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Kay [80]
3 years ago
10

(a) The distance to a star is approximately 8.64 ✕ 1018 m. If this star were to burn out today, in how many years would we see i

t disappear? years (b) How long does it take sunlight to reach Earth? minutes (c) How long does it take for a microwave radar signal to travel from Earth to the Moon and back? (The distance from Earth to the Moon is 3.84 ✕ 105 km.) s

Physics
2 answers:
spayn [35]3 years ago
7 0

Answer:

A. 922 yrs

B. 8.21mins

C. 2.56 seconds

Explanation:

Pls refer to attached file

Tamiku [17]3 years ago
4 0

Answer:

(a)  913.874 Years.

(b) 8 minutes.

(c) 2.56177 Seconds.

Explanation:

(a) Following Equation relates Distance to Velocity and time.

$$S = Vt$$ (S is for Distance,

in it we have.

S = 8.64*10^{18}m, V = 299792458m/s (Velocity of Light).

t = ?

Rearranging and solving for time gives us. t = 2.881993788*10^{18}seconds.

which when converted to years , we have t = 913.874years, that is how long it will take for us to notice that the star has been burned out.

(b) it take 8 minutes for sun light to reach earth, how.? because distance from earth to sun is 150.46 million Km. doing same calculations as in (a) gives us 8 minutes.

(c) Distance to moon and back is twice as much, because obviously it is round trip, so S = 2 x 3.84x10^5Km.

using same logic as in (a) and (b) we get t = 2.56177 seconds. that is how long light takes to reach moon and come back.

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It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0
VikaD [51]
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have 

rate = \frac{-(C_{new} - C_{old})}{t}

where C are the concentration values and t is the time taken for it to decompose.

14 = \frac{-(0.02 - 0.06)}{t}
t = 0.003 s

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0
3 years ago
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

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

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

7 0
3 years ago
Describe an object’s motion when balanced forces act on it
Alinara [238K]

Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal
3 0
3 years ago
Read 2 more answers
How many million kilometers is one astronomical unit.
Andrews [41]

Answer:

150 million kilometres

Explanation:

The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.

6 0
2 years ago
Energy that cannot be used to do useful work is referred to as Select one: a. potential energy. b. entropy. c. kinetic energy. d
topjm [15]

In thermodynamics, entropy (symbolized as S) is a physical magnitude for a thermodynamic system in equilibrium. It measures the number of microstates compatible with the equilibrium macrostate, it can also be said that it is the reason for an increase between internal energy versus an increase in system temperature.

The universe tends to distribute energy evenly; that is, to maximize entropy. Intuitively, entropy is a physical quantity that, by calculation, allows us to determine the part of energy per unit of temperature that cannot be used to produce work.

Therefore the correct answer is B.

5 0
3 years ago
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