Convert 87°F to Celsius scale?

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

3 years ago

How many significant digits are in the number 204.0920

monitta

Answer:

Seven

Explanation:

The rules for significant digits are:

Non-zero digits are always significant.
Zeros between significant digits are also significant.
Trailing zeros are significant only after a decimal point.

Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.

The first two 0s are significant because they are between significant digits.

The last 0 is significant because it is a trailing zero after a decimal point.

Therefore, all seven digits are significant.

3 0

3 years ago

Read 2 more answers

An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate

Bumek [7]

Answer:

a) n = 9.9 b) E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

In this case En = 19 eV let us reduce to the SI system

En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

n = √ (In 8 m L² / h²)

let's calculate

n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

n = √ (98) n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

In = (h² / 8mL2² n²

In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²

E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

E₁₀ = 1.925 101 eV

E₁₀ = 19.25 eV

the result with significant figures is

E₁₀ = 19.25 eV

3 0

3 years ago

A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy

snow_lady [41]

Answer:

The distance required = 16.97 cm

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

e = 16.97 cm

Thus the distance required = 16.97 cm

6 0

3 years ago

The additional product of the nuclear fission reaction shown in the

Arada [10]

The answer to your question is C.

8 0

3 years ago