Turquoise is what types of rocks sedimentary, metamorphic or igneous

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

2 years ago

Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t

Nezavi [6.7K]

Answer

Explanation:

As the three resistors are connected in series, the expression to be used for the

calculation of RT equivalent resistance

is:

RT = R1 + R2 + R3

We replace the data of the statement in the previous expression and it remains:

5 10 15 RT + R1 + R2 + R3 + +

We perform the mathematical operations that lead us to the result we are looking for:

RT - 30Ω

5 0

2 years ago

A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

kvv77 [185]

Answer:

v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

Em₀ = K = ½ m v²

final point. Higher

$Em_{f}$ = U = m g h

Let's use trigonometry to lock her up

cos 60 = y / L

y = L cos 60

Height is the initial length minus the length at the maximum angle

h = L - L cos 60

h = L (1- cos 60)

energy is conserved

Em₀ = Em_{f}

½ m v² = mgL (1 - cos 60)

v = 2g L (1- cos 60)

let's calculate

v² = 2 9.8 3.0 (1- cos 60)

v = 29.4 m / s

6 0

2 years ago

The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......

valina [46]

Answer:

s^ -1 ( or 1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec or m/sec

so b would have units of s^-1

(or perhaps a more general term would be time^-1)

8 0

1 year ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago