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LekaFEV [45]
2 years ago
10

HELP PLEASE ASAP!!! I CAN'T FAIL PHYSICS xoxo thank you!

Physics
1 answer:
Allushta [10]2 years ago
5 0
They are malleable and lustrous, and can conduct both electricity and thermal heat
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3 years ago
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How is velocity ratio of wheel and axle calculatedaad<br>​
iogann1982 [59]

Answer:

VR = \frac{Radius of wheel}{Radius of axle}

Explanation:

A machine is a device that can be used to overcome a load by the application of an effort through a pivot. Examples are: pulleys, wedge, screw jack, wheel and axle etc.

The wheel and axle is a simple device that can be used to lift a load through a height. Its velocity ratio (VR) can be determined by:

VR = \frac{Radius of wheel}{Radius of axle}

Note that for a practical wheel and axle, the radius of the wheel is greater than the radius of the axle.

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3 years ago
Atoms are very small. Which measurement is the approximate diameter of a helium atom?
stich3 [128]
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3 0
3 years ago
Richard Julius once made a model plane that could travel a max speed of 110 m/s. Suppose the plane was held in a circular path b
hjlf

Answer:

85.8 m/s

Explanation:

We know that the length of the circular path, L the plane travels is

L = rθ where r = radius of path and θ = angle covered

Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt

where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path

Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed  and t = time,

θ = θ₀ + ωt = 0 + ωt = ωt

So, v = rdθ/dt + θdr/dt

v = rω + ωtdr/dt

v = (r + tdr/dt)ω

v = (r + tdr/dt)v'/r

v = v' + tv'/r(dr/dt)

v = v'[1 + t(dr/dt)/r]

Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)

So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]

v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]

v = 110 m/s[1 + 11.0 s(-0.02/s)]

v = 110 m/s[1 - 0.22]

v = 110 m/s(0.78)

v = 85.8 m/s

8 0
2 years ago
Which letter BEST shows how to measure amplitude?
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Can u explain more like ,u have a pic or some?
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