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3 years ago

What is the most common kind of element in the solar wind?

1 answer:

8 0

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Given: G = 6.67259 × 10−11 N m2 /kg2 . A 438 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.94 × 105

photoshop1234 [79]

Answer:

449.37412 N

Explanation:

G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²

m = Mass of satellite = 438 kg

M = Mass of planet

T = Time period of the satellite = 24 h

r = Radius of planet = $1.94\times 10^8\ m$

The time period of the satellite is given by

$T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg$

The gravitational force is given by

$F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N$

The force acting on this satellite is 449.37412 N

3 0

3 years ago

A magnetic field is formed when electric charges move. When the electric charges stop moving, the magnetic field (which was form

Natali5045456 [20]

True

Explanation:

when electrons stop.moving it cause a stoppage

in formation of magnetic field

i hope.it helps u

7 0

3 years ago

An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Alexandra [31]

Answer:

1 / i + 1 / o = 1 / f thin lens equations

i = o f / (o - f) rearranging

Lens 1: object = 30 cm f = 15.2 cm

i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm

o2 = 40.2 - 30/8 = 9.4 cm distance of image 1 from lens 2

i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm

The final image is 24.6 cm to the left of lens 2

The first image is inverted

The second image is erect (as seen from the first image)

So the final image is inverted

M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69

7 0

3 years ago

During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz elec

erik [133]

Answer:

(a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

Explanation:

Given that,

Maximum potential = 4.00 mV

Distance = 0.300\ m

Frequency = 1.00 Hz

(a). We need to calculate the maximum electric field strength

Using formula of the potential difference

$\Delta V=Ed$

$E=\dfrac{\Delta V}{d}$

$E=\dfrac{4.00\times10^{-3}}{0.300}$

$E=0.0133\ V/m$

(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave

Using formula of the maximum magnetic field strength in the electromagnetic wave

$B=\dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{0.0133}{3\times10^{8}}$

$B=4.433\times10^{-11}\ T$

(c). We need to calculate the wavelength of the electromagnetic wave

Using formula of wavelength

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{1.00}$

$\lambda=3\times10^{8}\ m$

Hence, (a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

4 0

3 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago