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3 years ago

The cycle of the moon through its phases or the synodic month is___.

Physics

2 answers:

ivolga24 [154]3 years ago

5 0

Answer:

of 29.5 Days

Explanation:

As the Moon goes around the Earth, it shows different shapes as seen from the Earth which are called phases. These phases repeat in a particular interval of time which is called as Synodic month. It is basically the period between two consecutive Full Moons or New Moons. Synodic month for the Moon is 29.5 Earth days.

Sidereal Month for the Moon differs from synodic month. Sidereal month is the orbital period of the Moon around the Earth relative to fixed stars at infinity. This period is of 27.3 days.

ikadub [295]3 years ago

3 0

It is 29 and a half days long

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Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

Ksivusya [100]

Answer:

$A_c=87.73*10^{21}m/s$

Explanation:

From the question we are told that

$r=5\times 10^{-11}$

$T=1.5 \times 10^{-16}$

Generally the equation for velocity is mathematically given as

$Velocity (v)=\frac{2 \pi r}{t}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

Generally the equation for Centripetal acceleration is mathematically given as

$A_c=\frac{V^2}{r}$

$A_c=(\frac{20.944*10^5)}{r5*10^{-11}}$

$A_c=87.73*10^{21}m/s$

8 0

3 years ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

3 years ago

Which statement compares the strengths of electric forces between particles of matter?

Black_prince [1.1K]

Answer:

B. Metallic bonds are stronger than hydrogen bonds but weaker than ionic bonds.

Explanation:

a p e x , just took the quiz

4 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at

VARVARA [1.3K]

Answer:

μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

N-W = 0

N = W = mg

X axis

F -fr = 0

F = fr

The friction force has the formula

fr = μ N

fr = μ m g

At the limit point just before starting the movement

F = μ m g

μ = F / m g

calculate

μ = 34.8 / (19.8 9.8)

μ = 0.18

7 0

3 years ago

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