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tatiyna
3 years ago
5

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

ed away from the equilibrium position (x = 0) a distance of 17.5 cm and released. It then oscillates in simple harmonic motion with a frequency of 8.38 Hz. What is the mass’s maximum speed during this motion?
Physics
1 answer:
sdas [7]3 years ago
7 0

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

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0.22 L of pancake syrup has a mass of 33 g.
katrin2010 [14]

Answer:

a. 150 g/L

b. 75 g

c. 120 mL

Explanation:

a. 33g/0.22L=150 g/L

b. 33g/0.22L=150 g/L

150 g/L*0.5L=75g

c. 0.22L/33g=0.006667L/g

0.006667L/g*18g=0.12L

0.12L*1000=120mL

6 0
3 years ago
A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride
oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0
3 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
Secondary evidence is the basis for drawing scientific conclusions is this true ir false
Leto [7]
The statement: "secondary evidence is the basis for drawing scientific conclusions" is definitely false. Secondary evidence is the body of information obtained to prove the existence of unknown or missing primary evidence. In drawing scientific conclusions, primary evidence is needed.
7 0
3 years ago
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Debbie exerts applied force on the first cart. The first cart exert applied force on the second cart.



- Marlon Nunez

6 0
3 years ago
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