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3 years ago

Which physical property is shown by scratching one material with another

Physics

2 answers:

lina2011 [118]3 years ago

3 0

The hardness
nail or fingernail

vagabundo [1.1K]3 years ago

3 0

The physical property that is shown by scratching one kind of material over another is hardness, I believe.

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Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force

Darya [45]

I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.

7 0

2 years ago

is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N

ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force F acts on am object of mass m, it produces an acceleration a. The force is given by the expression,

$F=ma$

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

$F=ma\\ =(9.0kg)(3m/s^2)\\ =27N$

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

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3 years ago

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earnstyle [38]

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3 years ago

PLS ANSWER FAST TIMED TEST WILL GIVE BRAINLY!!!!!!!!!

lutik1710 [3]

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3 years ago

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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago