Ans: Beat Frequency = 1.97HzExplanation:
The fundamental frequency on a vibrating string is
<span> -- (A)</span>
<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>
Plug in the values in Equation (A)
<span>so </span>
<span> = 197.97Hz </span>
<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
a.
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N
By Newton's second law
b.By Newton's second law
Substitute the values
Hence, the force in the coupling between the dogs and the sled=1058 N
If it takes
seconds to reach the car, then the distance
is
.
The bear's distance from the tourist's starting point is
For maximum
, we set the equations equal to each other:
so the distance is
