Answer:
a) 30.84m/s
b) 348.32Hz
c) 32.34m/s
d) 289.69Hz
Explanation:
a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s
b) Based on Doppler effect:
/*I will take as positive direction the vector 
*/
c) 
d) 
<span>While
modern humans first evolved approximately 200,000 years ago,
the age of
the Earth, the Sun, and the rest of the solar system is believed
to be approximately 4.6 billion years.
That makes the Earth, the Sun, and the rest of the solar system
something like 23 thousand times as old as the human species !
</span>
Answer:
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Explanation:
The current flowing through your system is described by Ohm's law
V = I R
where I is the current, V the voltage and R the resistance
in this case three barateras are taken in series giving a total voltage of V = 4.5 V the typical resistance values of dry skin is R = 1000 000Ohm and the resinification of wet skin is R = 100000 ohm
let's calculate the current flowing
I = V / R
I = 4.5 / 1000000
I = 4.5 10⁻⁶ A
this is the current with dry hands, we see that much less than the value that allows to feel a painful response by the body
If the skin is
I = 4,5 / 100,000
I = 4.5 10⁻⁵ A
This value is small, but it is close to the pain threshold, but it is in the range of slight discomfort.
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
