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3 years ago

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The humidity you feel in the southern states of the U.S. is _____. part of the water cycle water vapor caused by the Sun's heat

all of the above

1 answer:

Likurg_2 [28]3 years ago

4 0

Hey there!

The answer is :
The <span>humidity you feel in the southern states of the U.S. is water vapor
water vapor is between a gas and liquid so that is what gives a humid feel in the hot sun

Hopes this helps u :D</span>

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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

4 years ago

Please help ASAP WILL GIVE BRAINLIEST

Westkost [7]

Answer:

D

Explanation: It makes the most sense. Plz mark brainliest

5 0

3 years ago

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A wall (of thermal conductivity 0.98 W/m Â· â—¦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside

disa [49]

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

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2 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

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3 years ago