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3 years ago

The droplets on a glass of ice water are a result of water vapor releasing energy and changing to water droplets.

2 answers:

4 0

That is true.

In the solid state the molecules of water have the lowest energy content.

To pass from solid state to liquid stated water needs to gain some energy to increase vibration and motion of the molecules and reach a higher energy states.

To pass from liquid state to vapor (gaseous state) the molecules of water have to gain more energy to increase more the motion of the molecules and reach a more energetic stated.

Then you can think that when the water vapor becomes liquid (droplets) it needs to release some energy.

The droplets formed on a glass of ice water is an example of water condensation, one of the processes of the water cycle.

The droplets on the external surface of a cold glass are the result of condensation of water.

Condensation is the pass from vapor state to liquid state. As whe have stated, when water passes from vapor state to liquid state it has to release energy because liquid state is lower in energy than vapor (the molecules in water experiment slower motion than molecules of vapor).

Rina8888 [55]3 years ago

3 0

the answer is condensation

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I WILL MARK YOU AS BRAINLIEST IF RIGHT The fastest pitch ever recorded in MLB was thrown by Aroldis Chapman at 5 points 105.1 mp

Goshia [24]

Answer: The force was 13.92 Newtons.

Explanation:

First, let's recall the second Newton's law:

The net force is equal to the mass times the acceleration, or:

F = m*a

where:

F = force

m = mass

a = acceleration.

When the player hits the ball with the bat, he applies a force that accelerates the ball for a small period of time, that increases greatly the speed of the ball.

In this case, we know that:

the mass of the ball is 0.145 kg

The acceleration of the ball is 96m/s^2

Then we can input those values in the above equation to find the force.

F = 0.145kg*96m/s^2 = 13.92 N

The force was 13.92 Newtons.

5 0

2 years ago

the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

dem82 [27]

Answer:

Time = 5 seconds

Distance = 50 meters

Explanation:

Constantly Accelerated Motion

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

$v_f=v_o+a.t$

And the distance s is

$\displaystyle s=v_o.t+\frac{gt^2}{2}$

Given the object starts from rest, vo=0 and vf=20 m/s at $a=4\ m/s^2$ . We compute t

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}$

$\boxed{t=5\ sec}$

Now we compute s

$\displaystyle s=0+\frac{4\times 5^2}{2}$

$\boxed{s=50\ m}$

5 0

2 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

Calculate the acceleration using the formula acceleration = (final velocity - initial velocity) / time.

PtichkaEL [24]

Answer:

10 km/hr/s

Explanation:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

For the car in this problem:

u = 0

$v=60 km/h$

t = 6 s

Substituting in the equation,

$a=\frac{60 km/h-0}{6s}=10 km/h/s$

6 0

3 years ago