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2 years ago

Which of the following statements describes an interaction between the geosphere and atmosphere?

Physics

2 answers:

german2 years ago

8 0

B is the answer, I’m really good at this subject

Misha Larkins [42]2 years ago

4 0

Answer:

Option (A)

Explanation:

Wind is created due to the pressure difference and it blows from a region of higher pressure to a region of low pressure.

Atmosphere refers to the envelope of gases that surrounds the earth. It is comprised of various types of gases which are present in various amount.

The wind blowing away sand in a desert environment describes the interaction between the geosphere and the atmosphere as the wind is a part of the atmosphere and the sand is a part of the geosphere.

Hence, the correct answer is option (A).

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Sasha is ordered Ampicillin 50mg/kg/day x 48 hours, to be given every 6 hours in 100mls of N/S run over 30 minutes. The tubing h

Anna007 [38]

Answer:

The correct dose = 1454.54 mg

and The jnfusion rate = 41.67 gitt/hr

Explanation: the correct dose will be 50mg/kg × kg/2.2 × 64lb

= 1454.54 mg

infusion rate will be

10 gtts/ml × 50mg/6 × 30/60

Infusion rate = 15000/360

= 41.67 gitt/hr

5 0

2 years ago

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0

2 years ago

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

avanturin [10]

6m at 0.9 radian means 6m at $51.57^0$

Since positive angles are counter clockwise

6m at $51.57^0$ can be written as 6*cos $51.57^0$ i+6*sin $51.57^0$ j = 3.73i+4.70j

5m at $-75^0$ can be written as 5*cos $(-75)^0$ i+5*sin $(-75)^0$ j = 1.294i-4.83j

4m at 1.2 radian means 4m at $68.75^0$

4m at $68.75^0$ can be written as 4*cos $68.75^0$ i+4*sin $68.75^0$ j = 1.45i+3.73j

6m at $-210^0$ can be written as 6*cos $(-210)^0$ i+6*sin $(-210)^0$ j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = $\sqrt{1.274^2+0.6^2} = 1.98 m$

c) Angle, $tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0$

7 0

2 years ago

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi

Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

$δ=1.18cm$

6 0

2 years ago

Read 2 more answers

A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d

AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

5 0

1 year ago