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katrin2010 [14]

3 years ago

15

In science class, students constructed an explanation on the composition of soil. Which student offered the BEST explanation? (A

KS 2e)
!!!(The file has the image)!!!

Answer choices:

A.

Student A

B.

Student B

C.

Student C

D.

Student D

2 answers:

Tpy6a [65]3 years ago

6 0

I would go with answer c) student c as it best describes all of the layers in the soil, giving it the most descriptive layer out of the others :)

Firdavs [7]3 years ago

4 0

STUDENT C would be the best answer

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An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

$V=0.235\times 2=0.47 V$

5 0

3 years ago

If you wanted to soundproof a room what insulator would you use

olasank [31]

Answer:

foam but thats what i heard from other people

Explanation:

4 0

3 years ago

Read 2 more answers

The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop

nalin [4]

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration $a= -4.20 m/s^2$

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

$0 =24.0 +(-4.20)\times t$

$t = \dfrac{-24.0}{-4.20}$

$t=5.71\ sec$

Hence, The time is 5.71 sec.

5 0

4 years ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

3 0

4 years ago

In a glider stunt at an air show, a towing airplane (motorized plane pulling the gliders) takes off from a level runway with two

irakobra [83]

Answer:

minimum length of runway is needed for take off 243.16 m

Explanation:

Given the data in the question;

mass of glider = 700 kg

Resisting force = 3700 N one one glider

Total resisting force on both glider = 2 × 3700 N = 7400 N

maximum allowed tension = 12000 N

from the image below, as we consider both gliders as a system

Equation force in x-direction

2ma = T -f

a = T-f / 2m

we substitute

a = (12000 - 7400 ) / (2 × 700 )

a = 4600/1400

a = 3.29 m/s²

Now, let Vf be the final speed and Ui = 0 ( as starts from rest )

Vf² = Ui² + 2as

solve for s

Vf² = 0 + 2as

2as = Vf²

s = Vf² / 2a

given that take of speed for the gliders and the plane is 40 m/s

we substitute

s = (40)² / 2×3.29

s = 1600 / 6.58

s = 243.16 m

Therefore, minimum length of runway is needed for take off 243.16 m

4 0

3 years ago