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FinnZ [79.3K]
3 years ago
14

How do I work out the percentage??

Physics
2 answers:
NeX [460]3 years ago
7 0
Percentage for which columns btw if its something as easy as percentage for 4/5... you divid 4/5= .80 then.... .80*100= 80%
tatiyna3 years ago
7 0
You just divide the percenage by the girst box and there is your answer
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The illustration represents one form of _________________, the process that enables all stars, including our sun, to continuousl
jonny [76]

We have no way to say what the illustration represents, mainly because
you haven't given us a way to see the illustration.

<span>However, the process that all stars, including our sun, use to continuously
produce energy is nuclear fusion.</span>

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3 years ago
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Why does the gravitational field strength increase with latitude
otez555 [7]
The second major reason for the difference in gravity at differentlatitudes is that the Earth's equatorial bulge (itself also caused by centrifugalforce from rotation) causes objects at the Equator to be farther from the planet's centre than objects at the poles.
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3 years ago
How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
aalyn [17]
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7 0
3 years ago
Balok diam diatas bidang miring pada sudut kemiringan 40° balok mulai bergerak,tentukan koefisien gesek statis antara balok dan
Pachacha [2.7K]

Answer:

0.84

Explanation:

m = Massa balok

g = Percepatan gravitasi

\theta = Sudut kemiringan

\mu = Koefisien gesekan statik antara balok dan bidang miring

Gaya balok karena beratnya diberikan oleh

F=mg\sin\theta

Gaya gesekan diberikan oleh

f=\mu mg\cos\theta

Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.

mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84

Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.

7 0
3 years ago
A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
3 years ago
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