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3 years ago

Describe and contrast inversion and eversion?

Physics

1 answer:

Kitty [74]3 years ago

7 0

Answer:

inversion would be going toward or inside internal and eversion is going outside or external

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Can someone please help me?? i’ll give brainilist

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Probably 90 j but im not sure I haven’t done any work like this in a while

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2 years ago

Imagine whirling a ball on a string over you head. Suppose the knot holding the ball comes loose and the ball is instantly relea

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3 years ago

What is the longest possible wavelength for a line in the balmer series

MariettaO [177]

Answer:

" In the Balmer series, the transitions happening in visible range are considered, which range from around 400 nm to 700 nm. The longest wavelength visible in the Balmer series is 656 nm."

Explanation:

Hope this is helpful :)

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3 years ago

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
<u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
<u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

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3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago