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dybincka [34]
3 years ago
8

An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an

Physics
1 answer:
JulsSmile [24]3 years ago
5 0

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

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A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0
3 years ago
What is an example of a invertebrate that doesn’t have an exoskeleton?
Tatiana [17]

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5 0
3 years ago
Read 2 more answers
Which of the following equations is balanced?
Gnesinka [82]

Answer: c

Explanation:

The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.

a. 2×2 atoms of Al ≠ 3×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!

b. 3×2 atoms of Al ≠ 3×1 atoms of Al

3×3 atoms of O ≠ 2×2 atoms of O

c. 2×2 atoms of Al = 4×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

BOTH ARE EQUAL, CORRECT ANSWER!!!

d. 2×2 atoms of Al ≠ 1×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

4 0
3 years ago
A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis
MaRussiya [10]

Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

According to it , pressure is transmitted in liquid from one point to another without any change .

25 cm diameter = 12.5 x 10⁻² m radius

Area = 3.14 x (12.5 x 10⁻²)²

= 490.625 x 10⁻⁴ m²

Pressure by vehicle

Force / area

13000 / 490.625 x 10⁻⁴

= 26.497 x 10⁴ Pa

5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

= 19.625 x 10⁻⁴ m²

If we assume required force F on this area

Pressure = F / 19.625 x 10⁻⁴ Pa

According to Pascal Law

F / 19.625 x 10⁻⁴  = 26.497 x 10⁴

F = 19.625 x 26.497

= 520 N

4 0
3 years ago
There is a 250-m-high cliff at half dome in yosemite national park in california. suppose a boulder breaks loose from the top of
lorasvet [3.4K]

Part A. For this part, we use two equations for linear motion:

<span>y = y0 + v0 t + 0.5 g t^2                   ---> 1</span>

<span>vf = v0 + g t                                         ---> 2</span>

First we solve for t using equation 1: y0 = 0 (initial point at top), y = 250 m, v0 = 0 (at rest)

250 = 0.5 (9.8) t^2

t = 7.143 s

Now we solve for final velocity vf using equation 2:

vf = g t

vf = 9.8 (7.143)

vf = 70 m/s

 

Part B. First we solve for the time it takes for the sound to reach the tourist.

t(sound) = 250 / 335 = 0.746 s

Therefore the total time would be:

t = 0.746 s + 0.300 s

t = 1.05 s

 

<span>Hence there is enough time for the tourist to get out before the boulder hits him.</span>

7 0
3 years ago
Read 2 more answers
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