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4 years ago

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how is a pure substance different from a mixture? A. mixtures can be separated by physical means B. Pure substances are made up

of mixtures C. A pure substance is heterogeneous D. A mixture is made of one substance

1 answer:

Nataliya [291]4 years ago

3 0

Well, a pure substance Is something that isn't mixed with anything else. it is it's own thing, and a mixture can be separated by physical means, so it is A.

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Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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4 years ago

The speed of sound in air on a certain stormy day is about 1130 feet per second. you see a flash of lightning striking some dist

Anna35 [415]

Answer 4,520 feet Explanation The light and the sound were produced at the same time but sound took long to reach since is speed is much lower than the speed of light. Speed is the rate of change of distance. So; Distance = speed × time = 1130 × 4 = 4,520 feet.

6 0

3 years ago

What are things that we can do to protect our climate for future generations?

Vilka [71]

<span>Reduce energy use.
Change the way you think about transportation. Walk or bike whenever possible.
Insulate your home. Insulate yourself and your home.
Make every drop count.
</span>Cool wash and hang to dry.
<span>Switch to "green power.
</span>Recycle.

6 0

3 years ago

A dynamite blast propels a heavy rock straight up with a launch velocity of 160ft/sec (about 109 mph). Write a position for the

Flura [38]

A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft

b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up

v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down

c) -32.174 ft/s², which is the acceleration due to gravity.

d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds

3 0

4 years ago

A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is

Studentka2010 [4]

Answer:-81.29 J

Explanation:

Given

mass of ball $m=2.35 kg$

Length of string $L=3.53 m$

height of Room $h=5.03 m$

Gravitational Potential Energy is given by

$P.E.=mgh$

where h=distance between datum and object

here Reference is ceiling

therefore $h=-3.53 m$

Potential Energy of ball w.r.t ceiling

$P.E.=2.35\times 9.8\times (-3.53)=-81.29 J$

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling

3 0

3 years ago