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3 years ago

Why do electrons move from the negative end of the tube to the positive end? please help

1 answer:

5 0

They are repelled from the negative end and attracted to the positive end because of its negative charge. Remember, opposites attract and like charged repel

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If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre

Dmitrij [34]

M represent mass in kg
v represents speed in m/s
r represents radius in m

Now, just substitute these into the formula:
 $F = \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }$

3 0

3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

baherus [9]

20 characters longer later and woah

5 0

3 years ago

2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially

Naddika [18.5K]

Answer:

E_particle = 1,129 10⁻²⁰ J / particle

T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

E_a = 2 5000 = 10000 J

Gas b

E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

E_total = E_a + E_b

E_total = 10000 + 24000 = 34000

The total mass is

M = m_a + m_b

M = 2 +3 = 5

The average energy among the entire mass is

E_averge = E_total / M

E_averago = 34000/5

E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

E_particle = 1,129 10⁻²⁰ J / particle

For find the temperature let's use equation

E = kT

T = E / k

T = 1,129 10⁻²⁰ / 1,381 10⁻²³

T = 8.175 102 K

T= 817.5 K

5 0

3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

2 years ago