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ivanzaharov
3 months ago
9

A rod 12.0 cm long is uniformly charged and has a total charge of -20.0 µc. determine the magnitude and direction of the electri

c field along the axis of the rod at a point 32.0 cm from its center.
Physics
2 answers:
Ostrovityanka [3.2K]3 months ago
7 0
<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

</span>
Ostrovityanka [3.2K]3 months ago
7 0

The equation for calculating the electric field resulting from a charged rod is:

E = k Q / [D (D + L)]

In this context,

E = electric field

k = Coulomb’s constant = 9 * 10^9 N m^2 / c^2

Q = total charge = -20.0 µc = -20.0 * 10^-6 c

L = length of rod = 12.0 cm = 0.12 m

D = distance from the center = 32.0 – 12.0 cm = 0.20 m

 

By substituting the known values into the formula:

E = (9 * 10^9 N m^2 / c^2) * (-20.0 * 10^-6 c) / [0.20 m (0.20 m + 0.12 m)]

E = -2,812,500 N

Since the value of E is negative, it indicates that E is directed toward the center of the rod.

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t
ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

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Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

T\cos\theta=mg

T=\dfrac{mg}{\cos\theta}

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T\sin\theta=m\omega^2 r

Substituting the tension value

mg\tan\theta=m\omega^2(d+L\sin\theta)

\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}

Substituting the value into the equation

\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}

\omega=1.34\ rad/s

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0
4 months ago
For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl
Yuliya22 [3333]

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

5 0
3 months ago
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