Answer:
magnitude of the resultant of forces is 11.45 N
Explanation:
given data
force F1 = 6N
force F2 = 8N
angle = 240°
solution
we get here resultant force that is express as
F(r) =
..............1
put here value and we get
F(r) = 
F(r) = 11.45 N
so magnitude of the resultant of forces is 11.45 N
Ok so use trigonometry to work out the vertical component of velocity.
sin(25) =opp/hyp
rearrange to:
30*sin(25) which equals 12.67ms^-1
now use SUVAT to get the time of flight from the vertical component,
V=U+at
Where V is velocity, U is the initial velocity, a is acceleration due to gravity or g. and t is the time.
rearranges to t= (V+u)/a
plug in some numbers and do some maths and we get 2.583s
this is the total air time of the golf ball.
now we can use Pythagoras to get the horizontal component of velocity.
30^2-12.67^2= 739.29
sqrt739.29 = 27.19ms^-1
and finally speed = distance/time
so--- 27.19ms^-1*2.583s= 70.24m
The ball makes it to the green, and the air time is 2.58s
Answer:
Smaller refractive power
Explanation:
The refractive power of an eye is the extent to which it can converge or diverge the light rays.
Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.
A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.
Thus the near point also increases and the refractive power becomes smaller.
Answer: B to C
Explanation: The line is curving inwards, practically calculating the stance that it had went down. If it went straight across, it stayed the same till a specific point, furthermore calculating the bent line bending upwards is actually a partial-raise, conclude points B to C is most likely an un-even balance, meaning it had went down; or decreasing. B to C is the decreasing segment of this equation/problem (question).
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz