The magnetic bearing of compass "north" is _____. 090 180 270 360

1 answer:

6 0

In navigation bearing refers to : 1. the direction of motion itself; 2. the direction of a distant object relative to the current course 3.the angle away from North of a distant point as observed at the current point.

There is a thing called absolute bearing, and it refers to the angle between the magnetic North/true North and an object. (e.g. an object of 0 degrees would be dead ahead, whereas and object of 180 degrees would be behind you)

You might be interested in

Pls help pls due today

BaLLatris [955]

The mutualism I believe. So sorry if I’m wrong

7 0

2 years ago

Read 2 more answers

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

7 0

2 years ago

Why did the potassium permanganate crystals start to dissolve in water without being stirred or shaken

KonstantinChe [14]

<span>because the substance in the potassium permanganate crystals are permeable to water, so that means it will dissolve instantly while poured into water </span>

7 0

2 years ago

Read 2 more answers

A ball is projected upward at time t=0.0s, from a point on a roof 90m above the ground. The ball rises, then falls and strikes t

chubhunter [2.5K]

As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .

also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)

4 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

2 years ago