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grin007 [14]
2 years ago
10

Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en

tire structure would sink by a certain amount for every additional floor that is added. If the maximum permissible limit for depression of the structure is 20 centimeters, how many floors can be safely added to the building?
Physics
1 answer:
trasher [3.6K]2 years ago
8 0
<span>If the maximum permissible limit for depression of the structure is 20 centimeters, the number of floors that can be safely added to the building is </span><span>C. 18</span>

depression = (depression/floor)(# floors) < 20

Here are the following choices:
<span>A. 14
B. 15
C. 18
D. 23</span>
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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
  2. Particle's speed and direction of motion brainly.com/question/2814900
  3. About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
2 years ago
Read 2 more answers
An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a
blsea [12.9K]

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0
2 years ago
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Which of the following most accurately represents John Dalton’s model of the atom? A. a tiny, solid sphere with an unpredictable
aleksley [76]
A and c are the answersss
6 0
2 years ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
2 years ago
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

5 0
2 years ago
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