Which of these is an example of force?

What did the scientists deduce from the fact that the ants eyes of the desert have multiple lenses

Evgesh-ka [11]

That the pupl is smaller than the nulian hope this helped

4 0

3 years ago

An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the

Bingel [31]

Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision

Explanation:

The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:

$\Delta p = F \Delta t$

where

$\Delta p$ is the change in momentum

F is the average force

$\Delta t$ is the duration of the collision

In this problem, neither side of the equation can be used to measure the change in momentum. In fact:

- The change in momentum (left side) is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

u is the initial velocity

v is the final velocity

Here the final velocity is not known, so it's not possible to use this side of the equation

- The impulse (right side) is given by

$F\Delta t$

here the average force is known, however the duration of the collision is not known, so it's not possible to use this side of the equation.

Learn more about momentum:

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago

Which road would exert the LEAST amount of friction on a car?

bearhunter [10]

Answer:

Icy roads

Explanation:

There is so little friction you slide on it way more than other roads. :)

6 0

2 years ago

What happens when an atoms losses an electron-

inysia [295]

It becomes positive

8 0

2 years ago

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