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3 years ago

Define What is Newton's third law of motion?

Physics

2 answers:

xenn [34]3 years ago

7 0

For every force in nature that there is an equal and opposite reaction

OlgaM077 [116]3 years ago

3 0

Answer:

I had that on 6th grade here u go

His third law states that for every action in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

3 years ago

Which is the SI base unit for distance?

stealth61 [152]

Yes you're correct. For distance, SI is based on meters.

7 0

3 years ago

Goldie Doldfish, a speed swimmer, loves to rece around the parks pond, which is 0.5 miles around. If she can swim around the tra

rosijanka [135]

Answer:

Her average speed is 0.25 miles / hour.

Explanation:

I did 0.5 divided by 2 = 0.25 miles / hour.

<em>Hope this helps! :)</em>

8 0

3 years ago

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

5 0

3 years ago

A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0

3 years ago