1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
eimsori [14]
2 years ago
11

What's true about the elliptical path that the planets follow around the sun?

Physics
1 answer:
sattari [20]2 years ago
5 0

Answer:

well, it is not a solid path and thats a teeny tiny fact

You might be interested in
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Alex787 [66]

Answer:

a) 1.092 m/s

b) 0.33 m

c) 0.25 m

Explanation:

To start with, from the formula of wave, we know that

v = f λ, where

v = velocity of wave

f = frequency of the wave

λ = wavelength of the wave

Again, on another hand, we know that

T = 1/f, where T = period of the wave

From the question, we are given that

t = 2.7 s

d = 0.66 m

λ = 5.9 m

Period, T = 2 * t

Period, T = 2 * 2.7

Period, T = 5.4 s

If T = 1/f, then f = 1/T, thus

Frequency, f = 1/5.4

Frequency, f = 0.185 hz

Remember, v = f λ

v = 0.185 * 5.9

v = 1.092 m/s

Amplitude, A = d/2

Amplitude, A = 0.66/2

Amplitude, A = 0.33 m

If the other distance travelled by the boat is 0.5, then Amplitude is

A = 0.5/2

A = 0.25 m

6 0
3 years ago
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel
insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0
3 years ago
Why does the pattern of colors repeat in a thin soap film? Please use 2 content related sentences. (ref: p.522-530)
Slav-nsk [51]
Becuz when you wash up in the tub you want layers of soap so you don’t stink
4 0
3 years ago
A train travelling at 18m/s accelerates uniformly at 2m/s² for 6 seconds. Calculate the final speed of the train.
kobusy [5.1K]

its going  to be 30 m/s

8 0
3 years ago
Calculate V1 and V2 ( V = voltage )
Kitty [74]

FORMULA:

  • V = IR, where V = P.D; I = Current; R = Resistance.

ANSWER:

Total equivalent resistance for circuit:

R(eq) = R1 + R2 [It is in series]

  • 330Ω + 470Ω
  • 800Ω

Now, Current passing through whole circuit:

I = V/R

  • 16/800
  • 1/50 ampere

We know that, In series combination current passing through whole circuit is same.

So, V¹ = IR¹

V¹ = 1/50 × 330

  • V¹ = 6.6 volt

And V² = IR²

V² = 1/50 × 470

  • V² = 9.4 volt
7 0
3 years ago
Read 2 more answers
Other questions:
  • a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista
    5·1 answer
  • How do most primary producers make their own food?​
    15·2 answers
  • Explain the difference in polarity between co2 and h2o by referring to the polarity of the bonds and the shape of the molecules
    8·2 answers
  • A bird watcher spots a sparrow in a tree. The sparrow sits in a nest that is 10.5 feet above the bird watcher's eye level, at a
    13·1 answer
  • Identifying Sources Used to Study Earth's Interie
    9·1 answer
  • A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12
    11·1 answer
  • A father wanted to explain how the moon shines to his 5-year old child by comparing it to an object that the child uses. Which s
    14·2 answers
  • A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
    5·1 answer
  • A uniform string of length 0.50 m is fixed at both ends. Find the
    15·1 answer
  • What is the difference between a low tide and a high tide
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!