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2 years ago

An object is thrown straight up into the air at 10 m/s. how fast is the object traveling after 2 seconds?

Physics

1 answer:

ziro4ka [17]2 years ago

7 0

First write down all the known variables:

vi = 10m/s

t = 1

a = -9.8m/s^2

vf = ?

Now choose the kinematic equation to relate these variables:

vf = vi + at

vf = (10) + (-9.8)(1)

vf = 0.2 m/s

The speed of the object travelling after a second would be 0.2m/s in reality, however, it is closest to the choice A, which would suffice as the answer for your question.

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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

I NEED HELP ASAP

34kurt

I think it’s gonna be b.

4 0

2 years ago

Thylakoids contain chlorophyl that absorb solar energy

steposvetlana [31]

Answer:

True

True statement:

Because pigment molecules absorb solar energy and thylakoids are pigment molecules

3 0

2 years ago

If the center of mass passes outside the area of support of an object, what will happen to it?

defon

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

6 0

3 years ago

Besides, the discovery that moons orbit Jupiter, what other discovery made by Galileo Galilei, with an early telescope, proved t

Thepotemich [5.8K]

Answer:

He made great advancements in developing a logical way to know more about the universe and celestial entities inside the space. And this theory is termed to be heliocentric in nature.

Explanation:

In early times most of the people believed that our planet Earth is the center of the universe or the solar system and rest of the celestial entities move around it in a given path, so, it confused the well known scientist named as Galileo Galilei. As, he observed the different dark patches or shadow like textures on the face of the Sun.

While, it is more obvious to known that any object having multiple small shadows means that it is present inside such a region that all of the celestial entities move or orbit around it in a given way.So, he concluded that planet Earth itself move around the red Giant in a given way rather then being the center of the universe.

5 0

3 years ago