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3 years ago

Displacement time squared graph

1 answer:

6 0

If distance versus time is a straight line then Free-Fall is at a constant velocity and the slope of the graph measures that velocity. If distance versus time-squared is a straight line then Free-Fall is at a constant acceleration and the slope of the graphmeasures one-half of that acceleration.

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If the valence electrons were removed, what would be the ion charge of the element?

kherson [118]

Answer:

I think C.

Explanation:

6 0

2 years ago

An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this

galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

$\gamma = \frac{Mt}{J}$

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

$J = \frac{1}{2} * m * r^2$

$Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2$

The wheel walls act like annular rings, for these the moment of inertia is:

$J = \frac{1}{2} * m * (re^2 - ri^2)$

$Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2$

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

$J = m * r^2$

$Jtread = 10 * 0.33^2 = 1.09 kg*m^2$

The axle acts like a rod, which is the same as the disk:

$Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2$

The drive shaft acts like a rod too:

$Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2$

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

$\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}$

4 0

3 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$