They are formed by evaporation
Answer:
0.21486 mm
Explanation:
The formula for the maximum intensity is given by;
I = I_o•cos²(Φ/2)
Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)
Where;
y is the distance from the central maximum
d is the distance between the slits
λ is the wavelength
L is the distance to the screen
Thus;
I = I_o•πdy/(λL)
We are given;
d = 0.05 mm = 0.5 × 10^(-3) m
λ = 540 nm = 540 × 10^(-9) m
L = 1.25 m
I/I_o = 50% = 0.5
From earlier, we saw that;
I = I_o•πdy/(λL)
We have I/I_o = 0.5
Thus;
I/I_o = πdy/(λL)
Plugging in the relevant values;
0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)
Making y the subject, we have;
y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))
y = 0.00021486 m
Converting to mm, we have;
y = 0.21486 mm
The object did not move / was kept from moving.
G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
To get the density, we divide mass by volume: 2.50 g / 3.7125 cm^3 = 0.6734 g/cm^3 = 0.6734 g/mL
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
