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3 years ago

Which interaction of the atmosphere with the lithosphere benefits the biosphere?

Physics

2 answers:

Natasha2012 [34]3 years ago

6 0

Answer:

B -Groundwater resources are replenished in different regions

Explanation:

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True [87]3 years ago

3 0

Answer:

Explanation: b

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The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?

NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, $B=2.5\ mT=2.5\times 10^{-3}\ T$

(A) Magnetic field of a current loop is given by :

$B=\dfrac{\mu_oI}{2r}$

I is the current in the loop

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}$

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}$

r = 0.00222 m

$r=2.2\times 10^{-3}\ m$

Hence, this is the required solution.

3 0

3 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

4 years ago

A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

5 0

3 years ago

An object is accelerating at a constant rate due to a force being applied.what can be done to the force to decrease the objects

Sliva [168]

Answer:

Explanation:

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6 0

3 years ago

a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th

Savatey [412]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

$v_x = v cos \theta$