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3 years ago

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An archer shot a 0.07 kg arrow at a target. The arrow accelerated at 4,000 m/s2 to reach a speed of 50.0 m/s as it left the bow.

During this acceleration, what was the net force on the arrow?
A) 280 N
B) .0000175 N
C) 57,142 N
D) 3.5 N

2 answers:

denpristay [2]3 years ago

3 0

Answer:

A. 280

Explanation:

saw5 [17]3 years ago

3 0

The answer is A ) 280 N

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The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf

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Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy

V0x = V0 cos(θ)

V0y = V0 sin(θ)

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant

Vx = Vocos(θ)

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2

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3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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lara31 [8.8K]

Answer:

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Explanation:

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The "D) People perceive objects as a whole" statement best describes an abductive reasoning. The abductive reasoning often has incomplete information as the base of its reasoning and the conclusion for this type of reasoning is not absolute. There will always be additional pieces of evidence and factors that could change the conclusion of this reasoning. Therefore<span> D statement is the most suitable answer.</span>

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The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P

defon

Answer:

0.938 m/s.

Explanation:

Given:

ω = 1 rev in 0.67 s

In rad/s,

1 rev = 2pi rad

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