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masha68 [24]
3 years ago
7

What is the answer to the question?

Physics
1 answer:
Masja [62]3 years ago
7 0

Answer:

Explanation:

The y component is measured by the horizontal component and the vertical component. Together they determine the magnitude of the vector. In this case, the y or vertical component is found by using the sine function.

<em>Formula</em>

Sin(angle) = vector resultant / y component component.

<em>Givens</em>

angle = 42 degrees.

vector = 419 degrees

<em>Solution</em>

sin(42) = y / 419                  Multiply both sides by 419

419 * sin(42) = 419  * y / 419

y = 419 * 0.6691

y = 280.37

<em>Note</em>

The vector is pointing downward so technically the vertical component should be negative.  I'm not sure what to tell you to answer. I would try - 280.37, but if the computer marks you wrong, try 280.37 (no minus sign).

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A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
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a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

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While the kinetic energy at any point is

K=\frac{1}{2}mv^2

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v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

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v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

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b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

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E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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