The law of interia applies to objects

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

Compare the gravity between these pairs, each consisting of an Earth-like planet and its star. You are given the mass of the pla

Maru [420]

Answer:

The answer is below

Explanation:

Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:

$F=G\frac{m_1m_2}{r^2} \\\\Where\ F=force,G=gravitational\ constant, m_1\ and\ m_1=mass\ of\ objects,r\ =distance\ between \ the\ two\ objects.$

The masses and distances for this question is in common units, Therefore the result would be in ratios

a) 4 MEarth / 2 MSolar / 3 AU

The force (F) = (4 * 3) / 3² = 4/3

b) 1 MEarth / 1 MSolar / 1 AU

The force (F) = (1 * 1) / 1² = 1

c) 1 MEarth / 2 MSolar / 2 AU

The force (F) = (1 * 2) / 2² = 1/2

6 0

2 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

2 years ago

HELLLPPP FAST PLS

r-ruslan [8.4K]

Answer:

Oh I had this question before! :)

Explanation:

Acid - citrus products - the classic one is lemon juice

Anything with a sour taste is acidic

Base - soap, oven cleaner, drain cleaner, baking soda

Basic things are bitter (please don't taste the drain cleaner and oven cleaner)

3 0

2 years ago

Why are scientific models important?

Wittaler [7]

Answer:

Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.

5 0

2 years ago