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ser-zykov [4K]
3 years ago
8

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

?
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
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Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

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Final after division

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Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

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b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

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    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

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The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

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d) impulse change

I₀ = Mv

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I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

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