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3 years ago

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find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo

nent

2 answers:

Ksivusya [100]3 years ago

7 0

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

katrin [286]3 years ago

4 0

Answer: x = 0.237

y = -0.0789

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2 years ago

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall

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Explanation:

Given:

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Solving for $v_{0y}$ ,

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b) We can solve for the horizontal component of the velocity $v_{0x}$ as

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3 years ago

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3 years ago

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miss Akunina [59]

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4 years ago

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k

skelet666 [1.2K]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ( $U_{e}$ ), in joules, is equal to maximum translational kinetic energy ( $K$ ), in joules:

$U_{e} = K$

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$A$ - Amplitude, in meters.

$m$ - Object mass, in kilograms.

$v$ - Speed of the object at equilibrium, in meters per second.

If we know that $k = 450\,\frac{N}{m}$ , $m = 0.25\,kg$ and $v = 0.3\,\frac{m}{s}$ , then the amplitude of the motion is:

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k\cdot A^{2} = m\cdot v^{2}$

$A = v\cdot \sqrt{\frac{m}{k} }$

$A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }$

$A \approx 0.274\,m$

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ( $E$ ), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ( $k = 450\,\frac{N}{m}$ , $A \approx 0.274\,m$ )

$E = U_{e}$

$E = \frac{1}{2}\cdot k\cdot A^{2}$

$E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}$

$E = 16.892\,J$

The total energy of the object at any point of its motion is 16.892 joules.

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3 years ago