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PolarNik [594]
3 years ago
7

When water vapor rises into the atmosphere and begins to condense, it releases ________, the energy of a thunderstorm.

Physics
2 answers:
ANEK [815]3 years ago
7 0

Answer:

A is the correct answer

zubka84 [21]3 years ago
5 0
The answer is latent heat. The specific latent heat of vaporisation, L_v, of a substance is the energy input required for each kilogram to be converted from liquid to gas by evaporation. The 'specific' means per kilogram, so more generally latent heat of vaporisation is the energy taken in during the process for a given mass.

Here we are not vaporising the substance. We are in fact condensing it, the reverse process. All this means is the latent heat is released as electrostatic potential decreases in the water, as opposed to being absorbed. I hope this helps you :)
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Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is someti
Vinvika [58]

Answer:

The heavier the load in a cart, the harder the cart is to turn.

5 0
3 years ago
Read 2 more answers
A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe
SVETLANKA909090 [29]

Answer:

T_s = 6.8 degree C

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)

here we know that

T = 34 degree C = 307 K

A = 1.38 m^2

e = 0.557

\sigma = 5.67 \times 10^{-8} W/m^2K^4

\frac{dQ}{dt} = 120 J/s

now we have

120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)

120 = (4.36 \times 10^{-8})(307^4 - T_s^4)

T_s = 279.8 K

T_s = 6.8 degree C

5 0
3 years ago
Give at least one fact about subatomic particles
victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

- These particles can be divided as Brayons and Leptons

- These particles are often held together by one of the four fundamental particles ( Weak force, strong force, electromagnetic force, gravitational force).

6 0
3 years ago
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
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