two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation
1 answer:
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Answer:
The rate of the boat in still water is 44 mph and the rate of the current is 4 mph
Explanation:
x = the rate of the boat in still water
y = the rate of the current.
Distance travelled = 120 mi
Time taken upstream = 3 hr
Time taken downstream = 2.5 hr
Speed = Distance / Time
Speed upstream
Speed downstream
Adding both the equations
The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>
According to Newton's 2nd law of motion:
F = m * a where F is the force applied in Newtons, m is the mass of the object in kg, and a is the acceleration of the object in m/
.
Therefore the force applied in this situation is simply:
F = 6 kg * 2.3 m/ = 13.8 N
Hope this helps!
F = m * a where F is the force applied in Newtons, m is the mass of the object in kg, and a is the acceleration of the object in m/
Therefore the force applied in this situation is simply:
F = 6 kg * 2.3 m/
Hope this helps!