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3 years ago

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If a = 7 × 10−6 C/m4 and b = 1 m, find E at r = 0.6 m. The permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in un

its of N/C

1 answer:

mart [117]3 years ago

7 0

Answer:

The electric field potential (E) is 1.75 X 10⁵ N/C

Explanation:

Electric field potential (E) is force per unit charge.

Electric field potential (E) = force/charge = f/q, unit is Newton (N) per Coulomb (C); N/C

$E = \frac{q}{4 \pi \epsilon r^2}$

where;

q is charge in coulomb (C)

ε is permittivity of free space = 8.8542 X 10⁻¹² C²/Nm²

$k = \frac{1}{4 \pi \epsilon} = \frac{1}{4\pi (8.8542 X 10^{-12})} = 8.99 X10^9 Nm^2/C^2$

if a = 7 X 10⁻⁶ C/m⁴ and b = 1m

a in "C" = $7 X 10^-{6} \frac{C}{m^4} X (1m)^4 = 7 X 10^-{6} C$

Electric field potential at r = 0.6m = $\frac{ka}{r^2}$

E = $\frac{(8.99 X10^9)(7X 10^{-6})}{0.6^2}$ = 1.75 X 10⁵ N/C

Therefore, the electric field potential (E) is 1.75 X 10⁵ N/C

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A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

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Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

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$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(30)^2}{2\times 4}$

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

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Does electrons have more mass than neutrons?

musickatia [10]

Answer:

No, neutrons have about the same mass as a proton, but both have more mass than electrons.

Hope this helps a bit,

Flips

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3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

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$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

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so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

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$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

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3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

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Let $p_{up}$ be the pressure of a point just above the wing.

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Given that area is $27ms^{2}$ .

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Stolb23 [73]

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