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Ivanshal [37]
3 years ago
12

An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

Calculate the possible values of vx.
Physics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

2 m/s and -2 m/s

Explanation:

The object travels with an angle of

60.0°

with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).

If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:

v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s

If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:

v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s

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The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
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Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

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If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

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The explosive force experienced by the shell inside the barrel is 23437500 newtons.

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Answer:

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