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melisa1 [442]
3 years ago
8

Assuming that the uncertainty in mass is negligible, and the uncertainty in velocity is 5 %, what is the uncertainty in the mome

ntum of your two-cart system?
Physics
1 answer:
Flura [38]3 years ago
8 0

Answer:

 Δp = 0.05 p

Explanation:

The moment is

      p = m v

The uncertainty of the moment is

       Δp = dp/dm Δm + dp/dv Δv

Like the uncertainty in the mass is zero

       Δm = 0

       ΔP = m Δv

We divide for the moment

       Δp / p = Δv / v

  They do not indicate that Δv / v = 0.05

       Δp / p = 0.05

       Δp = 0.05 p

In the case of a system consisting of two cars

      p = m₁ v₁ + m₂ v₂

     Δp = dp / dv₁ Δv₁ + dp / dv₂ Δv₂

     Δp = m₁ Δv₁ + m₂ Δv₂

     Δv₁ / v₁ = 0.05

     Δv₁ = 0.05 v₁

     Δv₂ / v₂ = 0.05

     Δv₂ = 0.05 v₂

We replace

       Δp = m₁ 0.05 v₁ + m₂ 0.05 v₂

       Δp = 0.05 p

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Answer:

R=803k\Omega

Explanation:

We have the following information,

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We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

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3 years ago
Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them
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Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

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Calculate the density of water in kg/m3. Express your answer in kilograms per cubic meter using three significant figures.
Svet_ta [14]

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, the density of water is generally known to be 997 kg/m³. The formula used to determine density is mass ÷ volume. In the case of the density of water, the mass is measured in kilograms (kg) while the volume is measured in cubic meter (m³).

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 17601760 × 103 seconds (ab
Yuliya22 [10]

Answer:

The value is a_r  = 3.81 *10^{-3} m/s^2

Explanation:

Generally the moon's radial acceleration is mathematically represented as

a_r  =  r *  w^2

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w =\frac{2 \pi }{ T}

substituting 1760 * 10^3 \ seconds for T(i.e the period of the moon ) we have

w =\frac{2  * 3.142 }{  1760 * 10^3}

=> w = 3.57 *10^{-6} \  rad/s

From the question r(which is the radius of the orbit ) is evaluated as

r =  R + H

substitute 3.60 * 10^6 m for R and 295.0 * 10^6  \  m H

       r =  295.0 * 10^6   +3.60 * 10^6

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So

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      a_r  = 3.81 *10^{-3} m/s^2

4 0
3 years ago
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