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IRISSAK [1]
3 years ago
5

Why are ionic solids poor conductors of electricity? The lattice is brittle. The ions are neutral. The ions are of equal but opp

osite charge. The ions are in fixed positions.
Physics
1 answer:
sveta [45]3 years ago
6 0

The ions are in fixed positions.

Explanation:

Ionic solids are poor conductors of electricity because their ions are fixed in position. Their ions are not free to move about. They are fixed in crystal lattices.

  • For the conduction of electricity, compounds must possess free mobile electrons and moving ions in solution.
  • Ionic compounds are formed by the electrostatic attraction between a metallic and non-metallic ion.
  • They actually contain ions but their ions are locked up.
  • They are not free to move about.
  • Electrical conduction involves ion mobility.
  • In molten and aqueous forms, they are able to conduct electricity because their ions are then mobile.

learn more:

Ionic compound brainly.com/question/6071838

#learnwithBrainly

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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o
dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching  the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

F\propto\frac{d}{dt} (p)

\Rightarrow F=\frac{d}{dt} (m.v)

since mass is constant

F=m\frac{d}{dt}v

when dt=10t

then,

F'=m.\frac{v}{10\times t}

F'=\frac{1}{10} \times \frac{m.v}{t}

F'=\frac{F}{10} the body will experience the tenth part of the maximum force.

where:

\frac{d}{dt} = represents the rate of change in dependent quantity with respect to time

p= momentum

m= mass of the person jumping

v= velocity of the body while hitting the ground.

7 0
2 years ago
Question 2: Start-Up
yuradex [85]

Answer:

The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.

Explanation:

7 0
2 years ago
The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2
Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0
3 years ago
Calculate the normal force pushing up from the table in each example 8 kg
garik1379 [7]

The correct answer to the question is: 78.4 N.

EXPLANATION:

As per the question, the mass of the object is given as m = 8 Kg.

We are asked to calculate the normal force of the object.

The object is resting on the table.

Hence, the object pushes the table downward with a force equal to it's weight.

The table, in turn, pushes the object with the same force as given by the object, but in the upward direction.

The force given by the table on the object is known as the normal reaction of the object which acts perpendicular to the surface of the object.

Hence, the normal reaction is equal to the weight of the body.

The weight of a body is the product of it's mass with acceleration due to gravity.

So, the weight of 8 Kg object is calculated as -

Weight = mg

= 8 × 9.8 N.

= 78.4 N.

Here, g is known as acceleration due to gravity.

Hence, the normal force pushing up from the table is 78.4 N.







8 0
3 years ago
A spaceship moves at a speed of 0.95 c away from the Earth. It shoots a star wars torpedo toward the Earthat a speed of 0.90 c r
wariber [46]

Answer:

0.345 c

Explanation:

v_{se} = velocity of spaceship relative to earth = 0.95 c

v_{ts} = velocity of torpedo relative to spaceship = - 0.90 c

v_{te} = velocity of torpedo relative to earth

Velocity of torpedo relative to earth is given as

v_{te}=\frac{v_{se} + v_{ts}}{1 + \frac{v_{se}v_{ts}}{c^{2}}}

v_{te}=\frac{0.95c + (- 0.90 c)}{1 + \frac{(0.95 c)(- 0.90 c))}{c^{2}}}

v_{te} = 0.345 c

5 0
3 years ago
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