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3 years ago

Which of these statements about a dipole are correct? Select all that are true.

Physics

1 answer:

krok68 [10]3 years ago

7 0

Answer:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

Explanation:

A dipole is a pair of magnetized, equal or oppositely charged poles the are being separated by a distance.

The statements about a dipole that are correct are:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

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Answer:

the higher the mass the faster it falls so you need to know the mass to know how fast or slow it will fall

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3 years ago

What is the state of an object accelerating due to the force of gravity alone?

Marysya12 [62]

Answer:

Its state is in uniformly accelerated motion

Explanation:

When an object is acted upon the force of gravity only, we said that the object is in free fall.

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$F=ma$

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In case of free fall, the rate of acceleration of the object is equal to $g=9.8 m/s^2$ , the acceleration due to gravity, and it is constant. So, the object is moving by uniformly accelerated motion.

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4 years ago

What properties does a loud, shrill whistle have? a.) high amplitude, high frequency

marysya [2.9K]

Answer:

a.) high amplitude, high frequency

Explanation:

Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.

While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.

High pitch means high frequency and high frequency is perceived to have a shrill sound.

The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.

Hence, the loud shrill whistle will have high frequency and high amplitude.

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3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago

The speed of a car cruising along a highway was 90.0km h^-1. What is this value in meters per second?

faust18 [17]

Answer:

25 m/s

Explanation:

1 km/h = .277778 m/s, so 90 x .277778 = 25

5 0

3 years ago