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Cerrena [4.2K]
2 years ago
6

A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the mot

orcycle tires and the walls is µ. How does the minimum µ needed to prevent the motorcycle from slipping downwards change with the motorcycle’s speed, s?
a) µ ∝ s0b) µ ∝ s−1/2c) µ ∝ s−1d) µ ∝ s−2e) none of these
Physics
1 answer:
igomit [66]2 years ago
7 0

Answer:

option D

Explanation:

given,

coefficient of friction between wall and tire = µ

speed of motorcycle = s

friction force = f = μ N

where normal force will be equal to centripetal force

N = \dfrac{mv^2}{r}

for motorcycle to not to slip weight should equal to the centripetal force

 now,

m g =\mu \dfrac{mv^2}{r}

\mu =\dfrac{rg}{s^2}

where "rg" is constant

\mu\ \alpha \ \dfrac{1}{s^2}

\mu\ \alpha \ s^{-2}

Hence, the correct answer is option D

   

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This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs
alekssr [168]

Here is the full question

A metal sphere with Radius  R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached:

b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = \frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}

The electric potential (V) at the surface of the first sphere = \frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}

= ER_1

b)

The total charge of the first sphere q_1 = \frac{Q_1R_1}{R_1+R_2}

The total charge of the second sphere q_2= \frac{Q_1R_2}{R_1+R_2}

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = R_1

The radius of the second sphere = R_2

Charge on the first sphere = Q_1

a) The electric field (E) at the surface is the first sphere = \frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}

The electric potential (V) at the surface of the first sphere = \frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}

= ER_1

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use q_1 \ and \ q_2 to represent their charges respectively

The potential on the surface of the first sphere;

V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}

The potential on the surface of the second sphere;

V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}

V_1=V_2

∴

\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}

Thus, we can say :

\frac{q_1}{q_2}= \frac{R_1}{R_2}

and Q_1 = q_1 + q_2

As such ;

The total charge of the first sphere q_1 = \frac{Q_1R_1}{R_1+R_2}

The total charge of the second sphere q_2= \frac{Q_1R_2}{R_1+R_2}

3 0
2 years ago
A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e
ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J

The energy extracted by the turbine every second is 2649600 Joules

8 0
2 years ago
The distribution of ________ across the globe provides the primary indicator of boundaries between all tectonic plates.
Fiesta28 [93]

The distribution of earthquake across the globe provides the primary indicator of boundaries between all tectonic plates. Earthquakes have a definite distribution pattern. There are three major belts in the world which are frequented by earthquakes of varying intensities and these belts are as under the circus pacific belt, mid Atlantic belt and mid continental belt. 

4 0
2 years ago
A fast Humvee drove from desert A to desert B. For the first 12
Zina [86]

Answer:

v = 172 km/h

Explanation:

For the first 12  hours, it traveled at an average speed of 185 km/h. Let d₁ is distance. So,

d_1=v_1\times t_1\\\\d_1=185\ km/h\times 12\ h\\\\d_1=2220\ km

For the  next 13 hours, it traveled at an average speed of 160 km/h. Let d₂ is the distance. So,

d_2=v_2\times t_2\\\\d_2=160\ km/h\times 13\ h\\\\d_2=2080\ km

Average speed = total distance/time taken

So,

v=\dfrac{d_1+d_2}{t_1+t_2}\\\\v=\dfrac{2220+2080}{12+13}\\\\v=172\ km/h

So, the average speed of the whole journey is 172 km/h.

4 0
2 years ago
A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W
dalvyx [7]

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

F_g = mg

now the acceleration of ball is given as

a = \frac{F_g}{m}

a = \frac{mg}{m} = g

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

5 0
2 years ago
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