Answer:
so what ever one is new here
Answer:
19320 kg/m³
Explanation:
density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.
The formula of density is given as,
D = m/v ......................... Equation 1.
Where D = Density of the gold, m = mass of the gold, v = volume of the gold.
Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.
Amount of water displace = 27.2 - 25 = 2.2 mL.
Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³
Substitute into equation 1
D = 0.042504/0.0000022
D = 19320 kg/m³
Hence the density of the piece of gold = 19320 kg/m³
Answer:
acceleration = 0.8181 m/s²
Explanation:
given data
mass = 1.1 kg
apart d = 1 m
charge q = 10 μC
to find out
What is the initial acceleration
solution
we know that acceleration is
acceleration = 
.................1
here force = 
here q1 q2 is charge and r is distance and Coulomb constant k = 9 × 
Nm²/C²
force = 
force = 0.9 N
so from equation 1
acceleration = 
acceleration = 0.8181 m/s²
Answer:
A) attached file
B) attached file
C) attached file
D) Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Explanation:
Given that the lifts are analogous to batteries, and the runs are analogous to resistors.
So from all the figures. The resistors represent the runs while the lift represents the battery.
Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Please find the attached file for the sketch
Answer:
5 fringes option C
Explanation:
Given:
- The wavelength of blue light λ = 450 nm
- The split spacing d = 0.001 mm
Find:
How many bright fringes will be seen?
Solution:
- The relationship between the wavelength of the incident light, grating and number of bright fringes seen on a screen is derived by Young's experiment as follows:
sin(Q) = n* λ / d
Where, n is the order of bright fringe. n = 0, 1, 2, 3, ....
- We need to compute the maximum number of fringes that can be observed with the given condition and setup. Hence we will maximize our expression above by approximating sin(Q).
sin(Q_max) = 1
Q_max = 90 degree
- Hence, we have:
n = d / λ
- plug values in n = 0.001 *10^-3 / 450*10^-9
n = 2.222
- Since n order number can only be an integer we will round down our number to n = 2.
- Hence, we will see a pair of bright fringes on each side of central order fringe.
- Total number of fringes = 2*2 + 1 = 5 fringes is total ... Hence, option C
