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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

The process of heat radiation is the ONLY method of heat transfer that can occur

lina2011 [118]

Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. so... electromagnetic waves ?

5 0

4 years ago

Read 2 more answers

A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w

balu736 [363]

Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
 185N = (mass) x (9.81 m/s²)
 Mass = (185N) / (9.81 m/s²) = 18.858 kilograms of frewium

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

 Volume of displaced water = 3,058 cm³

Finally, density of the frewium sample = (mass)/(volume)

 Density = (18,858 grams) / (3,058 cm³) = 6.167 gm/cm³ (rounded)

================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

 (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.

3 0

3 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide