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Delicious77 [7]
3 years ago
5

An object in projectile motion will follow which path? a]curved up from the ground b]curved down toward the ground c]straight do

wn toward the ground d]straight up from the ground
Physics
2 answers:
Lesechka [4]3 years ago
7 0
It's B because when you throw something it doesn't go up it slowly descends downward
aev [14]3 years ago
7 0

Answer: b] curved down toward the ground

Explanation:

When a body is thrown at an angle from the base, it takes a curved path. It is known as projectile motion. The body moves to the highest point and then falls under gravity.

In horizontal direction, the body has constant velocity as no acceleration acts in this direction. In the vertical direction, acceleration due to gravity acts. Because of the combined effect, the body moves in a curved path towards the ground. The path is curved down towards the ground.

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you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh
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Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

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4 years ago
A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa
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To solve this problem we will apply the concepts related to the electric field, linear charge density and electrostatic force.

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E = \frac{\lambda}{2\pi \epsilon_0 r}

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The linear charge density can be written as,

Linear charge density is given as

\lambda = \frac{q}{L}

Replacing,

E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}

E = \frac{q}{2\pi \epsilon_0 rL}

The initial and final electric Force can be written as function of the charge and the electric field as

F_i = E_i q

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F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})

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F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})

The ratio of the force is

\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}

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