frequency is equal to number of oscillations or vibrations upon time
therefore,
check picture
PLEASE HELP ME!!!
1 answer:
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Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
The anchor points are 78.37 ft and 111.99 ft
Explanation:
If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then
a) Sine (α) = b/c it means opposite side/hypotenuse
b) Cosine (α)= a/c, it means adjacent side/hypotenuse
c) Tangent (α) = b/a opposite side/adjacent side.
Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:
a² + b² = c²
As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).
Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.
To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x
⇒ x = 78.37 ft.
And finally, to find y we can apply Pythagoras theorem
(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft
Summarizing, the anchor points are 78.37 ft and 111.99 ft
Answer:
correct answer is C
Explanation:
The time constant of an RC circuit is
τ = RC
so to find the capacitance
C = τ/ R
C = 2.150 / 5.20 10³
C = 4.13 10⁻⁴ F
to find the error we use the worst case
ΔC = |
the absolute value guarantees that we find the worst case, we evaluate the derivatives
ΔC = 1 /R Δτ + τ/R² ΔR
the absolute values of the errors are
Δτ = 0.002 s
ΔR = 0.3 kΩ
we substitute
ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³
ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵
ΔC = 1.77 10⁻⁵ F
the uncertainty or error must be expressed with a significant figure
ΔC = 2 10⁻⁵ F
the percentage error is
Er% =
Er% =
Er% = 4.8%
the correct answer is C