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Helga [31]
3 years ago
12

Consider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST appli

es?
A) A book is at rest on a tabletop.
B) A physics student rests a backpack upon one shoulder.
C) A girl hangs by both hands, motionless, from a trapeze.
D) A girl falls slowly to Earth while strapped to a parachute.

Physics
2 answers:
Naily [24]3 years ago
8 0

C.)    A girl hangs by both hands, motionless, from a trapeze.

mestny [16]3 years ago
4 0

Answer:

C) A girl hangs by both hands, motionless, from a trapeze.

Explanation:

The free body diagram in the figure gives us several pieces of information:

- There are two forces upward and one force downward. --> these could correspond to the forces of the two harms of the girl (upward) and the weight of the girl (downward)

- The sum of the tension forcs is equal to the force gravity --> this tells us that the net force is zero, therefore this could corresponds to the girl hanging motionless from the trapeze.

Let's see why the other options are not acceptable:

A) A book is at rest on a tabletop.   --> In this case we would only one have one force downward (the force of gravity) and one force upward (the normal reaction of the tabletop)

B) A physics student rests a backpack upon one shoulder.  --> this does not corresponds to two forces upward and one force downward

D) A girl falls slowly to Earth while strapped to a parachute. --> Again, in this case we would have one force downward (force of gravity) and one force upward (the air resistance on the parachute)

So, the only possible option is C.

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The answer is an ionic bond.
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the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

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Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

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Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

0.5x = 150 - 1.5x                                  Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x             Combine like terms  

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Answer:

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f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source  while swinging to and from before the source of sound .

3 0
3 years ago
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