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crimeas [40]
3 years ago
5

Two identical blocks are connected to opposite ends of a compressed spring. the blocks initially slide together on a frictionles

s surface with a velocity v to the right. at some instant, the left block is moving at v/2 to the left and the other block is moving to the right. what is the speed of the center of the mass of they system at that instant?
Physics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

v to the right.

Explanation:

<u>If there is no external force on the system, then the velocity of the center of mass is </u><u>constant</u><u>. </u>

The initial velocity of center of mass is v to the right. Therefore, the velocity of the center of mass at any point is v to the right.

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According to the graph above, how large of a force is needed in order to stretch the string 1.00 meters?
gulaghasi [49]

My guess for this one would be; 400 N

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Unit 5 activity 2 I need answer
saul85 [17]

Answer:

Can you explain more in detail what it is?

Explanation:

6 0
3 years ago
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0
3 years ago
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