Calculate the average acceleration of a car that changes speed from 0 m/s to 20 m/s in 5 s.

The prevailing winds that form in Hadley cells are _____.

dmitriy555 [2]

Answer = C

It blows towards the poles from the sub-tropical high pressure areas

8 0

3 years ago

Read 2 more answers

If given both the speed of light in a material and an incident angle. How can you find the refracted angle?

Basile [38]

Answer:

$r = Sin^{-1}\left ( \frac{v Sini}{c} \right )$

Explanation:

Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.

By using the definition of refractive index

refractive index of the medium is given by

n = speed of light in vacuum / speed of light in medium

n = c / v ..... (1)

Use Snell's law

n = Sin i / Sin r

Where, r be the angle of refraction

From equation (1)

c / v = Sin i / Sin r

Sin r = v Sin i / c

$r = Sin^{-1}\left ( \frac{v Sini}{c} \right )$

8 0

4 years ago

Two harmonic waves traveling in opposite directions interfere to produce a standing wave described by y

AnnyKZ [126]

The wavelength of the interfering waves is 3.14 m.

<h3>Calculation:</h3>

The general equation of a standing wave is given by:

y = 2A sin (kx) cos (ωt) ......(1)

The given equation represents the standing wave produced by the interference of two harmonic waves:

y = 3 sin (2x) cos 5t .......(2)

Comparing equations (1) and (2):

k = 2

We know that,

k = 2π/λ

λ = 2π/k

λ = 2 (3.14)/ 2

λ = 3.14 m

Therefore, the wavelength of the interfering waves is 3.14 m.

I understand the question you are looking for is this:

Two harmonic waves traveling in opposite directions interfere to produce a standing wave described by y = 3 sin (2x) cos 5t where x is in m and t is in s. What is the wavelength of the interfering waves?

Learn more about interfering waves here:

brainly.com/question/2910205

#SPJ4

5 0

2 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

Which of the following is an element

Ilia_Sergeevich [38]

D oxygen i believe is your answer

5 0

4 years ago