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GuDViN [60]
2 years ago
15

Một đĩa tròn đồng chất có khối lượng m1 và bán kính R, quay quanh trục cố định AB với vận tốc góc ωo.Vào một thời điểm nào đó mộ

t chất điểm M có khối lượng m2 bắt đầu chuyển động từ tâm đĩa ra ngoài vành theo một đường bán kính. Xác định vận tốc góc ω của đĩa, tại thời điểm chất điểm M ở cách tâm đĩa một khoảng bằng 0,5 R. Bỏ qua lực ma sát ở ổ trục quay.
help me pls!!!!
Physics
1 answer:
Alenkasestr [34]2 years ago
6 0

Answer:

Someone thinks I've sinned, but I don't know how.

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200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?
ale4655 [162]

That's "<em><u>insolation</u></em>" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

           (200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .

Note that this is the intensity of the <em><u>incident</u></em> radiation.  It doesn't say anything
about how much soaks in or how much bounces off.

Wait ! 
I just looked back at the choices, and realized that I didn't answer the question
at all.  I have no idea what  "1 sun"  means.  Forgive me.  I have stolen your
points, and I am filled with remorse.

Wait again !
I found it, through literally several seconds of online research.

           1 sun = 1 kW/m².

So 2/3 of a kW per m²  =  2/3 of 1 sun

That's between 0.5 sun and 1.0 sun.

I feel better now, and plus, I learned something.


7 0
3 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
what is the difference in the charges on a balloon rubbed in your hair and a glass  rod rubbed with silk? why?
lbvjy [14]
The balloon steals electrons from your hair, leaving the hair positively charged, and the balloon negatively charged. It causes the hair to be apart from each other, because they have the same charge. Glass has a weaker hold on electrons, and silk absorbs the lost electrons.
8 0
3 years ago
A bowling sphere is sitting on the ground. One of your classmates
ikadub [295]

normal force because it is perpendicular to the surface

4 0
3 years ago
A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he
RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

4 0
3 years ago
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