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barxatty [35]
3 years ago
13

A circular plastic rod with uniform charge q produces an electric field of magnitude e at the center of curvature (at the origin

). in figs. 22-28b, c, and d, more circular rods,each with identical uniform charges q,are added until the circle is complete.a fth arrangement (which would be labeled
e.is like that in d except the rod in the fourth quadrant has charge q.rank the ve arrangements according to the magnitude of the electric eld at the center of curvature,greatest rst
Physics
1 answer:
Olin [163]3 years ago
4 0
Electric eld at the center of
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Identify each of the quantities below as either vector or scalar
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Answer:

2nd answer 3.2m/s is a scalar l, all others are vectors.

Explanation:

In order to be considered as a vector a certain quantity must have both a magnitude and a direction.

The most common example is velocity. It has both magnitude and direction, like the 1st and the last answers.

The 3rd answer is an acceleration. It is also given with its direction. So we have to consider it also as a vector.

But 2nd answer has no direction mentioned, so it is considered as a scalar.

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Where is the Long Head located?<br> Outside of the Arm<br> Inside of the Arm<br> On the Shoulder
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Answer:

The biceps muscle is located at the front of your upper arm. The muscle has two tendons that attach it to the bones of the scapula bone of the shoulder and one tendon that attaches to the radius bone at the elbow. The tendons are tough strips of tissue that connect muscles to bones and allow us to move our limbs.

Explanation:

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Hunter wants to measure the weight of a dumbbell. He should
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Use a scale and record the weight in cm^3
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4 years ago
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NEED HELP BADLY!!
Charra [1.4K]

Answer:

Explanation:

v² = u² + 2as

v² = 0² + 2(50)(241)

v² = 2410

v = 49.0917508...

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3 years ago
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth.What is the acc
garri49 [273]

Answer:

(a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

Explanation:

Given that,

Radius of Titania R_{t}= \dfrac{1}{8}R_{e}

Mass of Titania M_{t}= \dfrac{1}{1700}M_{e}

We need to calculate the acceleration due to gravity at the surface of Titania

Using formula of the acceleration due to gravity on earth

g_{e}=\dfrac{GM_{e}}{R_{e}^2}

The acceleration due to gravity on Titania

g_{t}=\dfrac{GM_{t}}{R_{t}^2}

Put the value into the formula

g_{t}=\dfrac{G\times\dfrac{1}{1700}M_{e}}{(\dfrac{1}{8}R_{e})^2}

g_{t}=\dfrac{64}{1700}\times G\dfrac{M_{e}}{R_{e}^2}

g_{t}=0.004705 g_{e}

g_{t}=0.03764\times9.8

g_{t}=0.37\ m/s^2

The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). We assume Titania is a sphere

The average density of the earth is 5500 kg/m³.

We need to calculate the average density of Titania

Using formula of density

\rho=\dfrac{m}{V}

\rho_{t}=\dfrac{M_{t}}{\dfrac{4}{3}\pi\times R_{t}^2}

\rho_{t}=\dfrac{\dfrac{M_{e}}{1700}}{\dfrac{4}{3}\pi\times(\dfrac{R_{e}}{8})^2}

\rho_{t}=\dfrac{512}{1700}\times\rho_{e}

\rho_{t}=\dfrac{512}{1700}\times5500

\rho_{t}=1656.47\ kg/m^{3}

The average density of Titania is 1656.47 kg/m³

Hence, (a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

8 0
3 years ago
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