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3 years ago

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Susan plans to grow cabbages and broccoli in her vegetable garden. Cabbages and broccoli grow best in clay soil. Which of these

is the most likely soil sample that Susan would select to grow cabbages and broccoli?

1 answer:

irakobra [83]3 years ago

7 0

The correct option is B.
As we are told in the question, the cabbage and broccoli grow best in clay soil. From the options we are given, it can be seen that it is only option B that has the highest amount of clay soil. The soil that will be used should be one that has the highest content of clay.

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

3 years ago

What is the relationship between environmental science and public policymakers?

jekas [21]

In a simple sentence I would say using environmental science public policy maker would be able to create policies on environmental issues such as littering,recycling specific items/reusing,not wasting excess energy,ect.

4 0

2 years ago

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 x

Blizzard [7]

Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced $7.5 * 10^{-3}$ meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

Answer:

The wavelength is $\lambda = 56250 nm$

Explanation:

From the question we are told that

The distance of slit separation is $d = 7.5 *10^{-3} \ m$

The distance of the screen is $D = 4 \ m$

The distance between the bright spot closest to the center of the interference is $k = 1.5 \ cm = 0.015 \ m$

Generally the width of the central maximum fringe produced is mathematically represented as

$y = 2 * k = \frac{ D * \lambda}{d}$

=> $2 * 0.015 = \frac{ \lambda * 4}{ 7.5 *10^{-3}}$

=> $\lambda = 56250 *10^{-9} \ m$

=> $\lambda = 56250 nm$

7 0

3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat

forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, $a=9.8\ m/s^2$

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

$F=mg-\dfrac{mv^2}{r}$

$F=m(g-\dfrac{v^2}{r})$

$F=30\times (9.8-\dfrac{(7.3)^2}{15})$

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0

3 years ago