Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
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
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.
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.
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 />
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 />
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
Answer:
h = 4.04 m
Explanation:
Given that,
Mass of a child, m = 25 kg
The speed of the child at the bottom of the swing is 8.9 m/s
We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,
Put all the values,
So, the child is able to go at a height of 4.04 m.