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3 years ago

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A closely-guarded count shall not be started during: a. A dribble. B. An interrupted dribble. C. A dribble away from the basket.

D. All of the above.

1 answer:

Rzqust [24]3 years ago

3 0

Answer:

Option B

Explanation:

As the rules of basketball states, during an interrupted dribble, player control can't be committed while a team control foul is permitted. Most importantly, a closely-guarded count can't be started just as time-out request can't be granted. Therefore, option B, an interrupted dribble is the right answer.

Keywords;

Basketball, dribble, closely-guarded count.

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Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

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Explanation:

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

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Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

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3 years ago

Read 2 more answers

What is the kinetic energy of a bike with a mass of 24 kg traveling at 3 m/s?

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Answer:

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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

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

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

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

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

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3 years ago

Para que sirve la caida libre

masya89 [10]

Eh? i don’t understand huhu

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2 years ago