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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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Compare the two circuit diagrams If one of the resistors is turned off (a light bulb goes out), what happens to the other resist
ella [17]

Answer:

it’s C

Explanation:

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IF a particular galays has a total mass of 10 to the 46 grams and the average star in that galaxy has a mass of 10 to the 36 gra
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Answer:

1.0E46     grams in galaxy

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3 0
2 years ago
In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second project
Trava [24]

Answer:

  Second  projectile is 1.4 times faster than first projectile.

Explanation:

By linear momentum conservation

Pi = Pf

m x U + M x 0 = (m + M) x V

U= \dfrac{(m + M)\times V}{m}

Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

KEi = \dfrac{(m + M)\times V^2}{2}

PEf = (m + M) g h

So,

\dfrac{(m + M)\times V^2}{2} + 0 = 0+ (m + M) g h

V =\sqrt {2gh}

So from above value of V

Initial velocity of projectile =U

U=\dfrac{(M+m)\sqrt{2gh}}{m}

Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.

Which means

\dfrac{U_2}{U_1}=\sqrt{\dfrac{h_2}{h_1}}

U_2=\sqrt{\dfrac{h_2}{h_1}}\times U_1

U_2=\sqrt{\dfrac{5.2}{2.6}}\times U_1

U₂ = 1.41 U₁

Therefore we can say that ,Second  projectile is 1.4 times faster than first projectile.

4 0
3 years ago
A forklift holds a 500 kg box 4 m in the air for 100 s. what is the minimum power required by the forklift
sineoko [7]
First of all, we need to calculate the work done by the forklift to lift the box of 500 kg by 4 meters. The work done is equal to the variation of gravitational potential energy of the box, given by:
W=\Delta U=mg\Delta h
where m=500 kg is the mass of the box, g=9.81 m/s^2 and \Delta h=4 m is the variation of height of the box. Substituting these data into the equation, we find the work
W=mg\Delta h=(500 kg)(9.81 m/s^2)(4 m)=19620 J

And now we can calculate the minimum power required by the forklift, which is equal to the ratio between the work done and the time taken:
P= \frac{W}{t}= \frac{19620 J}{100 s}=196.2 W
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3 years ago
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Answer:

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