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aivan3 [116]
3 years ago
9

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?
14
15
18
23

Physics
1 answer:
antiseptic1488 [7]3 years ago
8 0
1st find the slope of the line:
(11-5)/(8.2-2.3) = 6/5.9 = 1.01

Find the equation of the line:

y - 5 = 1.01(x - 2.3)
y - 5 = 1.01x - 2.323
y = 1.01x + 2.667

Now set x to 20
y = 1.01(20) + 2.667
y = 20.2 + 2.667
y = 22.867 floors ~ 23 floors
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Identify two fields where physical quantities are used in motion calculations​
larisa86 [58]

The two fields were physical quantities are used in motion calculations are length and mass with time.

The physical quantity in a field is referred as every point in a particular space time.

<h3>How physical quantities are used in motion calculations?</h3>

 If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.

Length, mass and time  are the physical quantities used in motion calculations.

Learn more about motion calculations,

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4 0
2 years ago
Which of the following is an example of how pictures of a planet's surface can provide evidence about the planet's natural resou
Volgvan
I think it’s c but I’m not for sure
5 0
3 years ago
A 2-kg box is pushed by a force of 4 N for 2 seconds. It has an initial velocity vo = 2 m/s to the right. NOTE: Since this probl
IRISSAK [1]

Answer:

Kf= 36 J

W(net) = 32 J

Explanation:

Given that

m = 2 kg

F= 4 N

t= 2 s

Initial velocity ,u= 2 m/s

We know that rate of change of linear momentum is called force.

F= dP/dt

F.t = ΔP

ΔP = Pf - Pi

ΔP = m v  - m u

v= Final velocity

By putting the values

4 x 2 = 2 ( v - 2)

8 =  2 ( v - 2)

4 = v - 2

v= 6 m/s

The final kinetic energy Kf

Kf= 1/2 m v²

Kf= 0.5 x 2 x 6²

Kf= 36 J

Initial kinetic energy Ki

Ki = 1/2 m u²

Ki= 0.5 x 2 x 2²

Ki = 4 J

We know that net work is equal to the change in kinetic energy

W(net) = Kf - Ki

W(net) = 36 - 4

W(net) = 32 J

7 0
3 years ago
The following forces act on an object: 10 N north, 7 N south, and 4 N east. What is the magnitude of the net force?
spin [16.1K]

Given data;

 Fn = 10 N

 Fs = 7 N

 Fe = 4 N

           force in X direction (Fx) = 4 N

           force in Y direction (Fy) = 10-7 = 3 N

           Net force (Fnet) = Sq.root[(Fx)² + (Fy)²]

                                      = Sq root [ 4² + 3² ]

                                      = 25 N

        <em> Net force acting = 25 N</em>

                                   

6 0
3 years ago
In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:
fenix001 [56]

Acceleration is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) -20.5 m/s^2

Let's convert the quantities into SI units first:

u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s

v=1600 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =444.4 m/s

t = 4.0 min = 240 s

So the acceleration is

a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2

B) -3.8 m/s^2

As before, let's convert the quantities into SI units first:

u = 444.4 m/s

v=321 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =89.2 m/s

t = 94 s

So the acceleration is

a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2

C) -53.0 m/s^2

For this part we have to use a different formula:

v^2 - u^2 = 2ad

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2

3 0
3 years ago
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