about it, the object that is being thrown downward will reach the ground first. The object thrown upwards will hit the ground the last. If you don't believe me, use just the vertical component of velocity and use acceleration = 9.8 m/s^2. Find the time it takes for an arbitrary displacement each ball will travel. You will find that the one pointed downward will reach the ground in the shortest amount of time. Just do the same math over for the rest of the balls and you can find the time it takes for each ball to hit the ground.

B. This one is actually really simple if you think about it. Using the Work-Energy theorem, using the system consisting a ball and the earth there will be no external forces doing work on the system. Since each ball is at some height "h" and the same initial speed, they will all have the same amount of energy in the beginning. If we set the ground as our reference point for gravitational potential energy, then there will be no gravitational energy as the height at that point will be 0. They all must have the same kinetic energy at the ground.

C.

Mechanical Energy Final = Mechanical Energy Initial + Net Work from non-conservative forces

Kinetic Energy Final + Gravitational Potential Energy Final = Kinetic Energy Initial + Gravitational Potential Energy Initial

I am using the symbol " ' " to indicate final speeds and heights

0.5*m*v'^2 + mgh' = 0.5*m*v^2 + mgh

0.5*m*v'^2 + 0 = 0.5*m*v^2 + mgh

(cancel out the mass)

0.5*v'^2 = 0.5*v^2 +gh

Since the "v" is the same initially for all three balls and "g" and "h" are the same for all three balls. The final speed v' must be the same for all three situations regardless of mass.

If you have any questions feel free to comment again. I'll try to clarify.

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Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

  • An empty masuring cylinder has a mass of 500 g.
  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
  • The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

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The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

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