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:
The answer is b !!!! Hope it helps
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Answer:
These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.
Explanation:
