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Sedaia [141]
3 years ago
9

A 72.0-kg object hits the ground at a velocity of 79.0 m/s. Neglecting air resistance, which relationship allows you to calculat

e the object’s potential energy before the fall? A. PE = mgh B. (PE)beginning = (KE)end C. Mechanical Energy = PE + KE D. KE = 1/2 mv2
Physics
1 answer:
kondaur [170]3 years ago
8 0
<h3>Answer:</h3>

B. (PE)beginning = (KE)end

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of an object is 72.0 kg
  • Velocity of the body before it hits the ground is 79.0 m/s

We are required to determine the relationship between the potential energy and kinetic energy before the fall.

  • When an object is at the highest point, it has maximum potential energy and minimum kinetic energy.
  • This is because potential energy is directly proportional to the height of an object above the earth's surface.
  • On the other hand, when an object attains the highest speed it has maximum kinetic energy and minimum kinetic energy.

In this case;

  • The velocity of the object when hitting the ground is maximum and thus the object will have maxim,um kinetic energy.
  • As the object falls towards the ground the potential energy is being converted to kinetic energy.
  • Therefore, the potential energy at the beginning will be equal to the kinetic energy at the end when the object is on the ground.
  • We can therefore, conclude that, (PE)beginning = (KE)end
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<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 />

  • A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
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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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