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3 years ago

11

What is direct result of having greenhouse gases in the atmosphere?

1 answer:

11111nata11111 [884]3 years ago

4 0

Earths atmosphere heats up polars melt

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A rotating wheel requires a time Δt = [01]_____________________ to rotate 37.0 revolutions. Its angular speed at the end of the

Sidana [21]

Answer:

Explanation:

Initial angular velocity ω₁ = 0 , final angular velocity ω₂ = 75.9 rad /s

angle rotated = θ

= 37 x 2π

= 74 π

The formula for angular velocity

ω₂² = ω₁² + 2αθ , α is angular acceleration

75.9² = 0 + 2 α x 74 π

α = 75.9² / 2 x 74 π

= 12.396 rad / s²

8 0

2 years ago

How much time does it take for an eagle flying at a speed of 50 kilometers per hour to travel a distance of 2000 kilometers?

gtnhenbr [62]

Answer:40 hour

Explanation:

8 0

3 years ago

Lydia is 9kg underweight. Caroline has an ideal weight for her height. Viv is 8kg overweight. Who is most likely to have the hig

Crank

Answer: Viv

Explanation:

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2 years ago

Read 2 more answers

People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

People at the halfway experience g/2

hence, the correct answer is option B

7 0

2 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago