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-Dominant- [34]

3 years ago

6

C A B 2. What is the measure of side AB?

1 answer:

Otrada [13]3 years ago

8 0

Your question has no context, nobody’s going to be able to answer it.

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The Sun is approximately 1,400 million kilometers from the planet Saturn, and light from the Sun travels to Saturn at the rate o

irinina [24]

Answer:

Light takes 77 minutes to travel from the Sun to Saturn.

Explanation:

The equation for the average velocity can be used to estimate the time that light will take to travel from the Sun to Saturn. The average velocity is defined as:

$v = \frac{d}{t}$ (1)

Where v is the velocity, d is the covered distance and t is the time.

Equation 1 can be rewritten for t to get:

$t = \frac{d}{v}$

It is necessary to express the speed of light in terms of minutes:

$300000 \frac{Km}{s} . \frac{60 s}{1 min}$ ⇒ $1.8x10^{7}$ <em>Km/min </em>

$t = \frac{1.4x10^{9} Km}{1.8x10^{7} Km/min}$

$t = 77 min$

It is known that light takes 8,3 minutes to go from the Sun to the Earth, where the distance between each other is 150.000.000 Kilometers.

5 0

4 years ago

Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also

ozzi

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$ ,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$

For maximum B, cos(2πft) = 1. Hence,

$B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12}$ T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}$

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}$

The plot is given in the attachment.

5 0

4 years ago

The index of refraction for red light in water is 1.331 and that for blue light is 1.340. If a ray of white light enters the wat

kobusy [5.1K]

Answer:

Angle of refraction for red light is $43.01^{\circ}$

Angle of refraction for blue light is $42.68^{\circ}$

Explanation:

It is given refractive index for red light is $\mu _{red}=1.331$

Refractive index of blue light $\mu _{blue}=1.340$

Angle of incidence $i=65.30^{\circ}$

According to law of refraction $\mu =\frac{sini}{sinr}$

For red light $1.331 =\frac{sin65.30^{\circ}}{sinr}$

$1.331 =\frac{0.908}{sinr}$

$sinr=0.682$

$r=43.01^{\circ}$

Therefore angle of refraction for red light is $43.01^{\circ}$

Similarly for blue light $1.340 =\frac{sin65.30^{\circ}}{sinr}$

$1.340 =\frac{0.908}{sinr}$

$sinr=0.677$

r = $42.68^{\circ}$

Therefore angle of refraction for blue light is $42.68^{\circ}$

6 0

3 years ago

What two things must a measurement include?

astraxan [27]

The number and units. An example would be 3cm or 4kg.

7 0

3 years ago

Read 2 more answers

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0

3 years ago