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tamaranim1 [39]

2 years ago

5

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.60 cm from the axis to equal

300000 g (where g is the acceleration due to gravity).
Express your answer numerically in revolutions per minute.

1 answer:

iVinArrow [24]2 years ago

8 0

Answer:

Explanation:

300000(9.8) = ω²(0.0160)

ω = 13555 rad/s

13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min

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If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

mel-nik [20]

As per the question the volume of mercury is given as 0.002 m^3 at 20 degree Celsius.

We are asked to calculate the volume of the mercury at 50 degree Celsius.

This problem is based on thermal expansion of matter.

Let us consider the initial and final volume of the mercury is denoted as -

$v_{1} \ and\ v_{2}$

Let the initial and final temperature of the mercury is denoted as -

$T_{1}\ and \ T_{2}$

As per question

$v_{1} =0.002 m^3$ $v_{2} =?$

$T_{1} =20^0 C$ $T_{2} =50^0 C$

The change in temperature is

$T_{2} -T_{1}$

$= 50^0 C -20^0 C$

$= 30^0 C$

Mercury is a fluid.So we have to apply volume expansion of liquid .

The coefficient of of volume expansion of mercury $[ \gamma ]$ at 20 degree Celsius is 0.00018 per centigrade.

As per volume expansion of liquid,

$V_{T} = v_{1} [1 +\gamma [T_{2} -T_{1} ]]$

Here $V_{T}$ is the volume at T degree Celsius.

Hence volume at 50 degree Celsius is calculated as-

$v_{2} =v_{1} [1+\gamma[50-30]]$

$= 0.002[1+0.00018*30]$

$=0.0020108m^3$ [ans]

As per the options given in the question ,option A is close to the calculated value. So option A is right.

5 0

2 years ago

Read 2 more answers

If your wave comes into contact with a different medium and is returned to the original medium it is known as:

Strike441 [17]

The return of a wave back to its original medium is called an reflection

8 0

2 years ago

describe the forces acting on a car as it movies along a level highway in still air at a constant speed

Travka [436]

Answers

The car's forward motion is opposed by the friction between the road and the tires and by the resistance of the air.

3 0

2 years ago

Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

3 years ago

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

where

F is the magnitude of the force

d is the distance covered by the object

$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

$\theta=0$

And the distance covered is equal to the circumference of the circle, which is:

$d=2\pi r=2\pi (2.5 m)=15.7 m$

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago