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stiv31 [10]
3 years ago
10

What is the mass of an object moving at a speed of 9.0 m/s with a kinetic energy of 200 J?

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0
Ek= 1/2mv²
200=1/2 m(9 ²)
200=1/2(81)m
200=40.5m
m=200/40.5
m=4.93kg

Answer: 4.9 kg (second option)
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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
LekaFEV [45]

Answer:

   (h₁-h₂) = 2.30 10² m

Explanation:

The pressure depends on the height with the formula

          P = P_atm + rho g h

Let's apply this expression for the building

         P₁ = P_atm + rho_air g h₁

        P₂ = P_atm + rho_air g h₂

Subtract

        P₁ - P₂ = roh_air g (h₁ –h₂)

         

         

The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

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          (h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

         (h₁-h₂) = 13600 / 1.18 (695-675)

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Let's reduce to meter

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         (h₁-h₂) = 2.30 10² m

4 0
3 years ago
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11111nata11111 [884]

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t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

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k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

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(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

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