Answer:
A it is less dense than water
Explanation:
The density, of a substance is its mass per unit volume
"<span>The pointer will be above the zero mark" is the one statement among the following choices given in the question that is true. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your great help.</span>
Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
Answer:
The answer to your question is : vf = 15.18 m/s
Explanation:
Data
vo = 24 m/s
d = 120 m
vf = ? when d = 60.0 m
Formula
vf² = vo² + 2ad
For d =100m
a = (vf² - vo²) / 2d
a = (0 -24²) / 2(100)
a = -576/200
a = 2.88 m/s²
Now, when d = 60
vf² = (24)² - 2(2.88)(60)
vf² = 576 - 345.6
vf² = 230.4
vf = 15.18 m/s
Path for transmitting electric current. An electric circuit includes a device that gives energy to the charged particles constituting the current, such as battery or a generator; devices that use current, such as lamps, electric motors, or computers; and the connecting wires or transmission lines
