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Ede4ka [16]
3 years ago
15

What is a complete wave?

Physics
1 answer:
Rashid [163]3 years ago
5 0
One crest of a wave to another.
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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<span> k=</span>450N/<span>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:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>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:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>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:
</span>E=U_{max}= \frac{1}{2}k A^2
<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
</span>v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }=  1.2 m/s
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
</span>E=U+K
<span>we find the kinetic energy when the glider is at this position:
</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>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
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
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) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span>a(t)=\omega^2 x(t)
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find 
</span>a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
8 0
3 years ago
A wire with mass 50 g is stretched so that it’s ends are tied down at points 100 cm apart. The wire vibrates in its fundamental
ivolga24 [154]
I think the answer is 2500 N
6 0
3 years ago
Read 2 more answers
in which of the following collisions would you expect the kinetic energy to be conserved? a. a bullet passes through a block of
Scorpion4ik [409]

An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.

<h3>Explain about the Elastic Collision?</h3>

A collision between two bodies in physics is referred to as an elastic collision if their combined kinetic energy stays constant. There is no net conversion of kinetic energy into other forms, such as heat, noise, or potential energy, in an ideal, fully elastic collision

An example of an elastic collision is when two balls collide at a pool table. It is an elastic collision when you throw a ball on the ground and it bounces back into your hand because there is no net change in the kinetic energy.

If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.

To learn more about Elastic Collision refer to:

brainly.com/question/7694106

#SPJ4

8 0
1 year ago
A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8
fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h

The final velocity of the system formed by the ballistic pendulum and the bullet is:

v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

v\approx 0.78\,\frac{m}{s}

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )

v = 186.90\,\frac{m}{s}

5 0
3 years ago
Answer <br> A)<br> B <br> C <br> please
babymother [125]

Answer:

<h2>a is b is c is</h2>

Explanation:

brash this so hard

5 0
2 years ago
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