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2 years ago

14

Need help asap please

1 answer:

cestrela7 [59]2 years ago

6 0

Answer:

9:49 is the answer because question say ratio is square of small and big

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If an impulse travels 100 m/s, about how long will it take the impulse to travel 10 meters?

PtichkaEL [24]

Answer:

0.1 seconds

Step-by-step explanation:

10/100=0.1

6 0

2 years ago

Read 2 more answers

If tommy had 3 x 99 amount of cookies and gave 4x56 away how many does he have left?

Monica [59]

he has 73 cookies left

5 0

3 years ago

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Is 4/5 equal to 5/10?

vfiekz [6]

No, 4/5 does not equal to 5/10.

Because:- When you try to see if it is proportional, it does not equal it.

4/5 ≠ 5/10

Simplify to make sure.

4/5 ≠ 1/2

4/5 is NOT equal to 1/2.

Therefore, our final answer is that 4/5 does not equal 5/10.

4 0

3 years ago

Read 2 more answers

1PLSS NEED HELP.........

tatiyna

Answer: I believe it can be a right triangle

Step-by-step explanation:

7 0

2 years 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:

a)

Speed at Equator = 463.97 meters per second

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

b)

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,

a)

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

b)

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