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3 years ago

Can someone please ell me the answer to this, please? :)

Mathematics

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

- 328 / 9

Step-by-step explanation:

- 36 + ( - 4 / 9 )

= - 36 - ( 4 / 9 )

= - 36 ( 9 / 9 ) - 4 / 9

= ( - 36 x 9 ) / 9 - ( 4 / 9 )

= ( - 324 / 9 ) - ( 4 / 9 )

= ( - 324 - 4 ) / 9

= - 328 / 9

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Need this quick ! Correct answers appreciated

ludmilkaskok [199]

5

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1 year ago

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

4 0

3 years ago

Please answer for me as I need help ASAP

tamaranim1 [39]

Answer:

Step-by-step explanation:

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3 years ago