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3 years ago

Signal far enough ahead so other drivers in your vicinity can make adjustments to your change in speed and ___________.

Physics

1 answer:

3241004551 [841]3 years ago

8 0

Answer:

DIRECTION

Explanation:

the answer of the blank will be DIRECTION

As the signal is far ahead so driver can make the decision of what he have to do either he can change direction or he can be on his same path.

But for that driver has to change his or her speed.

so, from the statement given the blank will be filled with DIRECTION

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In the diagram, q1 = +6.39*10^_9 C and

ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

$q_1=6.39\times 10^{-9} \ C$

$q_2=3.22\times 10^{-9} \ C$

$AP=(0.150+0.250)$

$=0.40 \ m$

$BP=0.25 \ m$

Now,

At point P, the electric potential will be:

⇒ $V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}$

By putting values, we get

⇒ $=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]$

⇒ $=259.695 \ Volt$

4 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$