Answer:
The mass of the beam is = 29 kg.
Explanation:
A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.
Taking moment about Point S
40 × 21 =
× 29
= 29 kg
Therefore the mass of the beam is = 29 kg.
Answer:
(a) ω = 1.57 rad/s
(b) ac = 4.92 m/s²
(c) μs = 0.5
Explanation:
(a)
The angular speed of the merry go-round can be found as follows:
ω = 2πf
where,
ω = angular speed = ?
f = frequency = 0.25 rev/s
Therefore,
ω = (2π)(0.25 rev/s)
<u>ω = 1.57 rad/s
</u>
(b)
The centripetal acceleration can be found as:
ac = v²/R
but,
v = Rω
Therefore,
ac = (Rω)²/R
ac = Rω²
therefore,
ac = (2 m)(1.57 rad/s)²
<u>ac = 4.92 m/s²
</u>
(c)
In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:
Centripetal Force = Frictional Force
m*ac = μs*R = μs*W
m*ac = μs*mg
ac = μs*g
μs = ac/g
μs = (4.92 m/s²)/(9.8 m/s²)
<u>μs = 0.5</u>
Explanation:
(i)
O is the object and I is the image.
The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.
Refer above image. 1
(ii)
O is the object and I is the image.
The image formed is diminished and erect. So the magnification will be positive (+) and less than1.
Refer above image. 2
(iii)
The image will be formed as the 2F on the other side of the lens and it will be of same of the object.
According to the rules for significant figures, simple decimal subtraction is done. But since 15.54 is only accurate to the hundredths place, then the resulting difference should also be rounded off to the nearest hundredths. This is shown below:
508.9538 - 15.54 = 493.4138
Rounding off to the nearest hundredths:
= 493.41
Answer:
![\omega'=19.419\ rev.s^{-1}](https://tex.z-dn.net/?f=%5Comega%27%3D19.419%5C%20rev.s%5E%7B-1%7D)
Explanation:
Given:
angular speed of rotation of friction-less platform, ![\omega=5.1\ rev.s^{-1}](https://tex.z-dn.net/?f=%5Comega%3D5.1%5C%20rev.s%5E%7B-1%7D)
moment of inertia with extended weight, ![I=9.9\ kg.m^2](https://tex.z-dn.net/?f=I%3D9.9%5C%20kg.m%5E2)
moment of inertia with contracted weight, ![I'=2.6\ kg.m^2](https://tex.z-dn.net/?f=I%27%3D2.6%5C%20kg.m%5E2)
<u>Now we use the law of conservation of angular momentum:</u>
![I.\omega=I'.\omega'](https://tex.z-dn.net/?f=I.%5Comega%3DI%27.%5Comega%27)
![9.9\times 5.1=2.6\times \omega'](https://tex.z-dn.net/?f=9.9%5Ctimes%205.1%3D2.6%5Ctimes%20%5Comega%27)
![\omega'=19.419\ rev.s^{-1}](https://tex.z-dn.net/?f=%5Comega%27%3D19.419%5C%20rev.s%5E%7B-1%7D)
The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.