If a ball is given an acceleration of 3.0 meters/second^2 while being pushed with a force of 0.75 newtons, what is the mass of t
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To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as
c = Propagation speed of waves in the medium
= Speed of the receiver relative to the medium
= Speed of the source relative to the medium
Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
Velocity of car
velocity of motor
Velocity of sound
Frequency emited by the source
Replacing we have that
Therefore the frequency that hear the motorcyclist is 601.7Hz
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 = = 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 = 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:
= 20 liters
(b) 1 m³ = 1,000,000 cm³
0.02 m³ = how many cm³ ?
Cross-multiplying gives;
= 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
- in a displacement can
- The mass of liquid which overflow is 20 g
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; = 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