What are the basic building blocks of matter

saw5 [17]

Answer:

The 10 rules of badminton are as follows:

1. A game starts with a coin toss. Whoever wins the toss gets to decide whether they would serve or receive first OR what side of the court they want to be on. The side losing the toss shall then exercise the remaining choice.

2. At no time during the game should the player touch the net, with his racquet or his body.

3. The shuttlecock should not be carried on or come to rest on the racquet.

4. A player should not reach over the net to hit the shuttlecock.

5. A serve must carry cross court (diagonally) to be valid.

6. During the serve, a player should not touch any of the lines of the court, until the server strikes the shuttlecock. During the serve the shuttlecock should always be hit from below the waist.

7. A point is added to a player's score as and when he wins a rally.

8. A player wins a rally when he strikes the shuttlecock and it touches the floor of the opponent's side of the court or when the opponent commits a fault. The most common type of fault is when a player fails to hit the shuttlecock over the net or it lands outside the boundary of the court.

9. Each side can strike the shuttlecock only once before it passes over the net. Once hit, a player can't strike the shuttlecock in a new movement or shot.

10. The shuttlecock hitting the ceiling, is counted as a fault.

Explanation:

8 0

3 years ago

Read 2 more answers

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

4 years ago

What's something that I've done in my life that is related to physics

Allisa [31]

That's a very difficult question to answer, because you give us
no information regarding what you have done in your life.

We can only assume that you have most likely breathed on occasion,
floated on your back in the ocean, lake or pool, maybe fallen off of a
ladder or out of bed, felt the warmth of the sun on your cheek, seen
a rainbow after a rainshower, heard the sound of thunder during a
summer storm, taken a trip in an airplane, and waited for a cup of
hot chocolate to cool off. The richness of any of these experiences
is greatly enhanced when you understand some of the Physics involved.

6 0

3 years ago

How is voltmeter connected in the circuit to measure the potential difference between two points?

34kurt

Voltmeter is used to find the potential difference between two points.

We always connect it in parallel to the points where we need the potential difference.

Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.

7 0

4 years ago

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

Aloiza [94]

Answer:

a = ω^2 A formula for max acceleration (ignoring sign)

V = ω A formula for max velocity

V^2 = ω^2 A^2 = a A from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

6 0

2 years ago