Answer:
Explanation:
Within a duration of 60 seconds, six waves are observed.
With a total of 6 waves,
this equates to 3 wavelengths.
As a result,
the period for each wavelength is calculated as 60 divided by 3.
Thus, period = 20 seconds.
According to the frequency-period relationship,
f = 1 / T
f = 1 / 20
f = 0.05 Hz
Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
Answer:
2.5 m
Explanation:
Billboard worker's weight = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension present in rope = 550 N
The total torques will be

The worker is positioned at 2.5 m