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FinnZ [79.3K]
3 years ago
15

A 5.0-kg clay putty ball and a 10.0-kg medicine ball are headed towards each other. Both have the same speed of 20 m/s. If they

collide perfectly inelastically, what approximately is the speed of the blob of clay and ball immediately after the collision?
Physics
1 answer:
babymother [125]3 years ago
8 0

Answer:

- 3.33 m /s

Explanation:

m1 = 5 kg

m2 = 10 kg

u1 = 10 m/s

u2 = - 10 m/s

Let the velocity of the combination of the blob is v.

Use the conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v

5 x 10 - 10 x 10 = (5 + 10)v

50 - 100 = 15 v

v = - 3.33 m /s

Thus, the velocity of the blob after sticking together is - 3.33 m/s.

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I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app
Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

  • initial velocity of the sprinter, u = 18 km/h
  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0
3 years ago
A man travles at an average velocity of 12 m/s for 4 seconds. what is the total distance travled by the man?
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The sum of all forces acting on an object
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Answer:

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Explanation:

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