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2 years ago

FIRST ANSWER GETS BRAINLIEST (ANSWER HAS TO BE IN DETAIL)

Mathematics

2 answers:

Nonamiya [84]2 years ago

5 0

Answer:

2,975.978

Step-by-step explanation:

The number after the decimal is called "tenths" after that is "hundredths" and after is "thousandths"

The number before the decimal is called "ones" before that is "tens" before that is "hundreds" and before that is "thousands"

agasfer [191]2 years ago

3 0

Answer:

2975.978

Step-by-step explanation:

This is pretty easy, just keep the order going on.

Here is the CORRECT order:

2 Thousand(s)

9 Hundred(s)

7 Ten(s)

5 One(s)

decimal part because there is th after each type of number (ex. thousandth)

9 Tenth(s)

7 Hundredth(s)

8 Thousandth(s)

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Answer:

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There are 3 inches of snow on the ground when it begins to snow 0.5 inches per hour.

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3 years ago

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3 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

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3 years ago