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valkas [14]
3 years ago
10

Blocks A (mass 2.00 kg) and B (mass 10.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B

is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head on, so all motion before and after it is along a straight line. Find
(a) the maximum energy stored in the spring bumpers and the velocity of each block at that time;
(b) the velocity of each block after they have moved apart.
Physics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:dont care

Explanation:

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Answer:

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Which of the following is an example of an electromagnetic wave ?
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On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of
11111nata11111 [884]

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

          Acceleration = \frac{Final velocity - Initial velocity}{time}

  From the equation above, the unknown is final velocity:

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 since initial velocity = 0

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6 0
3 years ago
PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o
BabaBlast [244]

Answer:

Explanation:

Newton's Gravitation Law

\displaystyle \frac{GmM}{d^2}

where G is a constant, M and M the masses e d the distance betwen masses.

\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}

7 0
2 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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