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4 years ago

If an object in translational equilibrium is also in rotational equilibrium about a given rotational axis. True or False

Physics

2 answers:

Mariulka [41]4 years ago

7 0

Complete Question:

If an object in translational equilibrium is also in rotational equilibrium about a given rotational axis the object will be in rotational equilibrium about any other axis. True or false

Answer:

True

Explanation:

An object is in translational equilibrium when the sum of all the external forces acting on it equals zero. $\sum F = 0$

Such an object is also experiencing zero overall acceleration. Therefore, it is either not moving or moving at a constant velocity

An object is in rotational equilibrium when the sum of all the external torques acting on it equals zero. $\sum T = 0$

Such an object either will not be moving or moving at a constant angular velocity. This means the object is experiencing zero angular acceleration

If an object in translational equilibrium is also in rotational equilibrium about a given rotational axis the object will be in rotational equilibrium about any other axis

Gala2k [10]4 years ago

5 0

Answer: True

Explanation:

The object will be in rotational equilibrium about any other axis

Therefore, it is either not moving or moving at a constant velocity. Similarly to translational equilibrium, an object is in rotational equilibrium when the sum of all the external torques acting on it equals zero. In rotational equilibrium, an object either will not be moving or moving at a constant angular velocity.

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Which property of light is not explained by the wave model of light?

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Blue light can knock electrons off a plate, but red light can't

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3 years ago

Which of the following is true about Viscosity of liquids:

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Viscosity of liquids is essentially the 'thickness' of the liquid. For instance, honey and water have different viscosities. Honey has a higher one and therefore, liquids with high viscosity do not flow as well as liquids with low viscosity (water).

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3 years ago

Read 2 more answers

A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What

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Answer:

speed of star is 1.37 × $10^{5}$ m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest = velocity of source / speed of light

so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

velocity of source star = 1.37 × $10^{5}$ m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

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3 years ago

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the

SOVA2 [1]

Answer:

Answer is It was deduced from the rate at which it glimmers.

Refer below.

Explanation:

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. The diameter of Cygnus X-1 found:

It was deduced from the rate at which it glimmers.

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3 years ago

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0

3 years ago