How many electrons does eaach shell hold?

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a c

Aneli [31]

Answer

given,

x = (3.9 cm)sin[(9.3 rad/s)πt]

general equation of displacement

x = A sin ω t

A is amplitude

now on comparing

c) Amplitude =3.9 cm

a) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{9.3\pi}{2\pi}$

f = 4.65 Hz

b) period of motion

$T= \dfrac{1}{f}$

$T= \dfrac{1}{4.65}$

T = 0.215 s

d) time when displacement is equal to x= 2.6 cm

x = (3.9 cm)sin[(9.3 rad/s)πt]

2.6 = (3.9 cm)sin[(9.3 rad/s)πt]

sin[(9.3 rad/s)πt] = 0.667

9.3 π t = 0.73

t = 0.025 s

4 0

3 years ago

Read 2 more answers

A very long, straight wire has charge per unit length 3.20 x 10^-10 c>m. At what distance from the wire is the electric-field

vovangra [49]

at 2.304×10⁹m the electric field magnitude equal to 2.50 nc.

electric field due to a long uniformly charged wire is given as,

E= 2kλ/r

where,

k is a constant

λ is uniform charge density

r is the distance from the wire

λ = 3.20 × 10⁻¹⁰C/m

k= 9×10⁹

therefore r = 2kλ/E

on substituting value we get,

r= 2.304×10⁹m

learn more about electric field due to a uniformly charged wire here:

brainly.com/question/28166144

#SPJ4

5 0

2 years ago

A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m

Degger [83]

Answer:

a)t = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ and V₀y = 0

And at the end of the movement t = ? the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9 ⇒ t² = 2,04 s

t = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then 15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf = 0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

3 0

3 years ago

Cómo es útil el ejercicio Squat Side Kick para practicar deportes?? Dame dos ejemplos

svetlana [45]

Answer:

do you speak english

Explanation:

4 0

3 years ago

If the potential due to a point charge is 5.05 ✕ 102 V at a distance of 15.7 m, what are the sign and magnitude of the charge? (

aleksklad [387]

Answer:

The magnitude of charge is $8.809\times 10^{-7}\ C$ with a positive charge.

Explanation:

Given that,

The potential due to a point charge, $V=5.05\times 10^2\ V$

Distance, d = 15.7 m

The potential due to a point charge is given by :

$V=\dfrac{kq}{d}$

q is charge

k is electrostatic constant

$q=\dfrac{Vd}{k}\\\\q=\dfrac{5.05\times 10^2\times 15.7}{9\times 10^9}\\\\q=8.809\times 10^{-7}\ C$

Here, the potential is positive, then the magnitude of charge is $8.809\times 10^{-7}\ C$ with a positive charge.

5 0

3 years ago