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3 years ago

5

Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl

e. Rank in order, from largest to smallest, the magnitudes of the horizontal forces F1, F2, and F3 acting on the arrows. Some may be equal. State your reasoning.
Arrow 1: is 80g @ 10 m/s
Arrow 2: is 80g @ 9 m/s
Arrow 3: is 90g @ 9 m/s

1 answer:

timurjin [86]3 years ago

4 0

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

given,

Arrow 1 mass = 80 g speed = 10 m/s

Arrow 2 mass = 80 g speed = 9 m/s

Arrow 3 mass = 90 g speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

F = m a

If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

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3 years ago

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

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$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

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The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

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2 years ago

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

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Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

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Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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2 years ago

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