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3 years ago

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

Physics

2 answers:

german3 years ago

8 0

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

Misha Larkins [42]3 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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Aspect of the physical world

Zepler [3.9K]

Answer:

Not enough information given. Are there answer choices? Or more details given in the question?

Explanation:

6 0

3 years ago

The balls in the image above have different masses and speeds. Rank them in terms of momentum, from least to greatest.

N76 [4]

The momentum of an object is equivalent to the product of the object's mass and velocity. Computing the momentum for each ball:
A- 15 * 0.7 = 10.5
B- 5.5 * 1.2 = 6.6
C- 5.0 * 2.5 = 12.5
D- 1.5 * 5.0 = 7.5

Therefore, ball C has the greatest momentum.

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3 years ago

Read 3 more answers

As the number of resistors in a parallel circuit is increased, what happens to the

aleksandr82 [10.1K]

Answer:

resistance- decreases current-increases

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3 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

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3 years ago

Drag the titles to the correct boxes to complete the pairs.

PilotLPTM [1.2K]

Can you input a picture??

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3 years ago