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3 years ago

The diagram shows two balls before they collide.

Physics

2 answers:

mihalych1998 [28]3 years ago

6 0

Answer:

B. 0.2 kg x m/s

Explanation:

Ronch [10]3 years ago

4 0

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

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A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

One Newton is equivalent to A. 1 kg/s2 B. 1 kg*m/s C. 1 kg*m/s2 D. 1 kg/s

tiny-mole [99]

Answer:

Explanation:

7 0

3 years ago

A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,

Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

$W=ma$

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

$\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}$

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0

1 year ago

As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant

Alika [10]

Answer:

Isolated or Closed system, both are correct

4 0

3 years ago

Aristotle supported which of these views?

katrin2010 [14]

The answer is A. The Sun and all the planets revolve around Earth.

Aristotle believed that the Earth was the centre of the solar system, and the Sun and the planets orbited around it. He believed that the universe was composed of Earth-like bodies, which were at rest, and of heavenly bodies, which were in perpetual motion.

3 0

3 years ago