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3 years ago

What will happen in the particles of liquid after 12 hours inside the freezer?

Physics

1 answer:

snow_tiger [21]3 years ago

3 0

Answer:

When a liquid (like water) is frozen, all of the molecules start sticking to each other and holding on very tightly. ... If you were to put that water in a closed container in the freezer, then it would still get bigger.

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A sprinter runs for 6.45 s at 7.44 m/s. How far does she get?

Gwar [14]

Answer:

D is the answer

Explanation:

6.45×7.44= 47.98800

Which if we round of we get 48m

8 0

3 years ago

Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0

3 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

3 years ago

You might use it to test how cold the pool water is

ozzi

You should test FC and PH as soon as you take the sample

3 0

3 years ago

Does anyone know this

NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

$F_x = F_a - F_f = 20 - 4 = 16 N$ (to the right)

While the vertical force is

$F_y = F_N - F_g = 400 - 400 = 0$

So the net force is 16 N to the right.

In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

$F_x =F_a - F_f = 100 -2 = 98 N$ (to the left)

While the vertical force is

$F_y = F_g - F_N = 4 - 4 = 0$ (downward)

And so, the net force is 98 N to the left.

The force acting on the squirrel in this problem are:

$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

$F_y = F_g - F_f = 8 - 7.5 = 0.5 N$

And since the weight is larger than the air resistance, the direction of the net force is downward.

The forces acting on Monkey are:

$F_1=95 N$ is the force applied to the right by Bunny
$F_2 = 75 N$ is the force applied by Deer from the left (so, also on the right)
$F_g = 50 N$ is the weight of Monkey, downward
$F_N = 50 N$ is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

$F_x = F_1 + F_2 = 95+75=170 N$ (to the right)

While the net force in the vertical direction is

$F_y = F_N - F_g = 50 - 50 = 0$

And therefore the net force is 170 N to the right

The forces acting on Deer are:

$F_a = 100 N + 100 N = 200 N$ to the right, the combined force applied by Bunny and Monkey
$F_f = 25 N$ to the left, the force of friction
$F_g = 150 N$ downward, the weight of the deer
$F_N = 150 N$ upward, the normal reaction from the surface that balances the weight