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ANEK [815]
3 years ago
12

When an unbalanced force acts on an object the change in the objects ____or ____ depends on the size and direction of the force

Physics
2 answers:
timama [110]3 years ago
7 0

When an unbalanced force acts on an object the change in the object state of rest or motion depends on the size and direction of the force.

If a body is at state of rest or motion, when an unbalanced external force acts on it, its starts moving in the direction of force and magnitude of its velocity or acceleration depends on the magnitude of force applied.

tresset_1 [31]3 years ago
6 0

Answer:speed or direction

Explanation:

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In a weak acid solution, _____.
erastovalidia [21]

Answer:

A)

Explanation:

8 0
3 years ago
Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single freque
Yuliya22 [10]

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

6 0
3 years ago
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
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
3 years ago
Free coins be quick to get them
olga nikolaevna [1]

Answer:

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2 years ago
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