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2 years ago

1. An MR-2 starts from rest and accelerates westward to a speed of 27 m/s in

Physics

1 answer:

dangina [55]2 years ago

8 0

Answer:

a) 2.28 m/s²

b) 159.86 m

Explanation:

Part A:

u = 0 m/s

v = 27 m/s

t = 11.8 seconds

a = ?

Formula:

v = u + at

=> 27 = 0 + a(11.8)

=> 27 = a(11.8)

=> a(11.8) = 27

=> a = 27/11.8

=> a = 2.28 m/s²

Part B:

s = ?

Formula:

s = (v² - u²)/2a

s = (27² - 0²) /2(2.28)

= 729/4.56

= 159.86 m

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A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr

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The complete question is;

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) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)

Thus;

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

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

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Answer:

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