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3 years ago

Which evidence did Alfred Wegener’s original theory of continental drift have access to?

Physics

1 answer:

Sophie [7]3 years ago

4 0

Answer:

Evidence for continental drift

Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.

Explanation:

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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

3 years ago

Stars on the left of the diagram above are (View the picture to see the diagram)

omeli [17]

<em>D. Less luminous than those on the right []</em>

4 0

3 years ago

Read 2 more answers

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

7 0

3 years ago

On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 4.80 kg , and on this particular planet its weig

Oksanka [162]

Answer:

a = 1,375 m / s²

Explanation:

For this exercise we use Newton's second law where the force applied by the astronaut is F = 48.1, the body weight is W = 41.5 N and the body mass is m = 4.80 kg

F - W = m a

a = (F -W) / m

let's calculate

a = (48.1 - 41.5) / 4.80

a = 1,375 m / s²

this acceleration is directed upwards

3 0

3 years ago

A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It travels a distance d before coming

sergeinik [125]

Answer:

D) quadruple.

Explanation:

Assuming the same constant acceleration a in both cases, as we have as givens the acceleration a, the distance d, and the initial velocity v, we can use the following kinematic equations in order to compare the distances:

vf² - v₀² = 2*a*d

As the final state of the car is at rest, the final velocity vf, is 0.

⇒ - v₀² = 2*(-a)*d ⇒ d =v₀² / 2*a

1) initial velocity v₀

d₁ = v₀² / 2 a

2 ) initial velocity 2*v₀

⇒ d₂ = (2*v₀)² / 2*a = 4*v₀² / 2*a ⇒ d₂ = 4* (v₀² / 2*a)

⇒ d₂ = 4* d₁

As the equation shows, the distance required to stop, if the initial velocity were doubled, the distance required to stop would quadruple.

3 0

3 years ago