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stealth61 [152]
3 years ago
10

What are all the invertebrates with a large foot

Physics
1 answer:
lyudmila [28]3 years ago
5 0

Answer:

Explanation:

Bobbitt worm ( Eunice aphroditois). This segmented polychaete marine worm can attain lengths of 10 feet. It bristles...

Goliath beetle ( Goliathus species). African goliath beetle ( Goliathus giganteus ). Five species of goliath beetle...

atlas moth ( Attacus atlas). Stop and rest your eyes on this lovely...

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A 25.0 kg object moving at +15.0 m/s strikes a 15.0 kg
lesya692 [45]

The final velocity of the 15 kg mass is 18.33 m/s.

<h3>Conservation of linear momentum</h3>

The final velocity of the 15 kg mass can be determined by applying the principles of conservation of linear momentum as follows;

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\

Where;

  • m₁ is the mass of the first object = 25 kg
  • u₁ is the initial velocity of the first object = 15 m/s
  • m₂ is the mass of the second object = 15 kg
  • u₂ is the initial velocity of the second object = -10 m/s
  • v₁ is the final velocity of the first object = -2 m/s
  • v₂ is the final velocity of the second object

Thus, the final velocity of the 15 kg mass after the collision is 18.33 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

8 0
2 years ago
1. A bicyclist travels for 1.5 hours at an average speed of 23 km/h. How far does the bicyclist travel
allsm [11]

Explanation:

23 km/hr x 1.5 hrs = 34.5 km

34.5 km = 34500 meters

8 0
3 years ago
Read 2 more answers
Which of the following is the BEST example of a goal
kvasek [131]
I will run 2 miles three times a week for 4 weeks.
7 0
3 years ago
Is a
VikaD [51]

Answer:

I believe the answer is D: Conductor.

4 0
3 years ago
Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp
yuradex [85]

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

7 0
3 years ago
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