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ludmilkaskok [199]

3 years ago

11

Imagine that you set your watch at local noon in Kansas City on Monday and then fly to the Coast on Tuesday. You stick a pole in

to the ground on a sunny day at the beach, wait until its shadow is shortest, and look at your watch. The watch says 10:00 am. Are you on the East Coast or West Coast?

1 answer:

Sergeu [11.5K]3 years ago

7 0

Answer: East Coast

Explanation:

The problem states we need to find whether the person is on the East Coast or West Coast.

Now, we know that the shadow is shortest when the Sun is at the highest point in the sky. So it means the noon time is 12:00 P.M. Now this is the local noon time which you have set in Kansas City. Now you flew to the Coast and it says 10:00 am which means you have traveled to a place where noon comes early. And noon comes early on Eastern Coast so you are on the East Coast.

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A basketball comes down after it is shot because of __________

Vlada [557]

Answer:

gravity, friction, opposite, friction. hope that helps :)

Explanation:

7 0

3 years ago

Read 2 more answers

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago

1) At which lettered point or points is the object moving the fastest?

emmainna [20.7K]

Here we have a time-position graph with some points marked on it, and we want to analyze some points in it.

1) The object is moving the fastest in the point where the slope of the curve is the largest.

In this case, the point is D.

2) Remember that the graph evolves to the right, so, if at any point, the right side is above the left side, then at that particular point the object is moving to the left (negative x-axis)

In this case, the points are:

C, D, and E.

3) The object speeds up when the curvature of the curve changes and the slope increases, for example, from B onwards, the speed increases.

Then at point C the speed is increasing.

4) Opposite to what we said before, if the slope decreases then the object is slowing down.

In this case we can see that the slope decreases after A and after E.

5) When the object is turning around?

This is easy, we can see that the object goes to the right (positive x-axis) until it reaches B and then goes back, then at point B the object turns around.

If you want to learn more, you can read:

brainly.com/question/11290689

8 0

3 years ago

Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0

3 years ago

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the

Bad White [126]

Answer:

$d = 0.018 m$

Explanation:

Charge on the plates of the capacitor due to transfer of electrons is given as

$Q = Ne$

here we know that

$N = 2.1 \times 10^9$

so we have

$Q = (2.1 \times 10^9)(1.6 \times 10^{-19})$

$Q = 3.36 \times 10^{-10} C$

now we have electric field between the plates is given as

$E = \frac{Q}{A\epsilon_0}$

here we have

$1.5 \times 10^5 = \frac{3.36 \times 10^{-10}}{A(8.85 \times 10^{-12})}$

$A = 2.53 \times 10^{-4} m^2$

now we have

$A = \frac{\pi d^2}{4}$

$2.53 \times 10^{-4} = \frac{\pi d^2}{4}$

$d = 0.018 m$

3 0

3 years ago