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Juli2301 [7.4K]
3 years ago
7

Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

ass of 2 kg is at (0 m, 3 m). Where is the center of mass (or center of gravity) of the system?
Physics
1 answer:
enot [183]3 years ago
3 0

Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).                  

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction  is :

C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1

The center of mass in y direction is :

C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

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A 20-gram bullet traveling at 250 m/s strikes a block of wood that weighs 2 kg. With what velocity will the block and bullet mov
Shkiper50 [21]

Answer:

the velocity of the bullet-wood system after the collision is 2.48 m/s

Explanation:

Given;

mass of the bullet, m₀ = 20 g = 0.02 kg

velocity of the bullet, v₀ = 250 m/s

mass of the wood, m₁ = 2 kg

velocity of the wood, v₁ = 0

Let the velocity of the bullet-wood system after collision = v

Apply the principle of conservation of linear momentum to calculate the final velocity of the system;

Initial momentum = final momentum

m₀v₀ + m₁v₁ = v(m₀ + m₁)

0.02 x 250  + 2 x 0    =    v(2  + 0.02)

5 + 0 = v(2.02)

5 = 2.02v

v = 5/2.02

v = 2.48 m/s

Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s

6 0
3 years ago
An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?
sleet_krkn [62]

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

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F = f which states that the applied Force equals the frictional force, choice a.

3 0
2 years ago
A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How
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0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>
  • Mathematically, the work done to strech a string is given as 1/2 ×K×x².
  • K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
  • The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
  • Then K = Force / x.
  • Here x = 15 - 10 = 5 cm = 0.05 m
  • K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
  • Work done = 1/2×k×x²
  • Here x= 18-15=3cm or 0.03 m
  • So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

#SPJ4

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1 year ago
I'm not sure what to do on this work sheet
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