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Leno4ka [110]
3 years ago
14

Which term does this explain?

Physics
1 answer:
AlladinOne [14]3 years ago
5 0
It is a fact because I just want to post a question man so I have to answers something pp poop
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If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot
Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

I=I_{1}+I_{2}

I=(m+m_{1})\times r^2+(m+m_{2})\times r^2

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2

I=117.45\ kg-m^2

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0
3 years ago
On the surface of Earth, the force of gravity acting on one kilogram is: *
Murrr4er [49]

Your answer would be:

Option (B) : 2.2 Pounds

Anyways stan LOONA

4 0
3 years ago
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0
3 years ago
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0
3 years ago
The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and
8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at 0.

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>
  • A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s

The velocity is positive at when the time of motion is as follows;

0.

The total distance traveled in the first 10 seconds is calculated as follows;

2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

Learn more about motion of particles here: brainly.com/question/11066673

4 0
2 years ago
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