The work that is required to increase the speed to 16 knots is 14,176.47 Joules
If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;
5.44×10^3 kg = 12 knots
For an increased speed to 16knots, we will have:
x = 16knots
Divide both expressions
To get the required work done, we will divide the mass by the speed of one knot to have:
Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules
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Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
Answer:
the amplitude of the subsequent oscillations is 0.11 m
the period of the subsequent oscillations is 1.94 s
Explanation:
given Information:
the mass of air-track glider, = 750 g = 0.75 kg
spring constant, k = 13.0 N/m
the mass of glider, = 200 g = 0.2 kg
the speed of glider, = 170 cm/s = 1.7 m/s
the amplitude of the subsequent oscillations is A = 0.11 m
according to mechanical enery equation, we have
where
A is the amplitude and is the final speed.
to find , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.
= 0, thus
(0.75+0.2) = (0.75)(0)+(0.2)(1.7)
0.95 = 0.34
= 0.36 m/s
Now we can calculate the amplitude
A = 0.11 m
the period of the subsequent oscillations is T = 1.94 s
the equation for period is
T = 2π
T = 2π
T = 1.94 s