Any one their to help

Examples of angular motion<br>

ohaa [14]

Answer:

A figure skater doing a double axle

The swing of a baseball bat

The leverage on a hockey stick

hope it helps

3 0

3 years ago

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A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

DENIUS [597]

Answer:

C = 4,174 10³ V / m^{3/4} , E = 7.19 10² / ∛x, E = 1.5 10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

V = C $x^{4/3}$

C = V / x^{4/3}

C = 220 / (11 10⁻²)^{4/3}

C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

V = E dx

E = dx / V

E = ∫ dx / C x^{4/3}

E = 1 / C x^{-1/3} / (- 1/3)

E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

E = 3 / C (0- (-1 / x^{1/3}))

E = 3 / 4,174 10³ (1 / x^{1/3})

E = 7.19 10² / ∛x

for x = 0.110 cm

E = 7.19 10² /∛0.11

E = 1.5 10³ N/C

6 0

3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

10) If the mass 2m, the left mass

Romashka-Z-Leto [24]

Answer:

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

Explanation:

Gravitational force between two objects of masses $m_{1}, m_{2}$ kept at a distance r is given by the formula

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$

Here , $m_{1}$ = 2m

$m_{2}$ = $\frac{m}{2}$

Thus , F = $\frac{-G.2m.\frac{m}{2} }{r^{2} }$

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

7 0

3 years ago

How do two sublevels of the same principal energy level differ from each other

Colt1911 [192]

Each energy sublevel corresponds to an orbital of a different shape.

Explanation:

Two sublevels of the same principal energy level differs from each other if the sublevels corrresponds to an orbital of a different shape.

The principal quantum number of an atom represents the main energy level in which the orbital is located or the distance of an orbital from the nucleus. It takes values of n = 1,2,3,4 et.c
The secondary quantum number gives the shape of the orbitals in subshells accommodating electrons.
The number of possible shapes is limited by the principal quantum numbers.

Take for example, Carbon:

1s² 2s² 2p²

The second energy level is 2 but with two different sublevels of s and p. They have different shapes. S is spherical and P is dumb-bell shaped .

Learn more:

Quantum number brainly.com/question/9288609

#learnwithBrainly

7 0

3 years ago