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Lemur [1.5K]
3 years ago
11

The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your

answer using two significant figures.
Physics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

v = 3×10^8 m/s

s= 384,400 km= 3.84×10^8 m/s

t = ?

v = s/t = 2s/t

t = 2s/v

t = (2×3.84×10^8) ÷ 3×10^8

t = 2.56 seconds

Explanation:

Earth's moon is the brightest object in our

night sky and the closest celestial body. Its

presence and proximity play a huge role in

making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.

The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696

km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."

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2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr
svp [43]
The acceleration of the particle at time t is:
a(t)=24t^2 ft/s^2
The velocity of the particle at time t is given by the integral of the acceleration a(t):
v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s
and the position of the particle at time t is given by the integral of the velocity v(t):
x(t)=\int v(t) = \int (8t^3)=8  \frac{t^4}{4}=2t^4 ft

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
x(2 s)=2 t^4 = 2 (2s)^4=32 ft
5 0
3 years ago
a 15 kg tv sots on a shelf at a height of 0.3 m. how much gravitational potential energy is added to the television when it is l
nata0808 [166]
Gravitational potential energy can be described as m*g*h (mass times gravity times height).

Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.

After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.

To find how much energy was added, we subtract final energy from initial energy:

147 J - 44.1 J = 102.9 Joules.
6 0
3 years ago
The velocity of sound apparatus is used in an investigation to determine the frequency of an unknown tuning fork. The temperatur
Hitman42 [59]

Answer:

Explanation:

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength.

6 0
3 years ago
What is another name for the breathing mechanism?
aleksandrvk [35]
Aspirate or inhale or respire or
8 0
3 years ago
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