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jok3333 [9.3K]
3 years ago
10

What type of charge is used as the test charge to determine the direction of electric fields?

Physics
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

B small positive charge

Explanation:

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Which statement is true?
vladimir2022 [97]
<span>a.current varies throughout a parallel circuit. 

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8 0
3 years ago
Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
Sanjay is making plans for his Saturday afternoon. He can choose a health-enhancing activity like lifting weights, or he can cho
Bezzdna [24]

Answer: Sanjay can burn 100 more calories every 30 minutes if he chooses to lift weights instead of watching tv

Explanation: 133-33= 100 calories (says in article and i just answered it)

7 0
3 years ago
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0
3 years ago
Read 2 more answers
2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal
Aleksandr [31]

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is

<em>x</em> = (7.2 m/s) <em>t</em>

The object's height <em>y</em> at time <em>t</em> is

<em>y</em> = 9.4 m - 1/2 <em>gt</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

<em>v</em> = -<em>gt</em>

(a) The object hits the ground when <em>y</em> = 0:

0 = 9.4 m - 1/2 <em>gt</em>²

<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)

<em>t</em> ≈ 1.92 s

at which time the object's vertical velocity is

<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

8 0
3 years ago
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