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2 years ago

The law of inertia states that a moving object will:.

Physics

1 answer:

Paha777 [63]2 years ago

5 0

Answer: will stay in motion unless acted upon by a force

Explanation: I gotchu

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The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle

kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0

2 years ago

A mass of gas under constant pressure occupies a volume of 0.5 m3 at a temperature of 20°C. Using the formula for cubic expansio

Tanzania [10]

Without a change in pressure

V/T = V/T
Before change = after change - expansion

x = volume at m3

(0.5)/20 = x/45

.025 = x/45

.025 * 45 = x

1 = x

The volume of this gas is 1 m3.

Hope this helps :) also the gas expanded .5 of m3.

4 0

3 years ago

Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

iogann1982 [59]

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

$\large{\boxed{\bold{\sum F=0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0\\\\ \sum F_y = 0$

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

$\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g$