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Tamiku [17]
4 years ago
5

How to find velocity right before impact?

Physics
1 answer:
lisabon 2012 [21]4 years ago
6 0
1. The problem statement, all variables and given/known data A person jumps from the roof of a house 3.4 meters high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 meters. If the mass of his torso (excluding legs) is 41 kg. A. Find his velocity just before his feet strike the ground. B. Find the average force exerted on his torso by his legs during deceleration. 2. Relevant equations I can't even seem to figure that part out. Help please? 3. The attempt at a solution I don't know how to start this at all
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When a car turns a corner on a level road, which force provides the necessary centripetal acceleration?
Vinil7 [7]

Answer:

The frictional force between the tire made with the road

Explanation:

This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

5 0
3 years ago
An important difference between waves and classical particles is that A. waves cannot interfere. B. particles cannot interfere.
pishuonlain [190]
<span>C. waves can only interfere destructively. </span>
5 0
3 years ago
_____ life cycle models of systems development assume that the scope of the project can be articulated clearly and the schedule
Natalka [10]

Answer:

Predictive

Explanation:

 There is a wide variety of life cycles applicable to project management. In this regard, we currently have as a reference two somewhat extreme approaches that mark two very different ways of dealing with projects. The predictive approach and the agile approach.

  A predictive approach involves a great effort in initial planning and re-planning every time changes are accepted in the project. Therefore, this approach is recommended for changing but not highly changing environments. Although this approach is applicable to any type of project, clear examples of application would be the construction of a subway line, a bridge, the development of critical software. That is, projects where correct and detailed planting is key.

4 0
3 years ago
A large object las Less of an attraction on a smaller <br><br> True <br> False
leva [86]
Do you mean ‘Does a larger object have less attraction than a smaller object?’ If so, then the answer is objects with a larger mass exert more attraction while objects with a smaller mass have less attraction.
6 0
3 years ago
A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g
Alexus [3.1K]

Answer:

\Delta v= 4.66\frac{m}{s}

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system  the total momentum is constant if no external forces act in the system. The formula is:</em>

m_1v_1+m_2v_2=m_1u_1+m_2u_2

<em>Where:</em>

m_1: Mass of the first object.

m_2: Mass of the second object.

v_1: Initial velocity of the first object.

v_2: Initial velocity of the second object.

u_1: Final velocity of the first object.

u_2: Final velocity of the second object.

In <u>this problem</u> we have:

m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}

u_1=0\frac{m}{s}

Observation: v_1_2: Is because the system has the same initial velocity.

First we have to find u_2,

m_1v_1+m_2v_2=m_1u_1+m_2u_2

We can rewrite it as:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2

Replacing with the data:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}

Then, the resulting change in the cart speed is:

\Delta v= 4.66\frac{m}{s}

5 0
3 years ago
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