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Alex17521 [72]
3 years ago
12

An ant takes 5 secs to complete one round

Physics
1 answer:
saveliy_v [14]3 years ago
4 0

Answer:

hi

Explanation:

hlo hlo hloxhdydydhfjci

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In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a
scoundrel [369]

Answer:

a)  Total mass form, density and axis of rotation location are  True

b)   I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

       I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

        ρ = m / V

        dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

- Linear speed       False

- Acceleration angular False

-  Total mass form True

-  density True

- axis of rotation location   True

b) we calculate the moment of inertia of a particle

For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

          I = m r²

4 0
3 years ago
Plz help urgent <br> i will give brainliest
In-s [12.5K]

Answer:

The answer is D. Balanced forces

Hope it helps.......... pls mark as brainliest

5 0
3 years ago
Chemistry Sem 2
3241004551 [841]

Answer:

Solids - Bricks , wood , Pottery, Bucket

Liquid - Water, soap, Sanitizers.

Gases - Aerosol in Deodorants, Chlorofluorocarbons in Fire extinguishers , Butane in lighters.

4 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
4 years ago
A 600 kg car is at test and then accelerated to 5m/s , what is its original kinetic energy
ycow [4]

Answer:

0 J

Explanation:

Kinetic energy is defined as:

KE = 1/2 m v²

where m is mass and v is velocity.

The car starts at rest, so it has zero velocity.  Therefore, its initial kinetic energy is 0 J.

3 0
3 years ago
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