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2 years ago

How do you calculate a bearing angle and its equivalent angle?

Physics

1 answer:

denis-greek [22]2 years ago

4 0

Explanation:

A bearing if an angle is measured clockwise from north direction.

e.g Below the bearing of B from A is 025. (3 figures are always given). the bearing of A from B is 205°.

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The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

7 0

3 years ago

A light bulb uses 0.5 amperes from a source of 120 volts. How much power is used by the bulb?

Sladkaya [172]

Power = voltage(V) * current(I)

= 120 * 0.5

Power = 60 watts

8 0

2 years ago

Help yall 13 points!!

Naya [18.7K]

Answer:

Explanation:

12.)

A. Opposite poles attract

B. Same poles repel

13.)

IDK

3 0

3 years ago

What is the frequency of a tuning fork that vibrates 518 times in 3.50 s?

Serga [27]

Given:

The number of vibrations of the tuning fork is,

$51\text{8 times}$

in,

$t=3.50\text{ s}$

To find:

The frequency of the fork

Explanation:

The frequency is the number of vibrations in 1 second

The frequency for the given fork is,

$\begin{gathered} \frac{The\text{ total vibrations}}{The\text{ time}} \\ =\frac{518}{3.50} \\ =148\text{ Hz} \end{gathered}$

Hence, the frequency is 148 Hz.

7 0

1 year ago

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by

scoray [572]

Answer:

<em>The initial charges on the spheres are </em> $6.796\ 10^{-6}\ c$ and $-3.898\ 10^{-6}\ c$

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

$\displaystyle F=\frac{K\ q_1\ q_2}{d^2}$

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

$\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537$

Rearranging

$\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}$

Solving for q1.q2

$\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]$

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

$\displaystyle q=\frac{q_1+q_2}{2}$