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SVETLANKA909090 [29]
3 years ago
13

What type of particles flow between objects when they are charged by friction or conduction??

Physics
1 answer:
Sedaia [141]3 years ago
4 0

Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. Three ways electrons can be transferred are conduction, friction, and polarization.

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Initially at rest a single-stage rocket is launched vertically from the ground. The rocket’s thrust overcomes gravity and provid
Lelechka [254]

Answer:

Explanation:

Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds

s = 1/2at²

= .5 x a x 100

=50a

Velocity after 10 s

u = at = 10a

Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.

s = ut + 1/2 gt²

50a = -10ax20 + .5 x 10x400

250a = 2000

a = 8 m s⁻² .

Velocity after 10 s

= at = 80 m/s

further height reached with this speed under free fall

h = v² / 2g

= 80 x 80 / 2 x 10

= 320 m

height achieved under acceleration

= 50a

= 50 x 8 = 400m

Total height

= 320 + 400= 720 m

velocity after falling from 720 m

v² = 2gh = 2 x10 x 720

v = 120 m/s

7 0
3 years ago
Read 2 more answers
What’s the purpose of life?
Alborosie
Purpose of life is to explore and get experience. Another reason is to love and let what makes you happy guide you
5 0
4 years ago
Compare these two collisions of a PE student with a wall.
Stolb23 [73]

1) The variable that is different in the two cases is \Delta t, the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

1)

The variable that is different in the two cases is \Delta t, the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, \Delta t, will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision \Delta t will be smaller.

2)

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

\Delta p = m(v-u)

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

\Delta p = (75)(0-8)=-600 kg m/s

3)

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, \Delta t:

J=F \Delta t

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

F=ma

So the impulse is

J=ma\Delta t

The acceleration can be rewritten as rate of change of velocity:

a=\frac{\Delta v}{\Delta t}

So the impulse becomes

J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

4)

The force in the collision is related to the impulse by

J=F\Delta t

where

J is the impulse

F is the force

\Delta t is the duration of the collision

The equation can be rewritten as

F=\frac{J}{\Delta t}

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time \Delta t is longer, while in case B (unpadded wall) the time \Delta t is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

3 0
4 years ago
The joule is an example of units of work. Which of the following is not a valid unit of work? ft-lb (kg-m2)/s2 inch-ton lb/s2
Dahasolnce [82]

The "<em>lb/s² </em>" is not a valid unit of work.

Each of the others is (a unit of force) x (a unit of distance) which is pretty much the definition of work.

5 0
4 years ago
A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.
OLga [1]

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

  • <em>Mass of the block, m = 2.5 kg</em>
  • <em>Spring constant, k = 56 N/m</em>
  • <em>Extension of the spring, x = 0.75 m</em>

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5  \ m/s

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here:  brainly.com/question/166559

3 0
2 years ago
Read 2 more answers
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