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erik [133]
3 years ago
11

A measure of randomness or disorder

Physics
2 answers:
erik [133]3 years ago
5 0
A measure of randomness or disorder is entropy
vodka [1.7K]3 years ago
3 0
It is simply called Entropy.

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Which substance is a greenhouse gas that is emitted by human activities, is the most widespread, and remains in the air the long
e-lub [12.9K]

Answer:

I am not really sure, but it is probably Carbon Dioxide

Explanation:

3 0
4 years ago
Read 2 more answers
The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec
PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

5 0
3 years ago
A vector has an x component of 25.0 units and a y
sammy [17]
Here is the answer to your question

8 0
3 years ago
Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus
ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N  (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0
2 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

7 0
3 years ago
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