Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer:
Both balls have the same speed.
Explanation:
Janelle throws the two balls from the same height, with the same speed. Both balls will have the same potential and kinetic energy. Energy must be conserved. When the balls pass Michael, again they must have the same potential and kinetic energy.
Answer:
31.831 Hz.
Explanation:
<u>Given:</u>
The vertical displacement of a wave is given in generalized form as
<em>where</em>,
- A = amplitude of the displacement of the wave.
- k = wave number of the wave =
= wavelength of the wave.- x = horizontal displacement of the wave.
= angular frequency of the wave = 
.- f = frequency of the wave.
- t = time at which the displacement is calculated.
On comparing the generalized equation with the given equation of the displacement of the wave, we get,
therefore,
It is the required frequency of the wave.
