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GuDViN [60]
3 years ago
8

I need some help plz!

Physics
1 answer:
hram777 [196]3 years ago
7 0
B is electrons and A is neutrons
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if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop
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The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
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3 years ago
jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
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  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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The theoretical line perpindicular to the surface where a light ray hits a mirror is called the
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Zolol [24]

Answer:

the first one is Primary

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