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Yakvenalex [24]
3 years ago
14

The strength of the force of friction depends on which two factors?

Physics
2 answers:
Katarina [22]3 years ago
7 0

Answer: How hard the surfaces push together and the types of surfaces involved

Explanation:

irga5000 [103]3 years ago
4 0

Answer:

coefficient of friction (μ) and normal force (N)

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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
3 years ago
A laser emits light of frequency 4.74 x 1014 hz. what is the wavelength of the light in nm?
Ulleksa [173]
<span>Use this formula:Wavelength=c/v, where c is the speed of light, and v the frequency. 6.33 X 10^-7=3 X 10^8/v v=3 X 10^8/6.33 X 10^-7 v=4.74 X 10^14 Hertz,</span>
5 0
3 years ago
Explain the difference between contact and non-contact forces. Give an example of each
Amiraneli [1.4K]

pushing on something that pushes back = action and reaction.

Gravity is non contact. Action at a distance. Hence stones wil fall if dropped.

7 0
3 years ago
Read 2 more answers
Define constant speed
mash [69]
It is the speed at which a body moves without any interuptions (external forces etc)
7 0
3 years ago
Read 2 more answers
What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?
sammy [17]

Answer:

Q = 29.4 x 10⁻⁹ C.

Explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

E=\frac{k\times Q}{d^2}

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}[/tex]

Q = 29.4 x 10⁻⁹ C.

5 0
3 years ago
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