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3 years ago

15

Which wave interaction is responsible for echoes?

2 answers:

pogonyaev3 years ago

6 0

Answer:

reflection. :D!! hope it helps

IrinaVladis [17]3 years ago

5 0

Answer:Reflection

Explanation:

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Define the apparent cubic expansivity of a liquid

poizon [28]

Answer:

THE APPARENT CUBIC EXPANSIVITY OF A LIQUID : is defined as the increase in volume per unit volume per unit rise in temperature when the liquid is heated in an expansible vessel.

7 0

3 years ago

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$

$2.1 Kg.m/s(\hat i)=1.5 Kg.m/s(\hat j)+ 0.1*v_{fs}$

$0.1*v_{fs}=2.1 Kg.m/s(\hat i)- 1.5 Kg.m/s(\hat j)$

$v_{fs}=21 Kg.m/s(\hat i)- 15 Kg.m/s(\hat j)$

$|v_{fs}|=\sqrt {(v_{x}^{2}+v_{y}^{2})}$

Substituting 21 for $v_{x}$ and 15 for $v_{y}$

$|v_{fs}|=\sqrt {(21^{2}+15^{2})}=25.8 m/s$

To find direction

$tan\theta=\frac {v_{y}}{v_{x}}$

$\theta=tan^{-1}(\frac {15}{21})=35.5^{o}$

Therefore, magnitude is 25.8 m/s and direction is $35.5^{o}$

8 0

4 years ago

A) Both A and R are true and R is the correct explanation of the assertion.

avanturin [10]

Hey pls don’t. Report my answer I’m in a test and I rlly want someone to answer my question

5 0

3 years ago

Read 2 more answers

Which is not an example of work being done?

mezya [45]

C. Carrying a backpack to class

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3 years ago

Read 2 more answers

A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his cl

Andre45 [30]

Answer:

B. 560 J

J. 1.2 m

Explanation:

v = Final velocity = 0

u = Initial velocity = 4 m/s

$\mu$ = Coefficient of friction = 0.7

m = Mass of runner = 70 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Kinetic energy is given by

$K=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow K=\dfrac{1}{2}\times 70\times (0^2-4^2)\\\Rightarrow K=-560\ \text{J}$

The mechanical energy lost is 560 J

Acceleration is given by

$a=-\mu g\\\Rightarrow a=-0.7\times 9.81\\\Rightarrow a=-6.867\ \text{m/s}^2$

From kinematic equations we get

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-4^2}{2\times -6.867}\\\Rightarrow s=1.165\approx 1.2\ \text{m}$

The runner slides for 1.2 m

4 0

3 years ago