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UkoKoshka [18]
2 years ago
11

What are three characteristic properties of matter?

Physics
2 answers:
olya-2409 [2.1K]2 years ago
5 0
Density,solubility,and melting point.

Licemer1 [7]2 years ago
4 0
Physical characteristics of matter include its mass<span>, weight, volume, and </span>density<span>. It also specifically describes its odor, shape, texture, and </span>hardness<span>. In addition, physical properties describe whether the object is a solid, a liquid, or a gas – its phase of matter at room temperature.</span>
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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
mihalych1998 [28]

Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=0.52\ m/s^2

b.\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. t=9.9 \ sec

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

\displaystyle a(0 )=8.133\ m/s^2

For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

\displaystyle c=-17.1482\approx -17.15

Now we complete the equation for the distance

\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

7 0
3 years ago
A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound
Yakvenalex [24]
Speed = distance/time
= 16.5/2.5 = 6.6 ms^-1
3 0
2 years ago
In which general compass traction is this hurricane moving
Ksenya-84 [330]

Answer:

It looks like its moving north.

Explanation:

3 0
2 years ago
A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth
Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

W=mg

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center


The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

800 N= W=mg=m \frac{GM}{R^2} (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

200 N = W' = mg' = m \frac{GM}{(R+h)^2} (2)

If we divide eq.(1) by eq.(2), we get

\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}

4=\frac{(R+h)^2}{R^2}

By solving the equation, we find:

4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0

which has two solutions:

h=-3R --> negative solution, we can ignore it

h=R --> this is our solution

Since the Earth's radius is R=6.4\cdot 10^6 m, the person should be at h=R=6.4\cdot 10^6 m=6400 km above Earth's surface.

5 0
3 years ago
A pool and stops at the
Bas_tet [7]

In Linear motion the swimmer swims

8 0
2 years ago
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