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labwork [276]
3 years ago
15

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 <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. 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.

5 0
3 years ago
A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.
earnstyle [38]

Answer:

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4 0
3 years ago
PLS ANSWER FAST TIMED TEST WILL GIVE BRAINLY!!!!!!!!!
lutik1710 [3]
The answer would be true
7 0
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
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