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pishuonlain [190]
3 years ago
6

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

frictionless surface and when the carts collide they stick together. What is the ratio of the initial kinetic energy of the two-cart system to the final kinetic energy of the two-cart system?
a. 2
b. 1.5
c. 2.5
d. 1.25
e. 3
Physics
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:b

Explanation:

Given

mass of first cart m_1=6 kg

mass of second cart m_2=3 kg

velocity of first cart v_1=3 m/s

conserving momentum

m_1v_1+m_2v_2=(m_1+m_2)v

6\times 3+3\times 0=(9)\cdot v

v=2 m/s

Initial kinetic Energy K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2

K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0

K.E._1=27 J

Final Kinetic Energy

K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J

Ratio of initial Kinetic Energy to the Final Kinetic Energy

=\frac{27}{18}=1.5

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An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =
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Answer:

a) 0, = -0.33 us

b) 140m

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Explanation:

a)

the lorentz factor expression is written as;

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where v  is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

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y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

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x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

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x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

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1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

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delltaX' = X2' - X1'

= 140m - 0

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c)

No, The event are not simultaneous i.e they did not occur at the same time.

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4 0
3 years ago
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Answer:

Given that,

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3 0
3 years ago
Anybody got any answers???
dexar [7]

Answer:

b? to a?

Explanation:

6 0
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