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3 years ago

The orientation of which of the following does not influence the phases of the moon?

Physics

2 answers:

Semmy [17]3 years ago

7 0

Answer:

So the correct answer would be the stars do NOT have an influence on the Moon phases.

klasskru [66]3 years ago

6 0

So we want to know what is not influencing the phases of the Moon. What influences Moon phases are celestial bodies that are close enough to have enough gravitational force to have an influence. Those celestial bodies would be the Earth and the Sun, since moon itself cant influence itself and all other stars are too far to exert gravitational force. So the correct answer would be the Moon and the stars do NOT have an influence on the Moon phases.

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Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require

strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

Em₀ = U = m g h

Final point. In the highest part of the loop

Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

I = m R²

v = w R

we substitute

Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

em_f = m v² + 2 m g R

Energy is conserved

Emo = Em_f

mgh = m v² + 2m g R

h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

h> 2R

3 0

2 years ago

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

irakobra [83]

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

Known:

R = 10.0 Ω

V = 12.0 V

Unknown:

I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

6 0

2 years ago

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0

2 years ago

An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for

VikaD [51]

We know that
g = LcosΘ
where g, L and Θ are centripetal gravity length, and angle of object
ω² = g/LcosΘ 
ω = √(g / LcosΘ)

8 0

3 years ago

Read 2 more answers

An apparatus like the one Cavendish used to ﬁnd G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.

Ber [7]

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ .

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$

So, the magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ . Hence, this is the required solution.

5 0

2 years ago