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3 years ago

Where are electrons with the lowest energy found?

Physics

2 answers:

Hatshy [7]3 years ago

6 0

D. They are closest to the nucleus. Therefore elections with the highest are found further away.

zloy xaker [14]3 years ago

6 0

It’s A closest to the nucleus

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What is current and si unit

Art [367]

Answer:

The SI unit of current is ampere which measures the flow of electric charge across a surface at the rate of one coulomb per second.

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3 years ago

to what part of the electromagnetic spectrum does the thermal radiation trapped in a greenhouse belong to?

IgorLugansk [536]

Answer:

Considerando, que al tope de la atmósfera llega un 100% de la radiación solar, de este total, sólo un 25% llega directamente a la superficie de la Tierra y un 25% es dispersado por la atmósfera como radiación difusa hacia la superficie, esto hace que cerca de un 50% de la radiación total incidente llegue a la ...

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3 years ago

A mass of 100kg is acted on by a 400N force acting vertically upward and a 600N force acting at a 45 degrees angle.Calculate the

slava [35]

Answer:

A 1.56 see below

Explanation:

I will assume the 45 degree force is upward

vertical component = 600 sin 45 = 424.26 N

added to the other vertical force will total 824.26 N

F = ma

824.26 = 100 * a shows a = 8.24 m/s upward

Now we have to assume the mass is ALSO acted on by gravity and this value is given as 9.8 ( so downward is positive)

9.81 - 8.24 = 1.56 m/s^2

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2 years ago

Hi!! This may sound really silly, but I am confused about resistors and currents. I am wondering how the current can have the sa

BartSMP [9]

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3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.