How many variables should there be in a well-designed experiment?
1 answer:
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Answer:
body position 4 is (-1,133, -1.83)
Explanation:
The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by
x_cm = 1 /M ∑ m_{i}
y_cm = 1 /M ∑ y_{i} mi
Where M is the total mass of the body, mi is the mass of each element
give us the mass and position of this masses
body 1
m1 = 2.00 ka
x1 = 0 me
y1 = 0 me
body 2
m2 = 2.20 kg
x2 = 0m
y2 = 5 m
body 3
m3 = 3.4 kg
x3 = 2.00 m
y3 = 0
body 4
m4 = 6 kg
x4=?
y4=?
mass center position
x_cm = 0
y_cm = 0
let's apply to the equations of the initial part
X axis
M = 2.00 + 2.20 + 3.40
M = 7.6 kg
0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)
x4 = -6.8 / 6
x4 = -1,133 m
Axis y
0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)
y4 = -11/6
y4 = -1.83 m
body position 4 is (-1,133, -1.83)
Answer:
Vc = 2.41 v
Explanation:
voltage (v) = 16 v
find the voltage between the ends of the copper rods .
applying the voltage divider theorem
Vc = V x ()
where
- Rc = resistance of copper =
(l = length , a = area, ρ = resistivity of copper)
- Ri = resistance of iron =
(l = length , a = area, ρ₀ = resistivity of copper)
Vc = V x ()
Vc = V x ()
Vc = V x ()
where
- ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
- ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter
Vc = 16 x ()
Vc = 2.41 v
Answer:
The vibrational frequency of the rope is 5 Hz.
Explanation:
Given;
number of complete oscillation of the rope, n = 20
time taken to make the oscillations, t = 4.00 s
The vibrational frequency of the rope is calculated as follows;
Therefore, the vibrational frequency of the rope is 5 Hz.