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MAXImum [283]
3 years ago
14

Help me with # 3 & #4 please

Physics
1 answer:
defon3 years ago
6 0

Answer:

DveC başarılar yardı olduysam sevinirim

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A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool
allochka39001 [22]

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m

Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus(Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F/(AY)

So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)

When expressed in percentage, it becomes 0.00729%

7 0
3 years ago
Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B
laiz [17]

Answer:

3

Explanation:

AP3X

5 0
3 years ago
You are holding a finishing sander with your right hand. THe sander has a flywheel which spins counterclockwise as seen from beh
ryzh [129]

Answer:

c. turn downward

Explanation:

From the information given:

To find the tendency of the sander;

We need to apply the right-hand rule torque; whereby we consider the direction of the flywheel, the direction at which the torque is acting, and the movement of the sander toward the right.

Since the flywheel of the sander is in counterclockwise movement, hence the torque direction will be outward placing on the wall. However, provided that the movement of the sander is toward the right, then there exists an opposite force that turns downward which showcases the tendency in the sander is downward.

3 0
3 years ago
if a moose were chasing you through the woods its enormous mass would be very threatening but if you zigzagged then it great mas
worty [1.4K]
Because you have the same mass as the mousse
3 0
3 years ago
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