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In-s [12.5K]
2 years ago
6

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

f it takes the cart 1.5 seconds to reach 8.2 cm/s, what is the acceleration of the cart? Round your answer to the nearest tenth. cm/s2
Physics
2 answers:
zheka24 [161]2 years ago
5 0
The first thing we are going to do for this case is write the equations of movement of the car.
 We have then:
 vf = a * t + vo
 Substituting values:
 8.2 = a * (1.5) + (3.5)
 Clearing the acceleration we have:
 a = (8.2-3.5) / (1.5)
 a = 3.1 m / s ^ 2
 Answer: 
 the acceleration of the cart is: 
 a = 3.1 m / s ^ 2
9966 [12]2 years ago
5 0

Answer:

3.1cm/s^2

Explanation:

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A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
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Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

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Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

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(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

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where
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v its tangential velocity
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Substituting the data of the problem, we find
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