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3 years ago

Would someone just answer this now?

Physics

1 answer:

sineoko [7]3 years ago

3 0

Angle = 90 - 39.4 = 50.6
cos50.6 = adjacent/47.3
(47.3)(cos50.6) = 30.02

The y component of the vector is 30.0 m.

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How many hydrogen atoms are there in one molecule of PH3?

galben [10]

Answer:

On molecules of PH3, there has to be 3 atoms.

Explanation:

The Element Hydrogen (H) has 3 attoms.

The Element Phosphorus (P) has 1 atom.

These two created into the molecule PH3, would have to have only 3 attoms because they are asking how many hydrogens atoms are in the molecule PH3.

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3 years ago

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

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Answer:

Answer

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3 years ago

What type of feeder is a snake? *

bija089 [108]

Answer:

slender-bodied non-stinging insect having iridescent wings that are outspread at rest; adults and nymphs feed on mosquitoes etc.

Explanation:

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3 years ago

To hide their loot, thieves put the 1.20 kg gold that they had robbed into a small box and threw it in a lake. The mass of the e

Dmitry [639]

Answer:

11.1 cm

Explanation:

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3 years ago

The radius of the earth is 6380 km and the height of mt.everest is 8848 m. if the value of of acceleration due to gravity on the

Bas_tet [7]

a) $9.80 m/s^2$

The acceleration due to gravity at a certain location on Earth is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

h is the altitude above the Earth's surface

At the top of Mt. Everest,

R = 6380 km = $6.38\cdot 10^6 m$

$h' = 8848 m$

$g'=9.77 m/s^2$

With

$g'=\frac{GM}{(R+h')^2}$ (1)

At the Earth's surface,

R = 6380 km = $6.38\cdot 10^6 m$

h = 0

g = ?

$g=\frac{GM}{R^2}$ (2)

By doing the ratio (2)/(1), we find an expression for g in terms of g':

$\frac{g}{g'}=\frac{\frac{GM}{R^2}}{\frac{GM}{(R+h')^2}}=\frac{(R+h')^2}{R^2}=\frac{(6.38\cdot 10^6+8848)^2}{(6.38\cdot 10^6)^2}=1.003$

And therefore,

$g=1.009g'=1.009(9.77)=9.80 m/s^2$

b) 519.3 N

The weight of an object near the Earth's surface is given by

$W=mg$

where

m is the mass of the object

g is the acceleration of gravity at the object's location

In this problem,

m = 50 kg is the mass of the object

g' = 9.77 m/s^2 is the acceleration of gravity on top of Mt Everest

Susbtituting,

$W=(50)(9.77)=519.3 N$

6 0

3 years ago