A negatively charged object has?

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

A pressure sensor inside of a mixing tank is designed to turn red when the pressure inside the tank exceeds 1.9 kPa. If the sens

spin [16.1K]

Answer:

19 N

Explanation:

From the question given above, the following data were obtained:

Pressure (P) = 1.9 kPa

Length (L) = 10 cm

Force (F) =?

Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:

1 KPa = 1000 N/m²

Therefore,

1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa

1.9 KPa = 1900 N/m²

Thus, 1.9 KPa is equivalent to 1900 N/m².

Next, we shall convert 10 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Thus, 10 cm is equivalent to 0.1 m

Next, we shall determine the area of the square. This can be obtained as follow:

Length (L) = 0.1 m

Area of square (A) =?

A = L²

A = 0.1²

A = 0.01 m²

Thus, the area of the square is 0.01 m².

Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:

Pressure (P) = 1900 N/m²

Area (A) = 0.01 m²

Force (F) =?

P = F/A

1900 = F / 0.01

Cross multiply

F = 1900 × 0.01

F = 19 N

Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.

3 0

3 years ago

Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0

3 years ago

Read 2 more answers

Pendulum clocks generally run fast in winter and slow in summer

Harman [31]

If the question is true or false then the answer is true

5 0

3 years ago

Read 2 more answers

What is the total charge of 1.0g of electrons?

Serhud [2]

<span>1.0 gram (g) of electrons would contain 10^27 electrons Electrons have an electric charge of −1.602×10−19 coulomb so total charge of 1 g electrons = -1.602 x 10^-19 x 10^27 = -1.602 x 10^8</span>

8 0

3 years ago