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Debora [2.8K]
3 years ago
14

Based on the data, which of the following explains the relationship between the distance of the planet from the Sun and its plan

etary motion around the sun?
A. The planets with the least distance from the sun experience a longer time of revolution and shorter days of rotation.

B. The planets with the least distance from the sun experience a shorter time of revolution and shortest days of rotation.

C. The planets with the greatest distance from the sun experience a longer time of revolution and longer days of rotation.

D. The planets with the greatest distance from the sun experience a longer time of revolution and shorter days of rotation.

Physics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

.

Explanation:

...

max2010maxim [7]3 years ago
3 0
Answer is B! Just think about mercury or Venus! Mercury takes 88 earth days to rotate around the sun and Venus takes 225. The farther you go out into space, the longer it takes to orbit the sun, because there’s a longer distance. It takes Pluto 248 years, and Neptune takes 165 years:)
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A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0
3 years ago
Read 2 more answers
Particles q1 = -53.0 uc, q2 = +105 uc, and
Nimfa-mama [501]

Answer:

-180.38\ \text{N}

Explanation:

q_1=-53\ \mu\text{C}

q_2=105\ \mu\text{C}

q_3=-88\ \mu\text{C}

r = Distance between the charges

r_{12}=0.5\ \text{m}

r_{23}=0.95\ \text{m}

r_{13}=1.45\ \text{m}

k = Coulomb constant = 9\times 10^9\ \text{Nm}^2/\text{C}^2

Net force is given by

F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}

The force on the particle q_1 is -180.38\ \text{N}.

8 0
3 years ago
2. Without changing the mass or height, what else do you think you could do to design a system in which GPE and KE values are mo
Andru [333]

Answer:

jgvvjlvcgkgc

Explanation:

5 0
3 years ago
Driving along a crowded freeway, you notice that it takes a time tt to go from one mile marker to the next. When you increase yo
ddd [48]

Answer:

38.6 mi/h

Explanation:

7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s

Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is

t = 1/v

Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means

t - 15 = 1/(v+0.00206)

We can substitute t = 1/v to solve for v

\frac{1}{v} - 15 = \frac{1}{v + 0.00206}

We can multiply both sides of the equation with v(v+0.00206)

v+0.00206 - 15v(v+0.00206) = v

-15v^2 - 0.0308v + 0.00206 = 0

v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}

v= \frac{0.03083\pm0.35}{-30}

v = -0.01278 or v = 0.01 0724 mi/s

Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h

6 0
4 years ago
¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *
lys-0071 [83]

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

  • Fuerza [force] (F) = 110 N
  • Metros [meters] (m) = 20 m
  • Trabajo [work] (W) = ?

Usar la fórmula (use formula):

  • \boxed{\bold{W = F*d}}

Reemplazar (replace):

  • \boxed{\bold{W = 110\ N*20\ m}}

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

  • \boxed{\boxed{\bold{W =2.200\ J}}}

Greetings.

7 0
3 years ago
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