When neither air mass displaces the adjacent one, their boundary is called a(n) _______

Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from

marin [14]

Answer:

v_2 = 2*v

Explanation:

Given:

- Mass of both charges = m

- Charge 1 = Q_1

- Speed of particle 1 = v

- Charge 2 = 4*Q_1

- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

F_net = m*a

- Equate the two expressions:

a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

v_i is the initial velocity = 0

v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

v_f^2 = 2*(Q*V / m*r)*r

v_f^2 = 2*Q*V / m

v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

v = sqrt (20*Q / m)

- The velocity of second particle Q = 4Q

v_2 = sqrt (20*4*Q / m)

v_2 = 2*sqrt (20*Q / m)

v_2 = 2*v

3 0

3 years ago

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

jek_recluse [69]

Answer:

$e=3367.2J$

% $e=1.43$ %

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

$v_{r}=10km/h=2.77m/s$

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

$v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s$

Knowing that we can calculate Kinetic energy for the real and experimental speed

$E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J$

$E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J$

Now, the potential error in her calculated kinetic energy is:

$e=E_{r}-E_{e}=(234023-230656)J=3367.2J$

% $e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43$ %

4 0

3 years ago

What are some of the downsides of using hydroelectric power?.

never [62]

Has an Environmental Impact. Perhaps the largest disadvantage of hydroelectric energy is the impact it can have on the environment.
It Displaces People.
It's Expensive.
There are Limited Reservoirs.
There are Droughts.
It's Not Always Safe

4 0

1 year ago

The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these

Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

$r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\$

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

$r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\$

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

$F = \frac{kq_{1}q_{2}}{r^2}\\$

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

$F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\$

5 0

2 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)