Answer:
θ = 60° or θ = 120°, which in radians is equivalent to θ = π/3 rad or θ = 2π/3 rad.
Explanation:
The given equation is sin θ = (√3)/2
Solving that equation is finding the value of the angle, θ, whose sine is (√3)/2.
The function that returns the value of the angle whose sine is (√3)/2 is the inverse function of the sine and it has a special name: arc sine or arcsin.
Hence, θ = arcsin (√3)/2.
You can use your knowledge of the notable angles to solve for that equation.
1) The function sine is positive in first and second quadrants.
2) The angles in the first quadrant go from 0° to 90°.
3) The sine of 60° (√3)/2, Hence the first value of θ is 60°
4) The angles in the second quatrant go from 90° to 120°. 60° is the reference angle in the second quadrant, and the angle searched is
- 180° - reference angle = 180° - 60° = 120°. So, 120° is the other solution of the equation.
5) You can convert both angles to radians using the equivalence
- π radians = 180° ⇒ 1 = π rad / 180°
- 60° = 60° × π rad / 180° = π/ 3 rad
- 120° = 120° × π rad / 180° = 2π/3 rad
6) You can verify that sin 60° = (√3)/2 = sin 120° .
1.0g/cm3 means that the mass of one cm3 is 1.0g
The easiest method to use is the rule of three, and let x be the mass of 10.0 cm3 of water
1g -- > 1.0 cm3
x --> 10.0 cm3
x= (10*1)/1
x=10.0 g
So the mass of 10.0 cm3 of water is 10.0g
Hope this Helps! :)
Answer:
Step-by-step explanation:
Not a clear list of options and/or reference frame
Probably 0.5 if angle t is measured from the positive x axis.
Answer:
(a) v = 1536640/ m^2.
(b) (i) 1186 m/s.
(ii) 52.4g .
Step-by-step explanation:
(a) The equation is v = k / m^2 where k is the constant of variation.
When m = 49, v = 640 so 640 = k / 49^2
k = 640 * 49^2
= 1,536,640.
Therefore the relation is v = 1536640/ m^2.
(b) (i)
When m = 36g, the speed v = 1536640 / 36^2
= 1186 m/s.
(ii) When the speed is 560 m/s :-
560 = 1536640 / m^2
m^2 = 1536640 / 560
= 2744
mass m = √2744
= 52.4g
<em>30, 20 and 10 are three numbers that are less than 40 and have 2 and 5 as factors.</em>
