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Lady_Fox [76]
3 years ago
15

A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500

g is attached. The spring-mass is then stretched 50 cm, and released. What is the top speed of the attached ball? 0 A. 4.47 m/s B. 447 m/s C. 0.14 m/s D. 6.32 m/s
Physics
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

E_e = E_k

kx^2/2 = mv^2/2

120*0.5^2/2 = 1.5*v^2/2

15 = 0.75v^2

v^2 = 15 / 0.75 = 20

v = \sqrt{20} = 4.47 m/s

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a truck drives up a hill with a 15 incline. if the truck has a constant speed of 22m/s, what are the horizontal and vertical com
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Answer:

speed of truck (v) =  22 m/s ,

angle of hill (Θ) =15°

Find

Vertical component (Fv) = ?

Harizontal component (Fh) =?

               Vertical component (Fh) = V cosΘ

                                                        = 22. cos 15

                                                        = 21.25 m/s.

               Harizontal component (Fv) = V sinΘ

                                                            = 22. sin 15

                                                            = 5.69 m/s.

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3 years ago
Match each measurement tool with what it measures.
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Thermometer-temperature
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3 years ago
Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4
arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

  • initial velocity of the car, u = 76 m/s
  • acceleration of the car, a = - 9 m/s²
  • time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

  • height above the ground, h = 500 m
  • velocity of spaceship, u = 150 m/s
  • time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) -  (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m

Learn more here: brainly.com/question/24527971

5 0
2 years ago
A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
blsea [12.9K]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest),  s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

1.30 = 0.36125a

a = 1.30/0.36125 = 3.6 m/s²

Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b.  Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

T = m(a + g)

substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg

T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N

c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.

α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

From the equation above, I = Tr/α

Substituting the variables we have

I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²

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KE=1/2mv²
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3 years ago
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