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3 years ago

Write an essay on basketball history and how the game has changed over time

1 answer:

5 0

Basket ball has gone through some changes and hence got into shape as what we play or watch nowadays. Overall rules have the same fundamental principles as set in 1891 by the founder of this game.

December 21st, in 1891, the introductory basketball game was played in Springfield, Massachusetts. It was first brought into shape by a Canadian-by birth, Dr. James Naismith. The basic idea of the new game was to keep the sports loving students in shape during the winters or in between the outdoor game seasons.

'The basket ball' which had a set of thirteen basic rules set up a strong foundation to the game and still played almost in the same style with few modifications.In the very year of 1891, after mixing and changing theme of several other played games of those days, basket ball was born.

Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

By comparing Naismith’s basketball and basketball of today we see a characteristic of the original game which had no dribbling, instead a player had to pass the ball to a another player of his team ,from his the very spot where he caught it. The second thing which has now changed is the fouls, in the original game if any team made 3 consecutive foul play ,the oponent received a goal in reward. Although this scoring technique doesn't exist in the modern basketball. Rather nowadays, if any team makes five fouls in one quarter, the offending team is in the penalty ,and shoot free throws are awarded against them.

All other rules are quiet intact in the modern play as well. As the game today,has extended to over 2 hundered countries and enjoyed by many on television screens, with modern umpiring and technologies being used in the game, but the original idea of the founder is still endured.

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A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/

kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 16.0 kg$ is the mass of the canoe

$u_1 = -12.5 m/s$ is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

$v_1$ is the final velocity of the canoe

$m_2 = 14.0 kg$ is the mass of the raft

$u_2 = +16.0 m/s$ is the initial velocity of the raft

$v_2 = -14.4 m/s$ is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

$v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s$

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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#LearnwithBrainly

5 0

3 years ago

A suitcase (mass m = 18 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures

Sindrei [870]

Answer:

P=2736 Pa

Explanation:

According to Newton we have that:

∑ $F=m*a\\$

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

∑ $F=m*g+m*a=m*(g+a)$

$F=205.2N$

We know that the pressure is given by:

$P=\frac{F}{A}\\P=\frac{205.2N}{(0.50m)*(0.15m)}\\P=2736N/m^2=2736Pa$

6 0

3 years ago

Which of the following physical properties would tell you about an element's mass per unit volume ratio?

Igoryamba

Well I can't see the following physical properties you talked about in the question.

Mass per unit volume ratio is called density.

6 0

3 years ago

A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniformly charged to - 8.00 nCnC ar

ch4aika [34]

Complete Question

A 10-cm-long thin glass rod uniformly charged to 8.00 nC and a 10-cm

long thin plastic rod uniformly charged to -8.00 nC are placed side by

side, 4.20 cm apart. What are the electric field strengths E_1 to E_3 at

distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line

connecting the midpoints of the two rods

a.) Specify the electric field strength E1

b.) Specify the electric field strength E2

c.) Specify the electric field strength E3

Answer:

$E_1=7.13*10^5 N/C$

$E_2= 2.95*10^{5} N/C$

$E_3= 3.84*10^5 N/C$

Explanation:

From the question we are told that

The length of the thin glass is $L = 10 cm$

The charge on the glass rod is $q_g = 8.00nC = 8* 10^{-9} C$

The length of the plastic rod is $L_p = 10cm$

The charge on the plastic rod is $q_p =- 8.00nC = -8.0*10^{-9}C$

The distance between the materials is $d = 4.20cm = \frac{4.2}{100} =0.042m$

The various distances to obtain electric field of are $r_1 = 1.0cm$

$r_2 = 2.0cm$

$r_3 = 3.0cm$

The objective of the solution is to obtain the electric field $E_1 , E_2 \ and E_3$ at distance $d_1 , d_2 \ and \ d_3$ from the glass rod along the line connecting its mid point

Generally electric field of a charge rod at a distance of r the line dividing the rod into half is mathematically represented as

$E = k \frac{2Q}{r\sqrt{L^2 + 4r^2} }$

For the $r_2 = 1.0cm = \frac{1}{100} = 0.01m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r \sqrt{L^2 + 4r^2_1} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_1) \sqrt{L^2 + 4r^2_1} }$

The net electric field is,

$E_{net} =E_1= E_l + E_r$

$= k \frac{2Q}{r_1\sqrt{L^2 + 4 r^2_1 } } + k \frac{2Q}{(0.04-r_1) \sqrt{L^2 + 4 (0.044 -r_1)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.01)\sqrt{(0.01^2 + 4(0.01)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.01)\sqrt{(0.01)^2 + (4) (0.042 - 0.01)^2} }$

$= 6.44*10^5 + 6.9*10^4$

$E_1=7.13*10^5 N/C$

For the $r_2 = 2.0cm = \frac{2}{100} = 0.02m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_2 \sqrt{L^2 + 4r^2_2} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_2) \sqrt{L^2 + 4r^2_2} }$

The net electric field is,

$E_{net} =E_2= E_l + E_r$

$= k \frac{2Q}{r_2\sqrt{L^2 + 4 r^2_2 } } + k \frac{2Q}{(0.04-r_2) \sqrt{L^2 + 4 (0.044 -r_2)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.02)\sqrt{(0.02^2 + 4(0.02)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.02)\sqrt{(0.02)^2 + (4) (0.042 - 0.02)^2} }$

$= 1.6*10^{5}+ 1.3*10^{5}$

$E_2= 2.95*10^{5} N/C$

For the $r_3 = 3.0cm = \frac{3}{100} = 0.03m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_3 \sqrt{L^2 + 4r^2_3} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_3) \sqrt{L^2 + 4r^2_3} }$

The net electric field is,

$E_{net} =E_3= E_l + E_r$

$= k \frac{2Q}{r_3\sqrt{L^2 + 4 r^2_3 } } + k \frac{2Q}{(0.04-r_3) \sqrt{L^2 + 4 (0.044 -r_3)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.03)\sqrt{(0.03^2 + 4(0.03)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.03)\sqrt{(0.03)^2 + (4) (0.042 - 0.03)^2} }$

$= 7.2 *10^{4} + 3.1*10^5$

$E_3= 3.84*10^5 N/C$

8 0

3 years ago

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

LenKa [72]

The question is incomplete. Here is the complete question.

A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

Answer: Net Force = $50.215.10^{-7}$ N

Explanation: Force and Magnetic field are related through the following formula:

F = I.L.B.sinθ

Magnetic field (B) in a straight long wire is given by

$B=\frac{\mu_{0}.I}{2.\pi.r}$

in which

$\mu_{0}$ is permeability of free space and is $4.\pi.10^{-7}$ T.m/A

I is current in the wire;

r is distance to the wire;

Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.

So, for the net force, the relevant forces will be on the sides parallel to the wire.

For the other forces, angle is 90°, sin(90°) = 1, then:

F = I.L.B

Replacing magnetic field:

F = $\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}$