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arlik [135]
3 years ago
12

Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance bet

ween the particles is reduced to one-third of the original separation
Physics
1 answer:
Elena-2011 [213]3 years ago
4 0

Answer:

243 N

Explanation:

The formula for electromagnetic force is F= Kq1q2/r^2

where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N

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shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

8 0
2 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

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You will accelerate down the slope.
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