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konstantin123 [22]

3 years ago

11

A 5.0 c charge is 10 m from a small test charge. what is the magnitude of the electric field at the location of the test charge

?

1 answer:

Inessa [10]3 years ago

7 0

Answer:

$4.50*10^8\frac{N}{C}$

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

$E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}$

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(a) Draw a Velocity - Time graph for an Object with Constant Acceleration.

den301095 [7]

Explanation:

answers to 1 and 2 above/below respectively but can't do 3 because I don't know what graph you're talking about

7 0

2 years ago

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

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

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

3 years ago

The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass

Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0

3 years ago

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

3 years ago

In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times

Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0

3 years ago