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3 years ago

a stone is thrown down off a bridge with a veloctiy of 5.6m/s. What is the velocity afer 3 seconds have passed?

Physics

2 answers:

Flura [38]3 years ago

6 0

I think it's 16.8 because you multiply 5.6x3=16.8

nexus9112 [7]3 years ago

5 0

Answer:

Final speed of the stone, v = 35 m/s

Explanation:

It is given that,

Initial speed of the stone, u = 5.6 m/s

Let v is the velocity of the stone after 3 seconds have passed. It is moving down under the action of gravity. So, using the first equation of motion as :

$v=u+at$

$v=u+gt$

$v=5.6+9.8\times 3$

v = 35 m/s

So, the final speed of the stone is 35 m/s. Hence, this is the required solution.

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trasher [3.6K]

Answer:

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Explanation:

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3 years ago

Amy wants to know whether or not an item will float when placed in a fluid. Which of the comparisons below, when true, will mean

o-na [289]

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B. The force of gravity is stronger than the buoyant force.

C. The buoyant force is stronger than the force of gravity.

D. The mass of the item is greater than its gravity

The answer is C

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3 years ago

Titan is the only moon in the solar system with a significant atmosphere. which planet does titan orbit?

sveticcg [70]

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6 0

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

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WITCHER [35]

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6 0

3 years ago