Ask question

Ask question

All categories

3 years ago

8

Which specialized science would you consult before cultivating your new field or farm for better and profitable yields? A. biote

chnology B. biochemistry C. integrated pest management D.soil science

2 answers:

krek1111 [17]3 years ago

5 0

Answer: D. soil science

Explanation:

Soil science is necessary field which is required to be consulted before cultivating the field for profitable yields. This is because of the fact that the plants grows over the soil. The soil is an integrated mixture of inorganic or organic particles, water, air and minerals which are necessary ingredients for the growth of new plants. The soil of good quality can yield profitable crops.

Zepler [3.9K]3 years ago

4 0

The answer is D soil cience

You might be interested in

For an object that floats in a liquid, which of the following statements is true? Group of answer choices The specific gravity o

SVEN [57.7K]

Answer:

Explanation:

Object is Floating in a liquid

weight of object is equal to weight of displaced fluid

i.e.

$\rho _oVg=\rho _mV'g$

i.e. $\rho _oV=mass\ of\ object$

$\rho _mV'=mass\ of\ liquid\ displaced$

where V and V' is the volume of object and volume of object inside the liquid.

therefore mass of liquid displaced must be equal to mass of object

8 0

4 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

The function carrying wire wrapped around an iron ore is called

fgiga [73]

The function is electromagnets.<span />

7 0

4 years ago

Colt1911 [192]

B-friction acts in direction opposite to objects motion

5 0

3 years ago

Read 2 more answers

Assume that Young’s modulus for a new material is known to be 2.25 × 1011 /2

Lady bird [3.3K]

Answer:

Maximum force, $F_{max} = 42 kN$

Explanation:

Maximum Stress, $\sigma_{max} = 2.1 * 10^{7} N/m^{2}$

The diameter of the material, d = 50.5 mm = 0.0505 m

The area of the material, $A = (\pi d^{2} )/4$

$A = \frac{\pi * 0.0505^{2} }{4} \\A = 0.002 m^{2}$

The formula for the maximum stress that the material can take is given by:

$\sigma_{max} = \frac{F_{max} }{A} \\2.1 * 10^7 = \frac{F_{max} }{0.002}\\F_{max} = 2.1 * 10^7 * 0.002\\F_{max} = 4.2 * 10^4 N\\F_{max} = 42 kN$

4 0

3 years ago