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Vladimir79 [104]
2 years ago
9

Would someone just answer this now?

Physics
1 answer:
Vitek1552 [10]2 years ago
6 0

Answer:

<em>x=36.5 m</em>

Explanation:

<u>Rectangular Components of a Vector</u>

A vector can be expressed as an ordered pair (x,y) where x is the x-coordinate, and y is the y-coordinate of a rectangular system.

It can also be given as a magnitude-angle pair (r,θ) where θ is measured from the positive x-axis. It's also known as polar coordinates.

The conversion from polar to rectangular coordinates is:

x=r\cos \theta

y=r\sin\theta

We are given the vector as a magnitude r=47.3 m and at an angle θ=39.4°. The x-component of the vector is:

x=47.3\cos (39.4^\circ)

Using a scientific calculator to find the trigonometric function:

x=47.3\cdot 0.7716

x=36.5 m

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Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

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