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3 years ago

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Consider a point located equidistant from point charges A and B, labeled C in the diagram. If A and B have the same magnitude ch

arge, but A is positive and B is negative, which way does the electric field vector point at C
a) up
b) down
c) right
d) left

1 answer:

klio [65]3 years ago

4 0

I'm pretty sure its' A. up, Hope this helps :)

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A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length

lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

λ = 4L / n n = 1, 3, 5, ... (2n-1)

How the third resonance is excited

m = 3

L = 192 cm = 1.92 m

λ = 4 1.92 / 3

λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

v = λ f

f = v / λ

f = 343 / 2.56

f = 133.98 Hz

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

λ = 2L / n

λ = 2 0.37 / 2

λ = 0.37 m

The speed of the wave is

v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

v = 0.37 133.98

v = 49.57 m / s

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v = √ T /μ

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3 years ago

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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

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