Explanation:
Formula for calculating the area of a rectangle A = Length *width
For statement A;
Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
Area of the rectangle = 2.536mm * 1.4mm
Area of the rectangle = 3.5504mm²
The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.
Area of the rectangle = 3.6mm² (to 2sf)
For statement B;
Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.
Area of the rectangle = 2.536mm * 1.41mm
Area of the rectangle = 3.57576mm²
Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.
Area of the rectangle = 3.58mm² (to 3sf)
Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.
No it won't. It'll vary inversely as the square of the separation.
Answer:
Q = 5 L/s
Explanation:
To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):
V: Volume (volumen) = 200L
t: time (tiempo) = 40 s
you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):
Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)
Answer:
Explanation:
Diffraction grating is used to form interference pattern of dark and bright band.
Distance between adjacent slits (a ) = 1 / 420 mm
= 2.38 x 10⁻³ mm
2.38 x 10⁻⁶ m
wave length of red light
= 680 x 10⁻⁹ m
For bright red band
position x on the screen
= n λD / a , n = 0,1,2,3 etc
D = distance of screen
putting n = 1 , 2 and 3 , we can get three locations of bright red band.
x₁ = λD / a
= 680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶
= .8 m
= 80 cm
Position of second bright band
= 2 λD / a
= 2 x 80
= 160 cm
Position of third bright band
= 3 λD / a
= 3 x 80
= 240 cm
Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite
