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3 years ago

Mr. Barcena reached a top speed of 5m/s in 2 seconds from rest, what was his average acceleration including units?

Physics

1 answer:

sashaice [31]3 years ago

4 0

Answer:

Explanation:

The equation for acceleration is

$a=\frac{v_f-v_0}{t}$ where vf is the final velocity and v0 is the initial velocity. Filling in:

$a=\frac{5-0}{2}$ so

a = 2.5 m/s/s

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The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s

Sholpan [36]

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds

use the energy conservation equation

$P_A +K_A=P_B+K_B$

Estimate the potential energy , component as position B as below

$P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft$

Estimate the kinetic energy , component as position A as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft$

Estimate the kinetic energy , component as position B as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2$

Substitute 20.5lb- ft for $P_A$

0.5lb-ft for $P_B$

0lb -ft for $K_A$

$0.0777V_2^2 for K_B$

$20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05$

= 16.05ft/sec

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