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3 years ago

1. The nearest distance of distinct vision of

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

The focal length of the lens is +30 and it is a convex lens.

Explanation:

Given that,

The nearest distance of distinct vision of a hypermetropial person is 60 cm, v = -60 cm

The distance is reduced by 20 cm, u = -20 cm

We need to find the nature and focal length of the lens.

Let f be the focal length of the lens. Using lens formula,

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{f}=\dfrac{1}{(-60)}-\dfrac{1}{(-20)}\\\\f=+30\ cm$

So, the focal length of the lens is +30 and it is a convex lens.

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two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?

Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force $F$ between two charged objects:

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}$ ,

where

$k$ is coulomb's constant. $k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}$ in vacuum.
$q_1$ and $q_2$ are the signed charge of the objects.
$r$ is the distance between the two objects.

For the two electrons:

$q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}$ .
$r = 1\times 10^{-10}\;\text{m}$ .

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}$ .

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3 years ago

How and where is old oceanic crust destroyed?

sladkih [1.3K]

Answer:

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Explanation:

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3 years ago

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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

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Answer:

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