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charle [14.2K]
3 years ago
11

the force shown in figure 7-15 moves an object from x=0 to x=0.75 m. How much work is done by the force?

Physics
1 answer:
Phantasy [73]3 years ago
6 0
Work is force multiplied by the distance the force moves the object
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When two objects collide their momentum after the collision is explained by what?
Gnesinka [82]
<span>When two objects collide their momentum after the collision is explained by</span> the conservation of momentum
5 0
3 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
3 years ago
If a person wants to move a heavy box across the room, what is required to get the box in motion? apply any force apply a force
garik1379 [7]
If a person want to move a <span>heavy box across the room, he must apply a force that is greater than starting frictional force acted on the heavy box in order to get the box in motion. the frictional force is equal to coefficient of friction times the normal force, and normal force is approximately the weight of the object. so the force that must be applied must be greater than the wieght of the object.</span>
4 0
3 years ago
Read 2 more answers
Sinisimulan ko<br> Kaya Ko<br> Gagawin ko<br> Nalaman ko<br> Natututo Ako
ozzi

Answer:

what!!!!!!

Explanation:

didn't got the point

7 0
3 years ago
The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o
kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

Y=Ae^{-kt}

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}

Y=Ae^{-kt}

\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years

hence time taken is equal to 14,156 years

5 0
3 years ago
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