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3 years ago

How does inertia affect the motion of an object?

Physics

1 answer:

Furkat [3]3 years ago

4 0

<u>Inertia affects the motion of an object as follows:</u>

When an object is in motion, it will continue to be in the same state unless otherwise some outside force is being applied to it. Thus, inertia affects the motion of an object. It restricts some other force being acted upon the object.

But mass of an object is directly proportional to inertia. So when the inertia is more on an object, it means that the object has more mass. For example, if there are two similar bricks, one that is made up of mortar and the other one is made of Styrofoam.

To identify which brick is made of Styrofoam without lifting the bricks, push both the bricks with equal force, the one that has less resistance tends to move faster. This means that it has less inertia and hence less mass.

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SOH-CAH-TOA is used to solve for the ________

Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

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3 years ago

How long did it take our planet to produce the fossil fuels we're using today?

Vesna [10]

Answer:

It takes millions sometimes hundreds of millions Explanation:

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3 years ago

Read 2 more answers

A motorbike is travelling with a velocity of 3m/s. It accelerates at a rate of 9.3m/s for 1.8s. Calculate the distance it travel

Misha Larkins [42]

Explanation:

Consider the kinematic equation,

$x = vt + \frac{1}{2} a {t}^{2}$

where x is the distance traveled, v is the initial velocity, a is the acceleration and t is time. By plugging in known values and solving for x,

$x = 3(1.8) + \frac{1}{2} (9.3) {1.8}^{2}$

through simple algebra we get

$x = 20.466$

where this is the distance traveled in meters.

4 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$