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Dovator [93]
1 year ago
9

A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

icycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Physics
1 answer:
mars1129 [50]1 year ago
8 0

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

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The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
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Felectric = q*E 
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</span> <span>q*E = m*a 
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<span> Our sign convention is "up is positive" 
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</span> answer
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3 0
1 year ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
2 years ago
A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
Reika [66]

Answer:

= 829.69 Watt

≅ 830 Watt

Explanation:

Given that,

Velocity of air flow = 12.5m/s

Rate of flow of air = 9m³/s

Density of air = 1.18kg/m³

power by kinetic energy = 1/2(mv²)

mass = density × volume

m = 1.18 × 9

  = 10.62 kg/s

power = 1/2 mV²

           = 1/2 (10.62 × 12.5²)

           = 829.69 Watt

           ≅ 830 Watt

Flow rate  

u

=

9

 

m

3

/

s

Velocity of the air  

V

=

8

 

m/s

Density of the air  

ρ

=

1.18

 

kg

/

m

3

5 0
2 years ago
American Football Field Uses A Field That Is 100.0 Yd Long, Whereas A Soccer Field Is 100.0m Long. Which Field Is Longer And By
postnew [5]
Note that
1 yd = 0.9144 m

Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m

Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) =  9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%

Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
4 0
2 years ago
When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw
Setler [38]

Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.

The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.

The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.

The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.

7 0
2 years ago
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