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joja [24]
3 years ago
6

Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2

Physics
1 answer:
Ann [662]3 years ago
6 0

Answer:

sqdqk3

Explanation:

qdjqưbdkq ưqdhud j

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Which most likely indicates a chemical change has occurred?
Law Incorporation [45]

green liquid becoming a red liquid

4 0
2 years ago
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What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?
Lisa [10]
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
7 0
3 years ago
A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc
alexdok [17]

Answer:

20.25 m

Explanation:

  • <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

         <em><u>ac = v²/r</u></em>

<em>         </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m</u></em>

Therefore;

r = v²/ac

  = 27²/36

  = 20.25 m

Hence the radius is 20.25 meters

5 0
2 years ago
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The Earth orbits around the sun because the gravitational force that the sun
kotykmax [81]
<h3>Question -:</h3>

The Earth orbits around the sun because the gravitational force that the sun

exerts on the Earth:

O A. causes Earth's acceleration toward the sun.

O B. is very small because the sun is so far from the Earth.

O c. is smaller than the force the Earth exerts on the sun.

O D. pushes the Earth away from the sun.

<h3>Answer -:</h3>

O A. causes Earth's acceleration toward the sun.

<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>

5 0
2 years ago
In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f
Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2  \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0
3 years ago
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