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3 years ago

How many times farther is Proxima Centauri from the sun than is earth from the sun

Physics

2 answers:

eduard3 years ago

8 0

When I work it out, I get roughly 270 thousand times.

Aleks04 [339]3 years ago

5 0

Answer:

Of the three stars in the system, the dimmest - called Proxima Centauri - is actually the nearest star to the Sun. The two bright stars, called Alpha Centauri A and B form a close binary system; they are separated by only 23 times the Earth - Sun distance.

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One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo

yuradex [85]

Answer:

μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

W₁ x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

cos 60 = x / L

where L is the length of the ladder

x = L cos 60

sin 60 = y / L

y = L sin60

the horizontal distance of man is

cos 60 = x2 / 7.0

x2 = 7 cos 60

we substitute

m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

fr = (735 + 2450) / 8.66

fr = 367.78 N

the friction force has the expression

fr = μ N

write the translational equilibrium equation

N - W₁ -W₂ = 0

N = m₁ g + W₂

N = 30 9.8 + 700

N = 994 N

we clear the friction force from the eucacion

μ = fr / N

μ = 367.78 / 994

μ = 0.37

3 0

3 years ago

Select the correct answer.

satela [25.4K]

Answer:

all of the above

Explanation:

muscular endurance is the ability to be able to do muscular activities for a long period of time. The longer you do it, the better you can handle it and for longer.

7 0

2 years ago

A drop

exis [7]

Answer:

$27.5\ m$

Explanation:

As we know that volume of cylinder is

$v=\pi r^{2} *h$

Where v=volume , h= height or thickness and r= radius

Here,

$v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5$

Putting these values in the previous equation , we get

$10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m$

Therefore thickness is 27.5 m

7 0

3 years ago

A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

Plug in the given values and solve for 'r'. This gives,

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

5 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago

How many times farther is Proxima Centauri from the sun than is earth from the sun￼

How many times farther is Proxima Centauri from the sun than is earth from the sun