Answer:
a) 
b) 
c) 
d) 
Explanation:
<u>Given equation of pressure variation:</u>
![\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%281.78%5C%20Pa%29%5C%20sin%5C%20%5B%280.888%5C%20m%5E%7B-1%7D%29x-%28500%5C%20s%5E%7B-1%7D%29t%5D)
We have the standard equation of periodic oscillations:
<em>By comparing, we deduce:</em>
(a)
amplitude:
(b)
angular frequency:
∴Frequency of oscillations:
(c)
wavelength is given by:
(d)
Speed of the wave is gives by:
Explanation:
Let us assume that the maximum allowable horizontal distance be represented by "d".
Therefore, torque equation about A will be as follows.
d = ![\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20%5Ctimes%2075%20%5Ctimes%20%280.7%2B0.15%2B0.15%29%20-%2060%20%5Ctimes%200.15%20-%20252%20%5Ctimes%200.15%20%5Ctimes%202%5D%7D%7B252%7D)
d = 0.409 m
Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
34. The element would have 34 protons. Also 34 electrons.
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
