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3 years ago

if i roll pieces of allium foil in a ball what property of aluminium foil changes as a result of my action

Physics

1 answer:

MArishka [77]3 years ago

6 0

Answer:

Malleability

Explanation:

The ability to roll aluminum foil in a ball is the property of metal called malleability.

Metallic bonds in metals makes it possible for them to take different shape without losing internal cohesion.
Metals are malleable and they posses a wide range of physical properties.
A malleable metal can be beaten into sheets and molded into many shapes.

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A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$

7 0

3 years ago

A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is: