Answer:
The positively charged particle will move towards the negatively charged particle.
Explanation:
According to Coulomb's Law, there are two types of charges--positive charge and negative charge. Coulomb said that they are either attractive or repulsive but the basic law in charges states that opposite attracts. Since they (the two charged objects) are oppositely charged, they will then move towards one another.
Explanation:
Here,
- Initial velocity (u) = 0 (as it starts from rest)
- Final velocity (v) = 60 km/h
- Time taken (t) = 15 min
<u>C</u><u>o</u><u>n</u><u>v</u><u>e</u><u>r</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>the</u><u> </u><u>quanti</u><u>ties</u><u> </u><u>into</u><u> </u><u>its</u><u> </u><u>standard </u><u>form</u><u> </u><u>:</u>
→ v = 60 km/h
→ v = ( 60 × ) m/s
→ v = ( 10 × ) m/s
→ v = m/s
→ <u>v = 16.66 m/s</u>
Also,
→ t = 15 minutes
→ t = (15 × 60) seconds
→<u> t = 900 seconds</u>
<h3><u>C</u><u>a</u><u>l</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>accele</u><u>ration</u><u> </u><u>:</u></h3>
★ v = u + at
- v is final velocity
- a is acceleration
- u is initial velocity
- t is time
→ 16.66 = 0 + 900a
→ 16.66 - 0 = 900a
→ 16.66 = 900a
→ = a
→ 0.0185 m/s² = a
∴ Acceleration is 0.0185 m/s².
<h2>_________________________</h2>
<h3><u>C</u><u>a</u><u>l</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>distan</u><u>ce</u><u> </u><u>cover</u><u>e</u><u>d</u><u> </u><u>:</u></h3>
★ v² - u² = 2as
- v is final velocity
- a is acceleration
- u is initial velocity
- s is distance
→ (16.66)² - (0)² = 2 × 0.0185 × s
→ 277.5556 - 0 = 0.037s
→ 277.5556 = 0.037s
→ = s
→ 7501.50 m = s
∴ Distance covered is 7501.50 m .
Answer:
PART A
It is always zero
PART B
Answer is : 0
Explanation:
PART A :It is always zero because according to my research, The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
PART B: You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will enclose zero net charge. This surface can be imagined around any point inside the conductor with the same result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting object, where there is an electric field.