A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

or

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

7 0

3 years ago

Why did Kenyatta want to distance himself from the Mau Mau? Check all that apply.

Savatey [412]

Answer:

B and C!

Explanation:

I just did it on edge. I hope this helps!

3 0

3 years ago

A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th

mr Goodwill [35]

elasticity stretches and can also return to it's normal size ..

8 0

2 years ago

3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second

jek_recluse [69]

Mass=0.25kg
Force=8N

$\\ \sf{:}\!\implies F=ma$

$\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}$

$\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}$

$\\ \sf{:}\!\implies Acceleration=32m/s^2$

5 0

2 years ago

‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak

denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, $\triangle \theta = 2\pi$

Torque, $\tau = 50 Nm$

Moment of Inertia, $I = 150 kg m^2$

If the wheel starts from rest, $w_{0} = 0 rad/s$

The angular displacement of the wheel can be given by the formula:

$\triangle \theta = \omega_0 t + 0.5 \alpha t^2$ ................(1)

Where $\alpha$ is the angular acceleration

$\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2$

To get t, put all necessary parameters into equation (1)

$2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s$

3 0

3 years ago