What property of matter determines how much inertia an object has

A person who has a job that requires working all night and sleeping during the day will end up fighting hos or her natural

LuckyWell [14K]

Answer:

A

The answer to this question is circadian rhythm. A person who has a job that requires working all night and sleeping during the day will end up fighting his or her natural circadian rythm. Hope it helps!

5 0

3 years ago

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What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?

alexira [117]

Answer: The correct answer is-15 Volts.

Explanation-

Voltage of a battery can be defined as the difference in electric potential that lies between the positive and negative terminals of a battery.

It can be calculated using Ohm's law, which states that the electric potential difference between two points on a circuit is equal to the product of the current that flows between the two points (I) and the total resistance that sis present between the two points. It can be mathematically depicted as-

ΔV = I • R

Putting the value of 'I' and 'R', we get-

ΔV = 5 X 3

= 15 V

8 0

3 years ago

A piston above liquid in a closed container has an area of 1m2. The piston carries a load of 350 kg. What will be the external p

grandymaker [24]

<span>Pressure = force / area</span>

I assume that 350kg is the mass
Therefore,

350 x 9.8 (gravity) = 3430N

3430 / 1 = 3430Pa

3.43 KPa

5 0

3 years ago

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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

5 0

3 years ago