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2 years ago

In one hour Earth receives enough solar energy to meet the energy needs of the world for

Physics

1 answer:

erastova [34]2 years ago

3 0

I have read that it is enough energy to power the needs of civilization for an entire year.

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calculate the time (in years) it would take to send a message to the Galactic Capitol located at the center of our Milky Way gal

Alinara [238K]

Answer:

The total time is $\bf{52850~Yr}$ .

Explanation:

Given:

The distance between Galactic Capitol and Earth is, $d = 2.5 \times 10^{20}~m$

The speed of light is, $c = 3 \times 10^{8}~m/s$

The time taken by the message to reach and come back to Earth is given by

$t &=& \dfrac{2d}{c}\\~~&=& \dfrac{2(2.5 \times 10^{20}~m)}{3 \times 10^{8}~m/s}\\~~&=& 1.6 \times 10^{12}~s\\~~&=& \dfrac{1.6 \times 10^{12}~s}{(86400 \times 365)~s}~Yr\\~~&\approx& 52849~Yr$

In addition they take 1 year to decide to reply.

Thus, the total time is $25850~Yr$ .

7 0

3 years ago

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l

goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

$\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}$

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

$\Delta t = \frac{0.18\ m}{340\ m/s}$

4 0

2 years ago

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle $F_{net}$ is given by $F_{net}=F-f$

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for $F_{net}$

$F_{net}=ma$ hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

6 0

3 years ago

could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

dexar [7]

Answer:

f = 4000 / 30 sec = 133.3 vibrations/sec

P = 1 / f = .0075 sec period of 1 vibration

4 0

2 years ago

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>

8 0

3 years ago