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Gwar [14]
3 years ago
14

A 1.5-kg ball is throw at 10 m/s. What is the balls momentum?

Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer : The momentum of ball is, 15 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

where,  

p = momentum  = ?

m = mass  = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

Therefore, the momentum of ball is 15 kg.m/s

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The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
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V = volume
T = temperature.

Note that
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1 L = 10⁻³ m³

Given:
At ground level,
p₁ = 752 mm Hg
     = (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
     = 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
     = 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
     = 300.8 K

At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
     = 9.7326 x 10³ Pa
T₂ = 235 K

If the volume at  36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
     = (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
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A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit
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Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

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Where Q(t) is the charge.

<em>The general solution of a linear equation is given as:</em>

<em>y(x) = c_1e^{-ax}+c_2e^{-bx}</em>

Applying this definiton in our differential equation we have that

Q(t) = C_1e^{at}+C_2e^{bt}

To find b and a we use the first equation and find the roots:

r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}

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Then we have

Q(t) = C_1e^{-1000t}+C_2e^{-500t}

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The we can replace at the equation and we have that the Charge at any moment is given by,

Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}

If we obtain the derivate we find also the Current, then

I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}

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