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3 years ago

9

How tilted is the axis of the earth's rotation as compared to the plane of the earth's orbit?

2 answers:

Phantasy [73]3 years ago

6 0

C. 23.5 degrees
This tilt in the axis causes the seasons.
Hope this helps.

ryzh [129]3 years ago

6 0

The Earth's axis is about 66.5 degrees out of the plane of the orbit.
It's about 23.5 degrees away from being "straight up and down".

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How does your body’s flexibility naturally change as you age?

chubhunter [2.5K]

<span>Your flexibility decreases. But if you exercise or stretch a few times a week you can slow down the process </span>

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3 years ago

Read 2 more answers

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

3 years ago

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

neonofarm [45]

Answer:

Electric force, $F=2.24\times 10^{-14}\ N$

Explanation:

It is given that,

Charge on an electron is $-1.6\times 10^{-19}\ C$

Electric field, $E=1.4\times 10^5\ N/m$

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N$

So, the electric force is $2.24\times 10^{-14}\ N$ .

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3 years ago

Un pendul este suspendat de un ax cu o tijă subțire foarte ușoară.

FromTheMoon [43]

Answer:

A)

B)

C)

Explanation:

Given that a pendulum is suspended by a shaft with a very light thin rod.

Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²

We can determine the answer to these questions using angular kinematics.

Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.

Here are the formulas for angular kinematics:

θ = ωt
∆w =
L [Angular momentum] = mvr [mass × velocity × radius]

A) What is the minimum speed required for the pendulum to traverse the complete circle?

We can use the formula v = √gL derived from

B) The same question if the pendulum is suspended with a wire?

C) What is the ratio of the two calculated speeds?

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2 years ago