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3 years ago

7

Red light Green light is used to work on what skill?

2 answers:

prisoha [69]3 years ago

6 0

Answer:

Dribbling

Explanation:

cuz yes, is answer took test

Talja [164]3 years ago

4 0

Running for sure ptttt

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Which of the following is most likely to be an observation made by a physiologist

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4 years ago

Analyze the problem of known:<br><br> Unknown:

garik1379 [7]

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3 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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4 years ago

the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a cur

cricket20 [7]

Answer:

0.03605 V/m is the electric field in the gold wire.

Explanation:

Resistivity of the gold = $\rho = 2.44\times 10^{-8} \Omega.m$

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = $0.45\times 0.001 m$

( 1mm = 0.001 m)

Area of the cross-section = $A=\pi r^2=\pi r^(0.45\times 0.001 m)^2$

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

$R=\rho\times \frac{L}{A}$

$V(voltage)=I(current)\times R(Resistance)$ ( Ohm's law)

$\frac{V}{I}=\rho\times \frac{L}{A}$

We know, Electric field is given by :

$E=\frac{dV}{dr}$

$E=\frac{V}{L}$

$E=\frac{V}{L}=\rho\times \frac{I}{A}$

$E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m$

0.03605 V/m is the electric field in the gold wire.

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4 years ago

Read the scenario and solve these two problems. When traveling at top speed, a roller coaster train with a mass of 12,000 kg has

andreev551 [17]

First answer 5400000
second answer 46

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3 years ago

Read 2 more answers