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11Alexandr11 [23.1K]

3 years ago

14

Which area of physics involves using a compass in the woods?

1 answer:

jarptica [38.1K]3 years ago

6 0

Magnetism. A compass uses the earth's magnetic field to point in the direction of the north geographic pole (south magnetic pole) the N point on a compass points to the S magnetic pole which is the north geographic pole. It's a little confusing, but it's just opposites.

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If you run your boat aground what should you do first

padilas [110]

If you run your boat aground, the first thing you should do is to calmly assess the situation. In case you have passengers on your boat, you should have them don PFDs (personal flotation devices). Afterwards, you should turn off the engine, check if there is any damage, and generally see if everything is okay.

8 0

3 years ago

an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m

Ksenya-84 [330]

Answer:

7.87x10^5m/s^2

Explanation:

3 0

2 years ago

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

5 0

2 years ago

Read 2 more answers

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = $\dfrac{2 \pi r}{T}$

$v = \dfrac{2\times \pi \times 15}{23}$

v = 4.10 m/s

b) The magnitude of the acceleration is:

$a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}$

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma

apparent weight =mg - ma

$\dfrac{apparent \ weight}{true \ weight} = \dfrac{(mg - ma)}{(mg)}$

$=1- \dfrac{1.12}{9.8}$

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

$\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}$

$\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}$

$= 1 + \dfrac{1.12}{9.8} \\ \\$

= 1.114 m/s²

6 0

3 years ago

Help please, i don't know what the answer is but i accidently clicked D .....

GaryK [48]

Answer:

You are right correct option is D.

4 0

2 years ago