3. A man hits a 0.2 kg golf ball with a force of 4

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago

A train is pulling four train cars and each car has a mass of 40,000 kg. The train is accelerating at 1.1 m/s^2. What is the for

IgorLugansk [536]

Answer:

176,000 N

Explanation:

Newton's second law:

∑F = ma

F = (4 × 40,000 kg) (1.1 m/s²)

F = 176,000 N

8 0

3 years ago

Read 2 more answers

Lexi is balancing equations. She is finding one equation to be very difficult to balance. Which explains how to balance the equa

Rina8888 [55]

Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer

Equation is already balanced.

Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4

6 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

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4 0

2 years ago

A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry

Sauron [17]

Answer:

a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

$F_{e}$ - $T_{x}$ = m a

Axis y

$T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

Fe = $T_{x}$

T_{y} = W

Let us search with trigonometry the components of the tendency

cos θ = T_{y} / T

sin θ = $T_{x}$ / T

T_{y} = cos θ

$T_{x}$ = T sin θ

We replace

q E = T sin θ

mg = T cosθ

a) the electric force is

$F_{e}$ = q E

E = $F_{e}$ / q

E = -0.032 / 80 10⁻⁶

E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

q E / mg = tan θ

θ = tan⁻¹ qE / mg

Let's calculate

θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

θ = tan⁻¹ 0.3265

θ = 18 ⁰

sin 18 = x/0.30

x =0.30 sin 18

x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

$F_{e}$ = m aₓ

aₓ = q E / m

aₓ = 80 10⁻⁶ 4 10² / 0.01

aₓ = 3.2 m / s²

Axis y

W = m $a_{y}$

a_{y} = g

a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

a = √ aₓ² + a_{y}²

a = √ 3.2² + 9.8²

a = 10.31 m / s²

The Angle meet him with trigonometry

tan θ = a_{y} / aₓ

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-9.8) / 3.2

θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0

3 years ago