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Nikitich [7]
3 years ago
7

A box, compressing a spring by 4 m, is loaded and ready. Moments later the 8.0 kg box has been launched with

Physics
1 answer:
Mrac [35]3 years ago
3 0

Answer:

50 N/m

Explanation:

Elastic energy = kinetic energy

EE = KE

½ kx² = ½ mv²

½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²

k = 50 N/m

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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
Distance between the two place is 35 kilometers . If two persons start running towards each other from these two place with a sp
pashok25 [27]

The first person = A

The second one = B

velocity of A = 3 km / h

velocity of B = 4 km / h

Distance of A = 3t

Distance of B = 35 - 4t

At what time do they meet?

3t = 35 - 4t

7t = 35

t = 5 hours

7 a.m. + 5 hours = 12 p.m.

7 0
3 years ago
Positive electric charges are always attracted to ________ charges.
Georgia [21]

Answer:

Negative electric charges

4 0
2 years ago
The SI unit of refractive index is ___<br>a)km/s. b)m/s. c)sec. d)no unit​
cupoosta [38]

Answer:

d)no unit

Explanation:

refractive index is a unit less quantity.

6 0
2 years ago
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 10 m from the takeoff point. you may want t
Stells [14]

<u>Answer:</u>

Takeoff speed of Kangaroo = 12.91 m/s

<u>Explanation:</u>

  The distance reached by Kangaroo =  10 meter

  Angle at which it jumps = 18°

  The motion of Kangaroo is like a projectile, the distance traveled is the range of projectile.

   Range of projectile = Time taken for the projectile to reach ground* Horizontal velocity

   Time taken for the projectile to reach ground:

       Time taken = Two times of the time taken for the projectile to reach maximum height

       Time taken for the projectile to reach maximum height = Vertical speed / Acceleration = u sin θ/g

      Time taken for the projectile to reach ground = 2 u sin θ/g

  So Range of projectile = ucos\theta*\frac{2usin\theta}{g} =\frac{u^2sin2\theta}{g}

 We have Range = 10 meter, θ = 18⁰

    Substituting

        10=\frac{u^2sin(2*18)}{9.8}\\ \\ u^2= 166.73\\ \\ u=12.91 m/s

 Takeoff speed of Kangaroo = 12.91 m/s

3 0
3 years ago
Read 2 more answers
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