Ask question

Ask question

All categories

3 years ago

11

A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.90 s . The particle has a charg

e-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.
What are the magnitude and direction of the electric field?Express your answer to two significant digits and include the appropriate units. Enter positive value if the electric field is upward and negative value if the electric field is downward

1 answer:

Ksju [112]3 years ago

7 0

Answer:

E = 5.5 10² N /C

Explanation:

For this exercise let's use Newton's second law

F = m a

the force is electric

F = q E

we substitute

q E = m a

E = $\frac{m}{q} \ a$ (1)

the acceleration can be found with kinematics,

v = v₀ + a t

as the particle starts from rest v₀ = 0

a = v / t

a = 160 / 2.90

a = 55.17 m / s²

the relationship

q / m = 0.100 C / kg

m / q = 10 kg / C

we use equation 1

E = 10 55.17

E = 551.7 N / C

E = 5.517 10² N / C

ask for the result with two significant figures

E = 5.5 10² N /C

You might be interested in

3.25 kcal is the same amount of energy as

PIT_PIT [208]

<span>1 cal = 4,185 J
1 kcal = 1*10^3 cal
or
=1000 cal</span>

8 0

3 years ago

Read 2 more answers

What had to be invented before we could discover Saturn and Neptune?

Ann [662]

The discovery of Uranus, Neptune, most asteroids, the moons of
any planet other than Earth, Pluto, and all the other dwarf planets,
required the invention and use of telescopes. Those objects are all
too dim to be seen with a bare naked human eye.

Saturn is bright and plainly visible with your eyes, if you know when
and where to look, and what you're looking at.

4 0

3 years ago

Read 2 more answers

Differences between pavement and<br>cuboidal epithelium tissue

pantera1 [17]

Answer:

Squamous epithelium has cells that are wider than their height (flat and scale-like). ... Cuboidal epithelium has cells whose height and width are approximately the same (cube shaped). Columnar epithelium has cells taller than they are wide (column-shaped).

8 0

3 years ago

Consider an ideal gas of 7 moles that is in contact with a thermal reservoir of temperature 475 K. The gas is enclosed in a cont

kozerog [31]

Answer:

(a). The initial pressure is $5.5\times10^{4}\ Pa$

(b). The final pressure is $1.8\times10^{4}\ Pa$

Explanation:

Given that,

Number of moles = 7

Temperature = 475 K

Initial volume = 0.50 m³

Expanded volume = 1.50 m³

We need to calculate the initial pressure

Using formula of pressure

$P_{i}=\dfrac{nRT_{i}}{V_{i}}$

Put the value into the formula

$P_{i}=\dfrac{7\times8.31\times475}{0.50}$

$P_{i}=55261.5\ Pa$

$P_{i}=5.5\times10^{4}\ Pa$

We need to calculate the final pressure

Using formula of pressure

$P_{f}V_{f}=nRT_{f}$

After expansion,

$\dfrac{P_{f}V_{f}}{P_{i}V_{i}}=\dfrac{nRT_{f}}{nRT_{i}}$

$P_{f}=\dfrac{T_{f}}{T_{i}}\times\dfrac{P_{i}V_{i}}{V_{f}}$

Put the value into the formula

For thermal process,

$T_{i}=T_{f}$

$P_{f}=\dfrac{5.5\times10^{4}\times0.50}{1.50}$

$P_{f}=18333.33\ Pa$

$P_{f}=1.8\times10^{4}\ Pa$

Hence, (a). The initial pressure is $5.5\times10^{4}\ Pa$

(b). The final pressure is $1.8\times10^{4}\ Pa$

3 0

4 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago