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Ne4ueva [31]
3 years ago
8

1. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to Newton's Third Law?  Action Force:

_______________ Reaction Force: _____________
Physics
1 answer:
Natali [406]3 years ago
7 0
The Action Force of this scenario is the pushing force of the Diver. The Reaction Force is the raft pushing back on the diver. 

The Third Law of Motion states that "For every action, there is an equal and opposite reaction." Now when the diver dives off the raft, the raft is also pushing the same amount of force as the diver did as he dives off. The diver will then move forward and the raft on the other hand will move backwards.

The movement of the raft shows the opposite force. It will move backwards depending on how strong the diver will push off on the raft. And the amount of force he pushes on it, the raft will exert the same force so the stronger the force of the diver, the farther he will go because the raft will push him in that same direction as it goes backwards. 
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6 0
3 years ago
Which can make 4 covalent bonds? Carbon, Silicon, or both?
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Answer:

I would say both

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5 0
3 years ago
Explain how the pressure at the bottom of a container depends on the container shape and the fluid height
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The pressure at the bottom of a column of fluid in a container
depends only on the depth of the fluid, not on the shape of the
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3 0
3 years ago
Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on
EleoNora [17]

Answer:

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = +40 nC, q₂ = +250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r  and the distance between q₁ and q₃ is  r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r +  kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r +  kq²/2r + kqQ/r = qk/r(2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = -ΔK

0 - qk(2Q + q/2)/r = -(1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk(2Q + q/2)/r = -1/2mv²

v = √[2qk(2Q + q/2)/mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 × 40 × 10⁻⁹ × 9 × 10⁹(2 × 250 × 10⁻⁹ + 40 × 10⁻⁹/2)/2 × 10⁻³ × 2 × 10⁻²]

v = √[360(500 × 10⁻⁹ + 20 × 10⁻⁹)/2 × 10⁻⁵]

v = √[720(520 × 10⁻⁹)/4 × 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

d.

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s

8 0
3 years ago
Question 10(Multiple Choice Worth 2 points)
Svetach [21]
P = mv

P momentum
m mass
v velocity
3 0
3 years ago
Read 2 more answers
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