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3 years ago

Which of the phases below are able to be compressed? A) Liquid only. B) Liquid and Gas. C) Solid only. D) Gas only

2 answers:

6 0

Answer - B
Liquid and Gas can be compressed because of the freely moving particles in them. Solids cannot be compressed as their structure is fixed.

amid [387]3 years ago

3 0

The phrase below that is able to be compressed is D. gas only.
Although it is possible to compress liquids, it requires a great deal of pressure to accomplish a little compression, which is why liquids and solids are often referred to as incompressible.

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A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a sp

cluponka [151]

Answer:

(a) Approximately $0.335\; \rm m$ .

(b) Approximately $1.86\; \rm m\cdot s^{-1}$ .

(c) Approximately $0.707\; \rm m$ .

(d) Approximately $0.228\; \rm m$ .

Explanation:

$v_i$ denotes the velocity of the object in the first diagram right before it came into contact with the spring.
Let $m$ denote the mass of the block.
Let $\mu$ denote the constant of kinetic friction between the object and the surface.
Let $g$ denote the constant of gravitational acceleration.
Let $k$ denote the spring constant of this spring.

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy $\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2$ .

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of $D$ and compressed the spring by the same distance.

Energy lost to friction: $\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D$ .
Elastic potential energy that the spring has gained: $\displaystyle \frac{1}{2}\,k\, D^2$ .

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

$\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

Assume that $g = 9.81\; \rm m \cdot s^{-2}$ . In the equation above, all symbols other than $D$ have known values:

$m =1.10\; \rm kg$ .
$v_i = 2.60\; \rm m \cdot s^{-1}$ .
$\mu = 0.250$ .
$g = 9.81\; \rm m \cdot s^{-2}$ .
$k = 50.0\; \rm N \cdot m^{-1}$ .

Substitute in the known values to obtain an equation for $D$ (where the unit of $D\!$ is $m$ .)

$3.178 = 2.69775\, D + 25\, D^2$ .

$2.69775\, D + 25\, D^2 + 3.178 = 0$ .

Simplify and solve for $D$ . Note that $D > 0$ because the energy lost to friction should be greater than zero.

$D \approx 0.335\; \rm m$ .

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

$\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J$ .

As the object moves to the left, part of that energy will be lost to friction:

$(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J$ .

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

$2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J$ .

Calculate the velocity corresponding to that kinetic energy:

$\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}$ .

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ( $1.91\; \rm J$ ) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is $\mu \cdot m \cdot g$ .

$\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m$ .

Similar to (a), solving (d) involves another quadratic equation about $D$ .

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) $1.91\; \rm J$ .

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

$\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

$25\, D^2 + 2.69775\, D - 1.90811\approx 0$ .

Again, $D > 0$ because the energy lost to friction is greater than zero.

$D \approx 0.228\; \rm m$ .

7 0

3 years ago

The electric force between electric charges is much larger than the gravitational force between the charges. Why, then, is the g

Mamont248 [21]

Answer:

Explanation:

The electric force between charges is much larger than the gravitational force but Gravitational force between earth and moon is dominant over Electric force because earth and moon are the electrically neutral body.

Electrically neutral bodies are those bodies that contain equal no of electron and proton in the body.

An example of electrically neutral bodies is a neutron.

5 0

3 years ago

Four identical balls are thrown from the top of a cliff, each with the same speed. The

Jlenok [28]

Answer:

the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Explanation:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

1) For the ball thrown straight up, we have;

θ = 90°

∴ v = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ K.E. = (1/2) × m × (u² + 2gh)²

2) For the second ball thrown 30° to the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

3) For the third ball thrown at 30° below the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

4) For the fourth ball thrown straight down, we have;

K.E. = (1/2) × m × (u² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Learn more about object kinetic energy of object if free fall here;

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3 years ago

can you suggest improvement that can be made towards the design of siphon so that the transfer of liquid is much higher.

ahrayia [7]

Answer:

A siphon is a tube that makes use of the potential energy of fluid at an elevated level to transfer the fluid to a lower level, due to pressure differences between the inlet and the outlet points of the tube, such that the pressure at the outlet is higher than the pressure at the inlet

The pressure energy is converted into velocity (kinetic) energy, and therefore, in other to increase the flow rate through the tube of a siphon, with constant diameter, the level of the fluid in the container at the inlet (supply) of the siphon is raised higher than the level at the outlet receiving) container or the outlet point of the siphon tube

The larger the difference between the inlet and outlet levels, the faster the transfer of fluid by the siphon

Explanation:

3 0

3 years ago

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone

Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, $a_c$ , is given as follows;