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3 years ago

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Fed [463]3 years ago

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Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago

What's the weight of a 30x30x50 cm body with the density of 1.8/cm cube?

grin007 [14]

Answer:

The weight of the body, W = 793.8 m/s²

Explanation:

Given,

The volume of the body, v = 45,000 cm³

The density of the body, ρ = 1.8 g/cm³

The mass of the body is given by the formula,

m = ρ x v

= 1.8 g/cm³ x 45,000 cm³

= 81,000 g

Hence, the mass of the body is m = 81 kg

The weight of the body,

W = m x g

= 81 kg x 9.8 m/s²

= 793.8 m/s²

Hence, the weight of the body, W = 793.8 m/s²

3 0

3 years ago

Solve the problems.

faltersainse [42]

Answer:

v= 1.911×10^8m/s

Explanation:

n=c/v

v=c/n

v= 3.0×10^8/1.57

v= 1.911×10^8m/s

7 0

3 years ago

A car travels around a curve. What can you say about the centripetal force if the velocity is tripled?

Lesechka [4]

Answer:

F = M a = M v^2 / R

If v is increased by three the force will be increased by nine,

C) is correct

4 0

2 years ago

A potter's wheel is a uniform disk of mass of 10.0 kg and radius 20.0 cm. A 2.0-kg lump of clay, roughly cylindrical with radius

Lera25 [3.4K]

Answer:

b. 29.2 rev/min

Explanation:

Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:

$L_{0} = L_{f} (1)$

The initial angular momentum L₀, can be expressed as follows:

$L_{0} = I_{0} * \omega_{0} (2)$

where I₀ = initial moment of inertia = moment of inertia of the disk +

moment of inertia of the cylinder and ω₀ = initial angular velocity =

30.0 rev/min.

Replacing by the values, we get: $I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2} = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)$ ⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/min
The final angular momentum can be written as follows: