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Kay [80]3 years ago

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It’s would be C.

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An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

So

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

2 years ago

Need 3 examples of unbalanced forces

jarptica [38.1K]

1. pulling tug of war
2. monkey swinging from vine
3. arm wrestle

4 0

3 years ago

Read 2 more answers

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the

Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0

3 years ago

How does tightening a string on a guitar affect its natural frequency?

iragen [17]

By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.

Hope this helps
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3 0

3 years ago

Read 2 more answers

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago