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snow_tiger [21]
3 years ago
6

What do I say when commenting on my result?

Physics
1 answer:
nevsk [136]3 years ago
8 0
Result for what exactly?
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A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
sladkih [1.3K]
I believe the answer is option C 1.8 kg•m/s to the east
6 0
3 years ago
The London Eye, which is a giant Ferris wheel, has a diameter of 135 m. It revolves at a constant rate and takes 30 minutes to c
Maurinko [17]

Answer:

Linear velocity of a rider in a capsule that is located at the perimeter of the wheel = 0.2355 m/s

Explanation:

Diameter = 135 m

Radius, r = 0.5 x 135 = 67.5 m

It revolves at a constant rate and takes 30 minutes

Period, T = 30 minutes= 1800 s

\texttt{Angular velocity, }\omega =\frac{2\pi }{T}=\frac{2\pi }{1800}=3.49\times 10^{-3}rad/s

Linear velocity, v = radius x angular velocity

                      v = 67.5 x 3.98 x 10⁻³ = 0.2355 m/s

Linear velocity of a rider in a capsule that is located at the perimeter of the wheel = 0.2355 m/s

6 0
3 years ago
2. What is meant by the statement that the MA of a simple machine is 3<br>​
Blababa [14]

Explanation:

MA of a machine is its mechanical advantage.

 Mechanical advantage is the ability of a machine to multiply force so as to get a work done.

  • This is done by minimizing effort and maximizing the load or output.
  • A MA of a simple machine with a value of 3 suggests that such a machine will multiply input force by a factor of 3.
  • Therefore, such a machine will amplify the force input into the system.
4 0
3 years ago
Human power can be used to generate electricity by connecting a bicycle to a generator. The efficiency of the bicycle and genera
NemiM [27]

Answer:

132 W

Explanation:

Efficiency = power out of generator / power into generator

0.6 = P / 220 W

P = 132 W

3 0
3 years ago
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions doe
kramer

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  N  =  109 \  rev      

Explanation:

From the question we are told that

    The speed of the car is  u  =  28.4 \  m/s

     The constant deceleration experienced is  a =  1.92 \  m/s^2

      The radius of the tire is  r =  0.307 \  m

     

Generally from kinematic equation we have that

      v^2 =  u^2 + 2as

Here  v is the final velocity which is  0 m/s

   So

         0^2 =  28.4^2 + 2 *  1.92 * s

=>      s = 210.04 \  m

Generally the circumference of the tire is mathematically represented as

         C =  2 \pi r

=>      C =  2 *  3.142 * 0.307    

=>      C = 1.929 \  m

Generally the number of revolution is mathematically represented as

         N  =  \frac{ s}{C}    

=>     N  =  \frac{210.04}{1.929}

=>     N  =  109 \  rev      

5 0
3 years ago
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