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spayn [35]
3 years ago
7

The relative hardness of a mineral can best be tested by

Physics
1 answer:
stellarik [79]3 years ago
6 0
<span>The relative hardness of a mineral can be tested by the Moh’s scale or the mineral hardness. The Moh’s scale can be characterized by the ability of the mineral to resist scratch resistance by scratching a harder or softer mineral. It has a scale of 1 to 10 where 1 being the softest and 10 being the hardest. The relative hardness of a mineral can best be tested by scratching the mineral across a glass plate. The answer is letter A. </span>
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Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

5 0
1 year ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
3 years ago
HELP ME!!
alexdok [17]

Answer:

Honey how can i draw you cant draw here

Explanation:

I wish i can help you

3 0
3 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
un resorte de 10cm de longitud recibe una magnitud de fuerza que lo estira hasta medir 15cm ¿cual es la magnitud de la tension u
Ber [7]

Answer: 0.5

Explanation:

The modulus of elasticity (called <em>"alargamiento unitario"</em> in spanish) \epsilon of a spring is given by the following formula:

\epsilon=\frac{\Delta L}{L}

Where:

L=10 cm  is the original length of the spring

\Delta L=L_{f}-L  is the elongation of the spring, being L_{f}=15 cm the length of the spring after a force is applied to it.

\epsilon=\frac{L_{f}-L}{L}=\frac{15 cm - 10 cm}{10 cm}

Then:

\epsilon=0.5

8 0
3 years ago
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