What is the position and kind of image produced by the lens below?

The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac

dybincka [34]

Answer:

The magnitude of charge on each is $5.707\times 10^{13} C$

Solution:

As per the question:

Mass of Earth, $M_{E} = 5.98\times 10^{24} kg$

Mass of Moon, $M_{M} = 7.35\times 10^{22} kg$

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

$F_{G} = \frac{GM_{E}M_{M}}{d^{2}}$ (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

$F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$ (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

$\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$

$(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}$

$\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q$

Q = $\pm 5.707\times 10^{13} C$

7 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

Why the tyres are in circular shape?

Nezavi [6.7K]

The circular shape allows the tire to roll easily with the least amount of bumps or jolts

3 0

3 years ago

A camera phone called the iPhone X has an image sensor size of 5 mm and a focal length of 4 mm. How long of a selfie-stick must

azamat

Answer:

length of selfie-stick is 1.62 m

Explanation:

Given data

image size h1 = 5 mm = 5 × $10^{-3}$ m

focal length = 4 mm = 4 × $10^{-3}$ m

distance h2 = 2.032 m

to find out

How long of a selfie-stick

solution

here we find first magnification

that is M = h1 /h2

M = 5 × $10^{-3}$ / 2.032

M = 2.46 × $10^{-3}$

and we know M = p/q

so p = Mq = 2.46 × $10^{-3}$ q

so we apply lens formula

1/f = 1/p - 1/q

1/ 4 × $10^{-3}$ = 1 / 2.46 × $10^{-3}$ q - 1/q

q = 1.622 m

so length of selfie-stick is 1.62 m

4 0

3 years ago

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)

katen-ka-za [31]

Answer:

Velocity will be equal to 7.31 m/sec

Explanation:

We have given mass of the student m = 61 kg

Height of the water slide h = 12.3 m

Acceleration due to gravity $g=9.8m/sec^2$

Potential energy is equal to $U=mgh=61\times 9.8\times 12.3=7352.94J$

Work done due to friction = -5800 J

So energy remained = 7352.94-5800 = 1552.94 J

This energy will be equal to kinetic energy

So $\frac{1}{2}mv^2=1552.94$

$\frac{1}{2}\times 61\times v^2=1552.94$

$v^2=50.91$

v = 7.13 m/sec

7 0

3 years ago