One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is uk = 0.400. You apply a constant force F to the block. F has magnitude F = 82.0 N and is directed toward the wall. At the instant that the spring is com-pressed 80.0 cm, what arc (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?

The speed of the combined object after collision is 0 m/s.

Total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)a

m₁ = object 1 mass = m, u₁ = velocity of object 1 before collision = v, m₂ = mass of object 2 = 3m, u₂ = velocity of object 2 before collision = -v/3, a = velocity after collision

mv + 3m(-v/3) = (m + 3m)a

mv - mv = 4ma

0 = 4ma

a = 0 m/s

The speed of the combined object after collision is 0 m/s.

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Answer:

the answer is 0 m/s

Explanation:

This question is describing the law of conservation of momentum

First object has mass =m

velocity of first object = v

second object = 3m

velocity of second object = v/3

the law of conservation of momentum is expressed as

m1V1 - m2V2 = (m1+ m2) V

substituting the parameters given;

making V as the subject of formular

V =

V =

V =

= 0 m/s

Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

So the resultant direction is 21.48 km 2.92° north of east.

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

Answer:

σ  = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

Therefore the surface tension σ  will be 0.8 N/m .

σ  = 0.8 N/m

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

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When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

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