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Sedbober [7]
3 years ago
11

When your vehicle is properly parked in a straight-in parking space,

Physics
2 answers:
Fofino [41]3 years ago
8 0

Answer:

option (a) is correct

Explanation:

There is lot of parking problems now these days. So to park a vehicle in a parking space, ensure that no part of your vehicle will extend out into the traffic lane, otherwise you may get fined or your vehicle may be damaged.

marin [14]3 years ago
3 0

The correct answer to this question would be:

<span><span>A. </span>No part of your vehicle will extend out into the traffic lane.</span>  

This kind of maneuver only shows your skill to handle the vehicle in tight spaces, ability to judge distance, and showing control of the vehicle as you turn into a straight-in parking space.  

<span> </span>

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4 years ago
A bus is designed to draw its power from a rotating flywheel that is brought up to 3000 rpm by an electric motor. The flywheel i
Arada [10]

To solve this problem we will apply the concept of rotational kinetic energy. Once this energy is found we will proceed to find the time from the definition of the power, which indicates the change of energy over time. Let's start with the kinetic energy of the rotating flywheel is

E_r = \frac{1}{2} I\omega^2

Here

I = moment of inertia

\omega = Angular velocity

Here we have that,

\omega = 3000\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1min}{60s})

\omega = 314.159rad/s

Replacing the value of the moment of inertia for this object we have,

E_r = \frac{1}{2} (\frac{MR^2}{2})\omega^2

E_r = \frac{1}{2} (\frac{2000(0.5)^2}{2})(314.159)^2

E_r = 1.233698*10^7J

The expression for average power is

P = \frac{E_r}{\Delta t}

\Delta t = \frac{E_r}{P}

\Delta t = \frac{1.233698*10^7}{20*10^3}

\Delta t = 616.8s \approx 620s

Therefore the correct answer is 620s.

3 0
4 years ago
Jayne lifts the barbell 120 cm upwards. She has a mass of 60kg. How much work does she do?
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4 years ago
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seraphim [82]
The liquid state into a gaseous state
3 0
3 years ago
Read 2 more answers
An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18
Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

eV=\dfrac{1}{2}mv^2

v is speed of the electron

v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m

Hence, this is the required solution.                                

7 0
3 years ago
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