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professor190 [17]
3 years ago
9

two small balls are suspended on parallel threads of the same length so that they touch each other in the vertical position. the

mass of the first ball is 0.4kg and the mass of the second ball is 200g. the first ball is deflected so that its centre of mass rises to a height of 6cm and it is then released. to what maximal height will the 200g ball rise if the collision is elastic?
Physics
1 answer:
Elden [556K]3 years ago
5 0

Answer: 12cm

Explanation:

Since first ball is raised to a height h1= 6cm, it gains a potential energy(PE1). On release, this PE1 is converted to kinetic Energy (KE1). This KE1 impact it energy on ball 2 ans it rise also. Since collision is elastic.

KE1 = KE2 ....> PE1 = PE2

m1 ×g × h1 = m2 × g× h2

h2 = m1×h1/m2

h2 = 0.4 × 0.06/0.2

h2 = 0.12m = 12cm

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Explanation:

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When the water in the barrel is poured overboard, the level of the swimming pool level would remain unchanged as the weight of the boat  with the water and barrel would remain unchanged ( as the density and volume of the whole system remains same) and hence, the weight of the water (of the swimming pool) displaced by the boat would remain same.

A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming pool level will <u>remain unchanged. </u>

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an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
muminat

Answer:

4°C

Explanation:

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6 0
3 years ago
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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
3 years ago
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