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2 years ago

D. 'g' vanishes at centre of

Physics

1 answer:

guapka [62]2 years ago

6 0

Answer:

a. 0.8 cm

Explanation:

The distance of the object from the lens, u = 1 cm

The magnification of the lens, m = 5

The focus of a lens formula is given as follows;

$f = \dfrac{1}{\dfrac{1}{v} + \dfrac{1}{v} }$

The magnification of the lens, m = -v/u

Where;

v = The distance of the image from the lens

Therefore, we have;

v = m × u

∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)

From which we get;

$f = \dfrac{1}{\dfrac{1}{1} + \dfrac{1}{5} } = \dfrac{5}{6} \approx 0.8$

The focal length ≈ 0.8 cm

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Which terrestrial planet exhibits retrograde rotation?.

Sati [7]

Answer:

Planets that are farther from the sun than the earth (all but Mercury and Venus) will exhibit retrograde motion.

If the position of the planet is observed relative to the background stars, the planet will appear to move backward relative to the stars when the earth is moving in an Eastward direction faster than the planet, and the planet appears to move backwards relative to the stars

(The planet will be on the side of the earth that is opposite that of the sun)

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2 years ago

A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric

MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

$J = \frac{I}{\pi R^2}$

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

$J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}$

This enclosed current is now to be used in Ampere’s Law.

$\mu_o I_{enc} = \int {B} \, dl$

Here, $\int \, dl$ represents the circular path of radius r. So we can replace the integral with the circumference of the path, $2\pi r$ .

As a result, the magnetic field is

$B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}$

5 0

2 years ago

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish

Snezhnost [94]

Answer:

7.1 m/s

Explanation:

First, find the time it takes for the fish to reach the water.

Given in the y direction:

Δy = 6.1 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.12 s

Next, find the velocity needed to travel 7.9 m in that time.

Given in the x direction:

Δx = 7.9 m

a = 0 m/s²

t = 1.12 s

Find: v₀

Δx = v₀ t + ½ at²

7.9 m = v₀ (1.12 s) + ½ (0 m/s²) (1.12 s)²

v₀ = 7.1 m/s

4 0

3 years ago

Read 2 more answers

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

2 years ago

A friend waves & yells at you. If the sound took 3.20 seconds to reach you and the speed of sound is 340 m/s,

xeze [42]

Answer:

1088 m

Explanation:

3.00 seconds times by 340 =1020+68=1088

5 0

2 years ago