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svetoff [14.1K]
3 years ago
6

Three 500-g point masses are at the corners of an equilateral triangle with 50-cm sides. What is the moment of inertia of this s

ystem about an axis perpendicular to the plane of the triangle and passing through one of the masses at a corner of the triangle?
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

0.25 kg m^2

Explanation:

mass of each , m = 500 g = 0.5 kg

distance, r = 50 cm = 0.5 m

Moment of inertia about the axis passing through one corner and perpendicular to the plane of triangle

I = mr^2 + mr^2

I = 2 mr^2

I = 2 x 0.5 x 0.5 x 0.5

I = 0.25 kgm^2

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Unstable atmospheric conditions lead to the formation of lightning and thunder in
Fantom [35]
Hi the answer is cumulonimbus clouds. Hope this helps.
6 0
3 years ago
A 20000 kg subway train initially traveling at 18.5 m/s slows to a stop in a station and then stays there long enough for its br
hodyreva [135]

Answer:

rise the air temperature is 0.179241 K

Explanation:

Given data

mass = 20000 kg

velocity  = 18.5 m/s

long = 65 m

wide = 20 m

height = 12 m

density of the air = 1.20 kg/ m³

specific heat = 1020 J/(kg*K)

to find out

how much does the air temperature in the station rise

solution

we know here Energy lost by the train that is calculated by  

loss in the kinetic energy that is = 1/2 m v²

loss in the kinetic energy = 0.5 × 20000 ×18.5²

loss in the kinetic energy  is 3422500 J

and

this energy is used here to rise the air temperature that is KE / ( specific hat × mass )

so here  

air volume = 65 ×20×12

air volume = 15600 m³

air mass  =  ρ × V = 1.2 × 15600

air mass = 18720 kg

so

rise the air temperature = 3422500 / ( 1020 × 18720)

rise the air temperature is 0.179241 K

7 0
4 years ago
Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
3 years ago
Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e
NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

Q=\frac{E}{Nr^2}

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

 Q=\frac{E}{Nr^2}

 

For more information on this visit

brainly.com/question/16517842

3 0
2 years ago
A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of th
cestrela7 [59]

Answer:

1. a_{rad}=17545.2\frac{m}{s^{2}}

2. a=4429.45 \frac{m}{s^{2}}

Explanation:

Radial acceleration is:

a_{rad}=\frac{v^2}{r} (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

v= \omega r (2)

By (2) on (1):

a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)

a_{rad}=17545.2\frac{m}{s^{2}}

To find the acceleration of the tube with the fall, we can use the expression:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (3)

Due impulse-momentum theorem:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (4)

with p the momentum and J the impulse. By (4) on (3):

\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)

And using Newton's second law (F=ma) and that (P=mv):

mv_f-mv_i=(ma)(\varDelta t) (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

\frac{mv_i}{2}=mgh

v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}

So (5) is:

-m(4.43)=(ma)(\varDelta t)

solving for a:

a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45

It’s negative because is opposed to the tube movement.

5 0
3 years ago
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