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andreev551 [17]

3 years ago

7

To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.5 times each persons weight - and t

herefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?

1 answer:

ANTONII [103]3 years ago

5 0

Answer:

The minimum coefficient of friction is 0.666

Explanation:

Suppose, In a classic carnival ride, patrons stand against the wall in a cylindrical shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 7.6 m and the room spins with a frequency of 20.9 revolutions per minute.

Given that,

The normal force does not exceed 1.5 times each persons weight

$N=1.5 mg$

We need to calculate the minimum coefficient of friction

Using balance equation

$\mu N=mg$

$\mu=\dfrac{mg}{N}$

Where, N = normal force

Put the value into the formula

$\mu=\dfrac{mg}{1.5 mg}$

$\mu=\dfrac{1}{1.5}$

$\mu=0.666$

Hence, The minimum coefficient of friction is 0.666

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The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 Km/s.

tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) the mass of the star Alpha is $M_{\alpha}$ = $7.80*10^{29} kg$

b) the mass of the star Beta is $M_{\beta}$ = $2.34*10^{30}kg$

c) the radius of the orbit of the orange star $R_{0}$ = $1.9*10^{9}m$

d) the radius of the orbit of the black hole $R_{B} = 34*10^{8}m$

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f) the orbital speed of the black hole $V_{B} =77 km/s$

Explanation:

The generally formula for orbital speed is given as $v =\frac{2\pi R}{T}$

where v is the orbital speed

R is the radius of the star

T is the orbital period

From the question we are given that

alpha star has an orbital speed of $V_{\alpha} = 36km/s = 36000m/s$

beta star has an orbital speed of $V_{\beta} = 12km/s = 12000m/s$

the orbital period is $T = 137d = 137(86400)sec$ 1 day is equal 86400 seconds

Making R the subject of formula we have for the radius of the alpha star as

$R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}$

$=6.78*10^{10}m$

for the radius of the Beta star as

$R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}$

$= 2.26*10^{10}m$

Looking at the value obtained for $R_{\alpha} and R_{\beta}$

Generally the moment about the center of the mass are equal then

$M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$

Thus $3M_{\alpha} = M_{\beta} -------(1)$

Generally the formula for the orbital period is given as

$T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }$

Then

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Where G is the gravitational constant given as $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(6.78*10^{10}m +2.26*10^{10}m)^{2}}{(137d*86400s/d)^{2}(6.67*10^{-11}m^{3}kg^{-1}s^{-2})}$

$M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)$

Thus solving (1) and (2) equations

Mass of alpha star is $M_{\alpha}=7.80*10^{29}kg$

and the Mass of Beta is $M_{\beta} =2.34*10^{30}kg$

Considering the equation

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Making $R_{\alpha} + R_{\beta}$ the subject

$R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)$

and considering this equation $M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$ from above

we have that $R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}$

Considering Question C

Let the Orange star be denoted by (0) and

Let the black-hole be denoted by (B)

And we are told from the question that

Mass of orange star $M_{0} = 0.67M_{sun}$ and

Mass of black hole $M_{B} =3.8M_{sun}$

And mass of sun is $M_{sun} = 1.99*10^{30}kg$

Then $R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)$

We are also given that the period is $T =7.75 days = 7.75 (86400s)$

Considering equation 3

$R_{0} + 0.176R_{0} = \sqrt[3]{\frac{(0.67+3.8)(1.99*10^{30}kg)(7.75(86400s))^{2} 6.67*10^{-11}Nm^{2}/s^{2}}{4\pi ^{2}} }$

Thus for V616 Monocerotis, $R_{0} =1.9*10^{9}m$

Considering equation 4

The black-hole is

$R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m$

From the formula for velocity of $V_{0} = \frac{2\pi R_{0}}{T} = 4.4*10^{9}km/s$

the velocity of $V_{B} = \frac{2\pi R_{B}}{T} =77km/s$

8 0

3 years ago