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3 years ago

9

Through which material would you expect sound waves to move fastest? O A. Iron O B. Water O c. Air O D. Milk

2 answers:

Alexxx [7]3 years ago

8 0

Answer:

Solids

Is this what your looking for, It might tell you the answer?

zheka24 [161]3 years ago

7 0

Answer:

iron

Explanation:

ap3x

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Day to night to morning, keep with me in the moment

krok68 [10]

ANSWER: What.....is this lyrics or smth WHY DONT YOU SAY SO

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3 years ago

Which are ways to improve the design of this experiment? Check all that apply.

Arte-miy333 [17]

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

allowing to know the scope of the initial assumptions

* Use a galvanometer.

The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

Since it allows to have greater magnetic fields and therefore expand the range of measurements

3 0

3 years ago

Read 2 more answers

What does the c in conptt stand for

Oliga [24]

the C stands for Consistent

4 0

3 years ago

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

3 years ago

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

So

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

So

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

5 0

3 years ago