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2 years ago

What is simple harmonic oscillation? express your answer verbally, mathematically, and visually.

1 answer:

3 0

Simple harmonic motion is the motion in objects where the restoring for d is directly proportional to the body displacement.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion a type of periodic motion where the force that restores moving object to it's right position is directly proportional to the magnitude of body's displacement and this normally acts towards the object's equilibrium position.

The formulate for simple harmonic motion are

,F = −kx, where F is the force, x is the displacement, and k is a constanst. An example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring.

Simple harmonic oscillator can be represented by the equation x ( t ) = A cos ⁡ ( 2 π f t ) x(t) = A\cos(2\pi f t) x(t)=Acos(2πft)x,

where t is period.

x is displacement

A is amplitude.

Learn more about simple harmonic motion here.

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Because Mars is further from the sun than Earth is, thus the gravitational pull is not as great on Mars as it is on Earth, making us lighter :)

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4 years ago

If the mass of an object is 35kg on venus what is its mass on the sun

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3 years ago

Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu

PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0

3 years ago

A rotating wheel requires 5 s to rotate 38 revolutions. Its angular velocity at end of time interval 5-s is 79 rad/s. What is th

Lera25 [3.4K]

"Rotating wheel" is meant to indcate that the wheel is already rotating at the start. Denote the initial angular velocity by ω₀. Then the angular displacement θ at time t is

θ = ω₀t + 1/2 αt ²

while the angular velocity ω is

ω = ω₀ + αt

It takes 5 s for the wheel to rotate 38 times, or turn a total of (2π rad/rev) (38 rev) = 76π rad, as well as to reach an angular velocity of 79 rad/s, so that

76π rad = ω₀ (5 s) + 1/2 α (5 s)²

79 rad/s = ω₀ + α (5 s)

Solve the second equation for ω₀ and substitute into the first equation, then solve for α :

ω₀ = 79 rad/s - α (5 s)

==> 76π rad = (79 rad/s - α (5 s)) (5 s) + 1/2 α (5 s)²

==> 76π rad = (79 rad/s) (5 s) - 1/2 α (5 s)²

==> 1/2 α (5 s)² = (79 rad/s) (5 s) - 76π rad

==> α = ((79 rad/s) (5 s) - 76π rad) / (1/2 (5 s)²)

==> α ≈ 12.5 rad/s²

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andrew11 [14]

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4 years ago