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Ahat [919]
3 years ago
15

A hammer falls from the top of a tall building. Ignoring air resistance, how far down will it fall in 3 secs

Physics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

pretty far if you ask me

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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)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

3 0
3 years ago
What is the distance traversed by the particle between 0 seconds and 6 seconds?
ad-work [718]

Answer:

I dont know to be honest

Explanation:

i dont know to be honest

5 0
3 years ago
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0
3 years ago
Read 2 more answers
a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin
Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0
3 years ago
Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref
Vitek1552 [10]

Answer:

hola me llamo bruno y tu?

Explanation:

pero yo soy de mexico

5 0
3 years ago
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