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loris [4]
3 years ago
15

Let two objects of equal mass m collide. Object 1 has initial velocityv, directed to the right, and object 2 isinitially station

ary.
If the collision is perfectly elastic, whatare the final velocities v_1 and v_2 of objects 1 and 2?
Physics
1 answer:
3241004551 [841]3 years ago
8 0
<span>Our equation 1 would be
m*v=M*V1+m*V2
v=V1+V2
 v-V1=V2 

the equation 2 would look like this
 </span>V^2=V1^2+V2^2  
V^2-V1^2=V2^2
(V-V1)*(V+V1)=V2^2Dividing with the 1
V+V1=V2 
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A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h
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v = 9 sin 60 - 9.8 t

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t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

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The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength
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8 0
3 years ago
Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .
lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² + 17 t - 0.049 = 0

now,

t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² - 17 t - 0.049 = 0

now,

t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0
3 years ago
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