6) A school bus moves at 2.2 m/s as it pulls away from the bus stop what is the speed of the bus 20 s later after its speed has increased by 5.225 m/s?
7) The height of a computer screen is 31.25 cm and its width is 47 cm. a) is the area of the screen known to one, two, three, or four significant figures?
b) calculate the area of the screen giving your answer with the correct number of significant figures
8) a parking lot is 144.3 m long and 47.66 m wide a) what is the perimeter of the lot b) what is the area?
6. V = 2.2 + 5.225 = 7.425 m/s.
7. a. 4 sig. fig.
b. A = 47 * 31.25 = 1469 cm^2.
8. a. P = 2L + 2W.
b. A = L*W.
6) After the speed of the bus increases by 5.225 m/s, it will be going faster than a toddler chasing after an ice cream truck. In fact, it will be moving at a speed of 2.2 m/s + 5.225 m/s, which equals 7.425 m/s. So, buckle up and hold on tight!
7) a) The area of the computer screen is known to three significant figures.
b) To calculate the area of the screen, we simply multiply the height (31.25 cm) by the width (47 cm), giving us an area of 1468.75 square centimeters. Since both the height and width were known to three significant figures, the answer should also be reported with three significant figures. Therefore, the area of the screen is 1460 cm². It's like a little canvas for digital masterpieces!
8) a) The perimeter of the parking lot is the sum of all its sides, just like the sum of all the times you've hit snooze on your alarm clock. For a lot that is 144.3 m long and 47.66 m wide, the perimeter would be 2(144.3 m) + 2(47.66 m), or 288.6 m + 95.32 m, which equals 383.92 m. That's quite a stroll.
b) To find the area of the parking lot, you simply multiply the length (144.3 m) by the width (47.66 m). The area is like a giant space waiting to be filled with cars doing questionable parking jobs. So, the area of the parking lot is 6888.798 m². Make sure to leave some space for clowns to juggle!
6) To find the speed of the bus 20 seconds later, we need to calculate the final speed.
Given:
Initial speed (u) = 2.2 m/s
Change in speed (Δv) = 5.225 m/s
Time (t) = 20 s
The final speed (v) is given by the equation:
v = u + Δv
Substituting the values:
v = 2.2 m/s + 5.225 m/s
v = 7.425 m/s
Therefore, the speed of the bus 20 seconds later is 7.425 m/s.
7)
a) The area of the screen is calculated by multiplying the height and width. As both dimensions are given to two significant figures, the area of the screen is also known to two significant figures.
b) To calculate the area of the screen, we multiply the height and width:
Area = height × width
Area = 31.25 cm × 47 cm
Calculating the area:
Area = 1468.75 cm^2
Since both the height and width were given with four significant figures, the area should also be reported with four significant figures.
Therefore, the area of the screen is 1468.8 cm^2 (rounded to four significant figures).
8)
a) The perimeter can be calculated by adding all the sides of the lot.
Perimeter = 2(length + width)
Perimeter = 2(144.3 m + 47.66 m)
Calculating the perimeter:
Perimeter = 2(191.96 m)
Perimeter = 383.92 m
Therefore, the perimeter of the lot is 383.92 m.
b) The area of the lot is calculated by multiplying the length and width.
Area = length × width
Area = 144.3 m × 47.66 m
Calculating the area:
Area = 6883.718 m^2
Therefore, the area of the lot is 6883.7 m^2 (rounded to one decimal place).
6) To find the speed of the bus 20 seconds later, we need to add the initial speed of the bus to the increase in speed. Given that the initial speed is 2.2 m/s and it increases by 5.225 m/s, we can add these two values together to find the final speed.
Initial speed = 2.2 m/s
Increase in speed = 5.225 m/s
Final speed = Initial speed + Increase in speed
Final speed = 2.2 m/s + 5.225 m/s = 7.425 m/s
Therefore, the speed of the bus 20 seconds later after its speed has increased is 7.425 m/s.
7) a) The area of the screen is calculated by multiplying the height and the width. The height is given as 31.25 cm, and the width is given as 47 cm.
Area = height x width
Area = 31.25 cm x 47 cm = 1468.75 cm²
Since both the height and the width are given with two decimal places, the area of the screen is known to two significant figures.
b) To calculate the area of the screen with the correct number of significant figures, we need to round the result to the appropriate number of significant figures. In this case, the area is known to two significant figures, so we need to round to two significant figures.
Area = 1468.75 cm² (to four significant figures)
Area = 1500 cm² (to two significant figures)
Therefore, the area of the screen, rounded to the correct number of significant figures, is 1500 cm².
8) a) The perimeter of the parking lot can be calculated by adding up the lengths of all its sides. Given that the length is 144.3 m and the width is 47.66 m, we can use these values to find the perimeter.
Perimeter = 2(length + width)
Perimeter = 2(144.3 m + 47.66 m)
Perimeter = 2(191.96 m)
Perimeter = 383.92 m
Therefore, the perimeter of the parking lot is 383.92 m.
b) The area of the parking lot is calculated by multiplying its length and width. Given that the length is 144.3 m and the width is 47.66 m, we can use these values to find the area.
Area = length x width
Area = 144.3 m x 47.66 m
Area = 6893.358 m²
Therefore, the area of the parking lot is 6893.358 m².