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Sergio039 [100]
3 years ago
6

I WIL A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. After the collision

, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? . .
Physics
1 answer:
pishuonlain [190]3 years ago
4 0
Using the following given values:

Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s 

Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span> 
</span>
Obtaining Va2 via law of conservation of momentum:

total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67

Total kinetic energy before collision:

KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J

</span>Total kinetic energy after collision:

KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:

Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
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3 years ago
A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?
lesantik [10]
<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>
  • Half life is the time taken for a radioactive isotope to decay by half of its original amount.
  • We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
  • A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

<em>1/4 = 1 × (1/2)^n</em>

<em>1/4 = (1/2)^n </em>

<em>n = 2 </em>

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0
3 years ago
Why are computer simulations useful in studying phenomena in the universe?
kenny6666 [7]

Computer simulation is useful because it helps in the prediction of what will likely happen in the future using data from past events.

<h3>What is computer simulation?</h3>
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5 0
2 years ago
A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on
makkiz [27]

Answer:

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh =  0.262 N

Now, using the law of conservation of energy

initial energy= final energy

Ki +Ui = Kf + Uf +Ea

1/2 mv₀² + 0 = 0 + mgh + fh

1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

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6 0
3 years ago
A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?
Anastaziya [24]

Answer: 1.089\times 10^5\ A

Explanation:

Given

Charge discharged Q=9.7\C

time taken t=8.9\times 10^{-5}\ s

Current is given as rate of change of discharge i.e.

\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A

Therefore, the average current is 1.089\times 10^5\ A

3 0
2 years ago
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