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iris [78.8K]
3 years ago
12

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

ciated with a uniform E field consist of a family of planes that are perpendicular to the E field.3-The electric potential difference between two locations can exist when there is only a source charge. Even when there is no TEST charge q, the potential difference can exist between two points in the space.
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

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Howler monkeys are the loudest land animal on the planet and a single one can be heard from as far as 2.6 km away.
Komok [63]

Answer:

Between 128dB and 149dB

Explanation:

When howling the Howler monkies has a volume between 128db and 149dB. This is as a result of them having a large hyoid bone in the neck.

6 0
4 years ago
A 2.35-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 22.0 N.
ahrayia [7]

Length of wire (L)=2.35m

Mass of wire (m)=0.100kg(m)=0.100kg

Tension in the wire (T)=0.042m

<h3>What is tension?</h3>

Any physical object that is in contact with another one can apply forces to that item. Depending on the sorts of objects in touch, we label these contact forces differently. We refer to the force as tension if a rope, string, chain, or cable is one of the things applying the force.

Ropes and cables can effectively convey a force over a long distance, making them valuable for exerting forces (e.g. the length of the rope). For instance, a team of Siberian Huskies can pull a sled by attaching ropes to them, allowing the dogs to move more freely than if they had to push against the sled's rear surface with their typical effort.

learn more about tension refer:

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8 0
2 years ago
A cylindrical glass that is 10cm high is partially filled with water. You see the glass in two positions. What is the height of
olga2289 [7]

Answer:

5 cm.

Explanation:

If the glass is located upright, the height of water is about 5 cm because the height of glass is 10 cm and it is partially filled with water and we know that partial means 50 percent of something and here 50 percent of glass is 5 cm. If the glass is slightly bend, so we can't find its right height so the proper height of water is attain if it is placed on flat table and present upright.

4 0
3 years ago
What happens to electron flow with a conductor of the voltage source is removed?
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The electrons stop flowing
4 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
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