All categories

3 years ago

NEED ANSWER FAST PLZ

Physics

2 answers:

Kryger [21]3 years ago

7 0

A. liquid and vapor in a confined area

DIA [1.3K]3 years ago

6 0

Answer:

Option (B)

Explanation:

The latent heat is often defined as the amount of temperature that is needed in the conversion of a phase, such as from solid to liquid (or gas) or from liquid to gas, and during this entire process, the temperature remains constant.

This latent heat plays an important role in generating intensive thunderstorms as well as hurricanes because it acts as a source of energy.

There are two types of latent heat and they are as follows-

(1) Latent heat of fusion- This involves a phase change from solid to liquid.

(2) Latent heat of vaporization- This involves a phase transformation from liquid to gas.

Thus, the correct answer is option (B).

You might be interested in

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

Earth, has a bigger mass than the Moon .

raketka [301]

Decreased it because you can float a lot

5 0

3 years ago

A leaf fell from a tree branch. The path it followed is shown in the diagram below.

prisoha [69]

The leaf fell at the crooked path instead of straight down because air currents and gravity applied changing and unbalanced forces to the leaf.

<h3>What is an air current?</h3>

An air current is defined as the changes in atmospheric pressure that causes the movement of air from one area to another.

When a leaf is detached naturally from the tree, it won't fall straight down to the floor but will fall a distance away from the tree due to the action of air current and some unbalanced forces.

Learn more about leaf here:

brainly.com/question/24234175

#SPJ1

4 0

2 years ago

A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0m. After what time interval does it s

aniked [119]

The acceleration is -9.8m/s^2. The initial velocity is -8m/s. The initial position is 30m. This describes a position function of
-(9.8/2)t^2-8t+30=0
Solve the quadratic equation for t to get t=1.789s

5 0

3 years ago

a circular loop is hanging on the wall. it has a radius of 33.3 cm and is comprised of 12 coils. there is a magnetic field perpe

Vika [28.1K]

Answer:

I = 2.19A, anticlockwise direction.

Explanation:

Given r = 33cm = 0.33m, N = 12, ΔB = 7.5 - 1.5 = 6.0T, Δt = 3s, R = 3.75Ω

By Faraday's law of electromagnetic induction when there is a change in flux in a coil or loop, an emf is induced in the coil or loop which is proportional to the time rate of change of the magnetic flux through the loop.

The emf E is related to the flux by the formula

E = – NdФ/dt

Where N = number of turns in the coil, Ф = magnetic flux through the loop = BA, B = magnetic field strength, A = Area

In this problem the strength of the magnetic field changes. As a result the flux too changes and an emf is induced in the coil.

ΔФ = ΔB×A = ΔB×πr² = 6×π×0.33² = 2.05Wb

E = -NΔФ/Δt = 12×2.05/3 = 8.2V

I = E/R = 8.2/ 3.75 = 2.19A

The direction of the current can be found by pointing the thumb of your right hand in the direction of the magnetic field and curling the remaining fingers around this direction. The direction of the curl of these fingers give the direction of current which in this case is anticlockwise.

5 0

3 years ago