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ANTONII [103]
3 years ago
10

Two balls collide in a head-on elastic collision and rebound in opposite directions. One ball has velocity 1.2 m/s before the co

llision, and −2.3 m/s after. The other ball has a mass of 1.1 kg and a velocity of −4.2 m/s before the collision. (a) What is the mass of the first ball? (b) What is the velocity of the second ball after the collision?
Physics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:

If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .

Explanation:

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What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the
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Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

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A weightlifter lifts a 1,250 N barbell 2 m in 3 s. How much power was used to lift the barbell?
OverLord2011 [107]
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3 years ago
In this case, lithium would be classified as a(n)
Black_prince [1.1K]

Lithium is to be considered as an ION

Atom is defined as its natural state where it will have all its electrons present in its shell

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So here answer must be

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3 years ago
Read 2 more answers
An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta
icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

v=\sqrt{\frac{T}{\mu} }

where T= 200 N is tension in the string , \mu=1.0 g/m is the linear mass density

v=\sqrt{\frac{200}{1\times 10^{-3} }

v=447.2 m/s

Wavelength of the wave in the string is

\lambda =2L=2\times 0.8=1.6 m

The frequency is

f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m

8 0
3 years ago
A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is
Wewaii [24]

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

4 0
3 years ago
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