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3 years ago

The layer of leaves that blocks most of the sunlight from reaching the ground in the rain forest is called the _____.

Physics

2 answers:

Zarrin [17]3 years ago

5 0

The layer of leaves that blocks most of the sunlight from reaching the ground in the rain forest is called the canopy.

love history [14]3 years ago

4 0

The answer should be <span>canopy.</span>

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Therefore, planetesimals are formed first.

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4 years ago

How can you use litmus paper to distinguish an acid from a base?

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3 years ago

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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.

4 0

3 years ago

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially

mixer [17]

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0

3 years ago