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Dmitrij [34]
2 years ago
9

Suppose astronomers discover a new planet farther away from the Sun than Earth. How would the day and year of this planet compar

e to Earth's?
A. The planet's day would be longer, and its year would be shorter.
B. The planet's day would be shorter, and its year would be longer.
C. The planet's year would be longer, but it is impossible to predict how its day would compare to Earth's.
D. The planet's year would be shorter, but it is impossible to predict how its day would compare to Earth's. PLEASE HELP ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Physics
2 answers:
Volgvan2 years ago
8 0
The year would definitely be longer, due to Keller’s 3rd law, but I know of several objects orbiting farther from the sun than earth with both much longer and much shorter day lengths.
posledela2 years ago
6 0
I think the answer would be C. because the orbit around the sun compared to the Earth's (Identifies the seasons change and the year) would be longer, but the day is unknown because that is where it would come down to the total mass of the planet and other important factors about the planets properties.
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Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, excep
lidiya [134]

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is  needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

8 0
2 years ago
Read 2 more answers
A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm
Alisiya [41]

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{12}ml^2+mx^2

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2

I=0.7500\ kg-m^2

Hence, The moment of inertia is 0.7500 kg-m².

5 0
2 years ago
What is the minimum value of force acting between two charges placed at 1 m apart from each other?
malfutka [58]

Answer:

Ke²

Explanation:

So,

q1 = e

q2 = e

r = 1m

By coulumb's law,

F = K (q1q2/r²)

F = K (e)(e)/(1)²  

F = Ke²  

 

Option(a)

5 0
1 year ago
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a
TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = \dfrac{4}{3}\pi r^3

                              =\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3

Density =\dfrac{mass}{volume}

Density =\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}

ρ = 2322 kg/m³

terminal velocity

v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}

v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}

v_t = 6693 x 10⁻⁷ m/s

t = \dfrac{d}{v_t}

t = \dfrac{0.15}{6693 \times 10^{-7}}

t = 224 s

7 0
2 years ago
A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict
madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0
2 years ago
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