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3 years ago

4. Which of the following statement describes the particle model.

Physics

2 answers:

m_a_m_a [10]3 years ago

8 0

The answer should be D!!!!!!!!!

Harlamova29_29 [7]3 years ago

7 0

I think it is A
Its not sure tho

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Why is it good that adolescents prune their brains

Aneli [31]

Answer:

Synaptic pruning is an essential part of brain development. By getting rid of the synapses that are no longer used, the brain becomes more efficient as you age.

5 0

3 years ago

A car drives 90 meters north, then turns around and drives 120 meters south. What was its displacement and distance?

Rufina [12.5K]

Answer:

displacement= 30 m towards south, distance= 210m

Explanation:

Distance (scalar quantity) how much ground an object has covered.

Displacement (vector quantity) refers to how far out of place an object is it is the object's overall change in position.

Basically meaning for displacement the directions will be very key

D for Displacement

D= D1+D2

D= 120 (S) + 90 m (N)

Must be in same direction

D= 120 (S) + (-90 m) (S)

D= 30 m (S)

and for distance you are simply just adding how much distance they have covered

so d= d1+d2

d= 90m + 120m

d= 210m

3 0

3 years ago

Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is sw

Mariana [72]

Answer:

Mutual inductance will be $M=1.8\times 10^{-3}Hennry$

Explanation:

We have given induced current $\Delta i=5A$

Time is given as $t=1ms=10^{-3}sec$

We have to find the mutual inductance between the coils

Induced emf is given as e = 9 volt

We know that induced emf is given by

$e=M\frac{\Delta i}{\Delta t}$

$9=M\times \frac{5}{10^{-3}}$

$M=1.8\times 10^{-3}Hennry$

7 0

3 years ago

a concrete cube of side 0.50 m and uniform density 2.0 x 103 kg m–3 is lifted 3.0 m vertically by a crane. what is the change in

AlladinOne [14]

Answer:

Change in potential energy = 7350 Joules

Explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube, $d=2\times 10^3\ kg/m^3$

The cube is lifted vertically by a crane to a height of 3 m

We know that, density $d=\dfrac{m}{V}$

So, m = d × V (V = volume of cube = a³)

$m=2\times 10^3\ kg/m^3\times (0.5\ m)^3$

m = 250 kg

We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.

Potential energy at height h is given by :

$PE=mgh$

PE = 250 kg × 9.8 m/s² ×3 m

PE = 7350 Joules

So, change in potential energy of the cube is 7350 Joules.

5 0

3 years ago

Read 2 more answers

Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie

liberstina [14]

You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless

6 0

3 years ago