A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when

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.

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

(c) Formula for the total energy is given by;

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

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

hope this : )

explanation:

i think the answer is a, hope this : )