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Goshia [24]
3 years ago
8

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

Physics
1 answer:
timofeeve [1]3 years ago
6 0

The solution would be like this for this specific problem:

 

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery,
snow_lady [41]
The capacitor is used to store electric charge.That is what makes capacitors special. <span>
The charge that flows into the capacitor is stored on the plate of the capacitor that the source voltage is connected to. </span>When current flows into a capacitor, the charges get “stuck” on the plates because they can’t get past the insulating dielectric. One plate is positively charged and the other negatively <span>The stationary charges on these plates create an </span>electric field. <span>When charges group together on a capacitor like this, the cap is storing electric energy just as a battery might store chemical energy.</span>
8 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
The first five questions refer to the following problem:
Kruka [31]

Answer:

<h2>Δd=d2−d1</h2>

Explanation:

<h2>good morning</h2>
8 0
3 years ago
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
Agata [3.3K]

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

5 0
3 years ago
How do scientist determine the the number of iso types?
mash [69]
If you are talking about Isotopes then I believe it is the amount of atoms that have different amounts of neutrons ,but the same number of protons.
3 0
2 years ago
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