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Murrr4er [49]
4 years ago
10

Can someone help me ?

Physics
1 answer:
hammer [34]4 years ago
4 0

Answer:

1)    Time interval                 Blue Car                      Red Car

          0 - 2 s                Constant Velocity           Increasing Velocity

          2 - 3 s                Constant Velocity           Constant Velocity

          3 - 5 s                Constant Velocity           Increasing Velocity

          5 - 6 s                Constant Velocity           Decreasing Velocity

2) For Red and Blue car y₂  = 120       v = \frac{y_{2}-y_{1}}{t_{2}-t_{1}} = \frac{120-0}{6-0} = 20 m/s

     We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3)   At t = 2s, the cars are the same position, and are moving at the same rate

                    Position - same

                    Velocity - same

The position-time graph shares the same spot for two cars.

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Define mechanics and describe its three major divisions
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4 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

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7 0
3 years ago
When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static
natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

Coefficient of static friction = Frictional force/Normal force

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0
3 years ago
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