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yaroslaw [1]
2 years ago
12

What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =

8.99 × 109 N • ___ V What is the electric potential at a distance of 99 m from the charge?___ V
Physics
2 answers:
Doss [256]2 years ago
6 0
The first part of the question is 3,100 V.

The second part of the question is 200 V.
const2013 [10]2 years ago
3 0

Electric potential due to a point charge is given by

V = \frac{kQ}{r}

here we know that

Q = charge

r = distance from the charge

V = \frac{9*10^9* 2.2* 10^{-6}}{6.3}

V = 3142.8 Volts

now again by the same equation

V = \frac{kQ}{r}

V = \frac{9*10^9* 2.2* 10^{-6}}{99}

V = 200 Volts


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A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the
Orlov [11]

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻²  ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻²   ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

3 0
1 year ago
From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

T^2\propto a^3\\\\T^2=ka^3

It is mentioned that, an asteroid with an orbital period of 8 years. So,

(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

7 0
1 year ago
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma
konstantin123 [22]

Answer:

a=2.5\ m/s^2

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2

So, the required angular acceleration of the pebble is equal to 2.5\ m/s^2.

8 0
1 year ago
When the volcano Krakatoa erupted in 1883, it was heard 5000 km away. Which statement about the sound from the volcano is not co
lina2011 [118]
Answer a) is incorrect as sound does not travel in a vacuum.
7 0
2 years ago
(b) Figure 4 shows a car travelling on a motorway.
Alik [6]

Answer:

To calculate anything - speed, acceleration, all that - we need <em>data</em>. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to <em>measure </em>it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring <em>time </em>and <em>distance</em>, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

  1. Press start on the stopwatch right as you pass a lamp post
  2. Each time you pass another lamp post, press the lap button on the stopwatch
  3. Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
  4. Record your data
  5. Calculate the time intervals for passing each lamp post using the lap data
  6. Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

5 0
1 year ago
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