All categories

3 years ago

You went 100 miles north and 14

Physics

1 answer:

skad [1K]3 years ago

5 0

Answer:

82 degrees

Explanation:

consider your staying point to be the center of a circle. this center has the coordinates (0, 0).

the radius of the circle is the distance you walked East (14 miles).

I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.

then the 100 miles North is tan(displacement angle)×14.

as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.

so, again.

100 = 14×tan(angle)

tan(angle) = 100/14 = 50/7 = 7.142857...

the displacement angle is then 82 degrees.

You might be interested in

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca

dolphi86 [110]

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 / 2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

8 0

3 years ago

In which phase of matter are the atoms closely packed but still able to slide past each other?

borishaifa [10]

The phase of matter is the Solid Phase.

3 0

4 years ago

Passenger side, rear-view mirrors on automobiles are generally ____.

Lesechka [4]

I believe it’s C. Plane .

8 0

4 years ago

Which option describes nonmetals?

jonny [76]

Answer:

D: Conducts electrcity poorly

Explanation:

Most conductors are metals, and non- metals are insulators, therefore they dont conduct electricity

8 0

3 years ago