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sergeinik [125]
3 years ago
5

A string is wound around a uniform disc of radius 0.68 m and mass 3.7 kg. The disc is released from rest with the string vertica

l and its top end tied to a fixed support. calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 1.2 m. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
Zolol [24]3 years ago
8 0

Answer:

3.962

Explanation:

mass = 1.2

gravity g = 9.81

velocity = ?

mgh = 1/2mv² + 1/2(1/2)mv²

= 9.81 x 1.2  = 3/4v²

11.772 x 4 = 3/4v²

47.088 = 3v²

divide through both sides by 3

15.696 = v²

take the square root of both sides

√15.696 = √v²

= 3.6962 = v

the speed is therefore 3.692

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Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0
3 years ago
If a car changes its velocity from 32km/hr in 8.0 seconds what is its acceleration
puteri [66]

We don't know the change in velocity, so can't answer.

3 0
3 years ago
The Trans-Siberian Railroad is the longest single railroad in the world. Starting in Moscow, the tracks stretch 9,354 km across
Lubov Fominskaja [6]
103.9 hours, if you never stopped for any reason.
4 0
4 years ago
Imagine that you pushed a box, applying a force of 30 Newton’s , over a distance of 5 meters. How much would you do have done ?
kati45 [8]
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7 0
3 years ago
Read 2 more answers
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

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m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0
3 years ago
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