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zubka84 [21]
3 years ago
14

A 5000 kg train is traveling at a velocity of 100 m/s and hits another train. The two trains stick together and the new velocity

is 50m/s. What is the mass of the second train?

Physics
1 answer:
Nadya [2.5K]3 years ago
4 0

Answer:

Explanation:

Conservation of momentum is used to solve

Unfortunately we have a missing piece of information such as the initial velocity of the unknown mass train.

If we ASSUME that the second train is at rest

5000(100) + m(0) = 5000(50) + m(50)

which means m = 5000 kg

However, I'll show you the importance of knowing that initial velocity by finding it assuming the other answers are valid

if m = 15000 kg

5000(100) + 15000(v₀) = (5000 + 15000)(50)

v₀ = 33 ⅓ m/s

if m = 10000 kg

5000(100) + 10000(v₀) = (5000 + 10000)(50)

v₀ = 25 m/s

if m = 8000 kg

5000(100) + 8000(v₀) = (5000 + 8000)(50)

v₀ = 18.75 m/s

So you can see why I had to assume an initial velocity. Any of the masses could work if the initial velocity is chosen correctly.

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A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

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