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AnnyKZ [126]
2 years ago
15

What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Physics
1 answer:
Kobotan [32]2 years ago
7 0

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

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When an airplane flies from Dallas - Ft. Worth International airport to Los Angeles International airport, the time in the air i
Elanso [62]

Answer:

7.99 or 8 depends where you round.

Explanation:

Distance divided by time so 1246/156=7.98717948718

7 0
3 years ago
Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A
ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif
Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

7 0
3 years ago
WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!!!
stepan [7]
*FRICTIONAL FORCE* in the opposite direction of the way Bobby is pushing.
Friction is a force which varies but it is always opposing the direction of motion.

*APPLIED FORCE* is the force that Bobby is pushing with.
An applied force is literally the force that is applied to an object.

*WEIGHT FORCE* is also called the force of gravity. It is straight downward.
It is the weight of the object multiplied by the force of gravity. If the TV weighed 100kg, acceleration is always 9.81 m/s^2, so the weight force would be 981 N.

*NORMAL FORCE* is the force which is holding the TV above ground. The ground supplies a force upward against the TV.
Normal force is just the force that prevents the TV from falling through the ground. We don't normally realize it in our everyday life, but the floor must hold everything up because gravity is always "pushing" against it.
3 0
3 years ago
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume
Svetach [21]

Answer:

9.56\cdot 10^{-7} C

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

A=L^2=(0.19)^2=0.036m^2

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

E=3\cdot 10^6 N/C

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C

7 0
3 years ago
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