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3 years ago

How do line symmetry differ from rotational symmetry?.

Physics

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

A line of symmetry is a line that separates a shape into two identical halves.
Rotational symmetry is the same thing except when you rotate the object, it has to have the exact same line of symmetry.

Hope this helps!!!

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A golf ball is rolling in the grass. What must happen to stop the ball from continuing to roll?

slamgirl [31]

The net force must be zero

This is in accordance to Newton's first law, which states that any object in motion will remain in motion and any object at rest will remain at rest unless acted upon by an unbalanced force. An unbalanced force is one where the net force is not zero. If no unbalanced force is applied to a moving object, it will keep moving forever. The reason that we do not observe this in our daily lives is due to friction acting as the unbalanced force.

4 0

3 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

What’s the kinetic energy of the roller coaster at the top and bottom of the hill? Use . A kiddie roller coaster car has a mass

serious [3.7K]

K.E1=1/2×100×3²
=50×9
=450J
K.E2=1/2×100×36
=50×36
=1800J

0 0

4 years ago

Read 2 more answers

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

-BARSIC- [3]

Answer:

4.2s

Explanation:

Given parameters:

Power = 2190W

Mass of box = 1.47 x 10⁴g

distance = 6.34 x 10⁴mm

Unknown:

Time = ?

Solution:

Power is the rate at which work is done;

Mathematically;

Power = $\frac{work done}{time}$

Time = $\frac{work done}{power}$

Work done = weight x height

convert mass to kg;

100g = 1kg;

1.47 x 10⁴g = 14.7kg

convert the height to m;

1000mm = 1m

6.34 x 10⁴mm gives 63.4m

Work done = 14.7 x 9.8 x 63.4 = 9133.4J

Time taken = $\frac{9133.4}{2190}$ = 4.2s

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3 years ago

A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu

Zielflug [23.3K]

-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero

7 0

3 years ago