Which two statements are true of electromagnetic waves?

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

Amiraneli [1.4K]

Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

Part A

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / R₂

G m M / R₂² = m v² / R₂

v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

K = ½ m v²

K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

U- U₀ = G m M ∫ dr / r²

U - U₀ = G m M (- r⁻¹)

We evaluate for

U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

6 0

3 years ago

Warm water is generally less dense than cold water. The mass of a ship increases as it is loaded with cargo. If a ship will be s

KengaRu [80]

If a ship will be sailing through warm and cold water, people think about making it less dense than the warmest water as they load the ship with cargo. I think you forgot to give the options along with the question. I hope that this is the answer that has actually come to your desired help.

6 0

2 years ago

Read 2 more answers

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done

blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=\frac{F}{m}$

Hence acceleration of the block is given by:

$\Rightarrow a=\frac{25}{2}=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+\frac{1}{2}at^2$

Substitute the values

$\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

3 0

2 years ago

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

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

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

8 0

2 years ago

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

lawyer [7]

Answer:

a) Yes, there are $10^8$ shakes in a second ( $1 s \frac{1 shake}{10^{-8}s}=10^8 shake$ ) in a year there are 31,536,000 seconds... that is 3.1536x $10^7$ ( $1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7$ )

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

$106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s$

That is about 2 and half hours.

3 0

3 years ago