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3 years ago

Emma a total of 25 seconds to crawl 4 feet south and then 6 feet east.

Physics

1 answer:

svlad2 [7]3 years ago

8 0

Answer:

what is the question exactly

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The acacia ant is a parasite on the acacia tree. True False

stiv31 [10]

The answer you are looking for is: 
True 
Hope that helps!! 
Have a wonderful day!!

4 0

3 years ago

The disk rotates about a fixed axis through point O with a clockwise angular velocity ω0 = 16 rad/s and a counterclockwise angul

Lapatulllka [165]

Answer:

V = 4.48m/s a = 1.57m/s²

Explanation:

ω₀ = 16rad/s

α₀ = 5.6rad/s²

r = 280mm = 0.28m

a = ?

v = ?

Angular velocity (ω₀) = velocity of acceleration / length of path

ω₀ = v / r

V = ω₀ * r

V = 16 * 0.28

V = 4.48m/s

Acceleration = ?

Angular acceleration α₀ = angular velocity (ω) / time take (t)

α₀ = ω / t .... equation i

But acceleration (a) = velocity (v) / time (t)

a = v / t

t = v / a

Put t = v / a into equation i

α₀ = ω / (v / a)

α₀ = ω * a / v

α * v = ω * a

a = (α * v) / ω

a = (5.6 * 4.48) / 16

a = 1.568m/s²

a = 1.57m/s²

3 0

3 years ago

Ken wants to jump and have some fun too! Barbie loans Ken her bungee cord. Is this a good idea? Explain with evidence and reason

skelet666 [1.2K]

Answer:

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4 0

3 years ago

Read 2 more answers

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

4 years ago

Read 2 more answers

Given a wind turbine with blades that sweep out a 10 m diameter circle, and a wind speed of 2 m/s, approximately what is the max

Mars2501 [29]

Answer:

223.55 W

Explanation:

v = Velocity of wind = 10 m/s

r = Radius of circle = 5 m

S = Swept area

$\rho$ = Density of air = 1.2 kg/m³

$C_p$ = Power coefficient = 0.593

$P=C_p\frac {1}{2}}\rho Sv^{3}\\\Rightarrow P=0.593\times \frac{1}{2}1.2 \pi \times 5^2\times 2^{3}\\\Rightarrow P=223.55\ W$

The maximum possible power that can be produced by the turbine is 223.55 W

8 0

4 years ago