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2 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?

Physics

1 answer:

sweet [91]2 years ago

4 0

Answer:

Speed of the electron will be $v=5.896\times 10^7m/sec$

Explanation:

We have given that charge on electron $e=1.6\times 10^{-19}C$

Mass of electron $m=9.11\times 10^{-31}kg$

Potential difference = $V=9.9KV=9.9\times 10^3volt$

Now according to energy conservation $eV=\frac{1}{2}mv^2$

$1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=5.896\times 10^7m/sec$

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. A little car has a maximum acceleration of 2.57 m/s2. What is the new maximum acceleration of the little car if it tows anothe

valkas [14]

Answer:

$a'=1.285\ m/s^2$

Explanation:

Let m be the mass of a little car and m' be the mass of another car.

We know that,

Force = mass × acceleration

ATQ,

m × a = 2m × a'

a = 2 × a'

$a'=\dfrac{a}{2}\\\\a'=\dfrac{2.57}{2}\\\\a'=1.285\ m/s^2$

So, the acceleration of another little car is equal to $1.285\ m/s^2$ .

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2 years ago

Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10

DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = 6.67 10⁻¹¹ 1 1 / 1²

F = 6.67 10⁻¹¹ N

Electric force

F = $k \frac{q_1q_2}{r^2}$

indicates that the charge is q = 10 C

F = 9 10⁹ 10 10 / 1²

F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

3 0

2 years ago

if forces acting on an object are unbalanced, which factors may result from an unbalanced force? Check all that apply. The net f

Alchen [17]

The net force is negative, and there is a change in motion.

8 0

2 years ago

Read 2 more answers

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

Over [174]

Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

$g=9.8 m/s^2$ is the gravitational acceleration

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

2 years ago

Read 2 more answers

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

natta225 [31]

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.

Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.

4 0

2 years ago