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taurus [48]
3 years ago
9

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

to barleycorns. (1barleycorn=1/3 inch.)
Physics
1 answer:
alukav5142 [94]3 years ago
3 0

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.  

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

1\ light\ year=3.72\times 10^{17}\ inches

So, 4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches

4.218\ light\ year=1.56\times 10^{18}\ inches

We need to convert 4.218 light-years barley corns.  

Since, 1 barleycorn = 1/3 inch  

1\ inch=3\ barleycorn

1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn

So, the nearest star to the Earth is at a distance of 4.68\times 10^{18}\ barleycorn. Hence, this is the required solution.

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Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h
PolarNik [594]

The kinetic energy is 9\cdot 10^{40}J.

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

m=5\cdot 10^{31} kg is its mass

v=216,000 km/h is the speed

First, we convert the speed  from km/h to m/s:

v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s

Therefore, the kinetic energy of the comet is

K=\frac{1}{2}(5\cdot 10^{31})(60,000)^2=9\cdot 10^{40}J

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?
miv72 [106K]
The brand of car wax
4 0
2 years ago
Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field
Xelga [282]

Answer:

6

Explanation:

The magnetic field inside a solenoid is given by the following formula:

B = \mu_{0}nI

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

B_{1} = \mu_{0}n_{1}I  ------------------- equation 1

For Solenoid 2:

n₂ = 6n₁

Therefore,

B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I  ----------------- equation 2

Diving equation 1 and equation 2:

\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6

Hence, the correct option is:

<u>6</u>

8 0
2 years ago
John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
3 years ago
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